Trigonometry. Version 4 ɛ. adapted by Katherine Cliff from OpenStax Algebra and Trigonometry, and Precalculus by Stitz and Zeager.

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1 Trigonometry Version 4 ɛ adapted by Katherine Cliff from OpenStax Algebra and Trigonometry, and Precalculus by Stitz and Zeager July 8, 0

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3 Table of Contents Preface v 4 Foundations of Trigonometry Angles Exercises Right Triangle Trig Exercises Trig in the Plane Exercises The Circular Functions The Unit Circle Exercises Other Circular Functions Exercises Graphs of Sine and Cosine Exercises Graphs of Other Circular Functions Exercises Inverse Trig Functions Exercises Trig Equations Exercises Analytical Trigonometry The Pythagorean Identities Exercises The Sum and Difference Identities Exercises Double and Half Angle Identities Exercises Solving Equations Involving Identities

4 iv Table of Contents Exercises A Algebra Review 61 A.1 Basic Set Theory and Interval Notation A.1.1 Some Basic Set Theory Notions A.1. Sets of Real Numbers A.1. Exercises A.1.4 Answers A. Real Number Arithmetic A..1 Exercises A.. Answers A. The Cartesian Plane A..1 The Cartesian Coordinate Plane A.. Distance in the Plane A.. Exercises A..4 Answers A.4 Linear Equations and Inequalities A.4.1 Linear Equations A.4. Linear Inequalities A.4. Exercises A.4.4 Answers A.5 Graphing Lines A.5.1 Exercises A.5. Answers A.6 Systems of Two Linear Equations in Two Unknowns A.6.1 Exercises A.6. Answers A.7 Absolute Value Equations and Inequalities A.7.1 Absolute Value Equations A.7. Absolute Value Inequalities A.7. Exercises A.7.4 Answers A.8 Polynomial Arithmetic A.8.1 Polynomial Addition, Subtraction and Multiplication A.8. Polynomial Long Division A.8. Exercises A.8.4 Answers A.9 Basic Factoring Techniques A.9.1 Solving Equations by Factoring A.9. Exercises A.9. Answers A.10 Quadratic Equations

5 Table of Contents v A.10.1 Exercises A.10. Answers A.11 Complex Numbers A.11.1 Exercises A.11. Answers A.1 Rational Expressions and Equations A.1.1 Exercises A.1. Answers A.1 Radical Equations A.1.1 Rationalizing Denominators and Numerators A.1. Exercises A.1. Answers A.14 Variation A.14.1 Exercises A.14. Answers B Geometry Review 441 B.1 Angles in Degrees B.1.1 Exercises B.1. Answers B. Right Triangle Trigonometry B..1 Exercises B.. Answers C Selected Proofs 467 C.1 The Density of the Rationals within the Reals C. Why Pi is Actually a Constant C. Cauchy s Bound C.4 Fermat s Penultimate Theorem

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7 Preface This is a pdf copy of the OpenSource Trigonometry texbook compiled for University of Colorado Colorado Springs by Katherine Cliff. This textbook is based off of material from OpenStax Algebra and Trigonometry and from Stitz and Zeager s Precalculus. Both of these texts are licensed under the Create Commons Attribution, NonCommercial, ShareAlike license, as is this revised and remixed copy of the text. If you have found the pdf floating around in cyberspace, may I humbly recommend that you check out the interactive, online version of the text which, frankly, I think is a lot more fun. I designed this text to support our students and teachers at UCCS in learning the discipline of Trigonometry, mainly with the goal of creating better-prepared Calculus I students. In my search for a textbook that was student-friendly and appropriately rigorous, I came upon the two texts mentioned above. Thanks to a grant from the Colorado Department of Higher Education, I was able to devote time in the summer of 00 (when nothing much else was happening, right?) to combining these texts and adding interactive content of my own that reflects the ways I prefer to teach Trigonometry. There are a few strategies and choices in this text which, I think, are fairly unique among Trigonometry textbooks. In particular I choose to approach graphing of trigonometric functions completely through the lens of graphical transformations (that section is very fun online, I think). I also choose to not teach the tangent sum/difference formulas, which are often presented as identities, but which contain major domain issues. The end result of this work is a text which is interactive, student-friendly, and completely free. My hope is that it will support all of our students in their education while relieving some of the financial burden of college. If you would like to make contributions to this text, or find an error or issue, please reach out to me at kcliff@uccs.edu. Katherine Cliff University of Colorado Colorado Springs Summer of 0

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9 Chapter 4 Foundations of Trigonometry 4.1 Angles Learning Objectives In this section you will: Draw angles in standard position. Convert between degrees and radians. Find coterminal angles. Find the length of a circular arc. Use linear and angular speed to describe motion on a circular path. A golfer swings to hit a ball over a sand trap and onto the green. An airline pilot maneuvers a plane toward a narrow runway. A dress designer creates the latest fashion. What do they all have in common? They all work with angles, and so do all of us at one time or another. Sometimes we need to measure angles exactly with instruments. Other times we estimate them or judge them by eye. Either way, the proper angle can make the difference between success and failure in many undertakings. In this section, we will examine properties of angles. Drawing Angles in Standard Position Properly defining an angle first requires that we define a ray. A ray is a directed line segment. It consists of one point on a line and all points extending in one direction from that point. The first point is called the endpoint of the ray. We can refer to a specific ray by stating its endpoint and any other point on it. The ray pictured below can be named as ray EF, or in symbol for EF.

10 Foundations of Trigonometry An angle is the union of two rays having a common endpoint. The endpoint is called the vertex of the angle, and the two rays are the sides of the angle. The angle shown below is formed from ED and EF. Angles can be named using a point on each ray and the vertex, such as angle DEF, or in symbol form DEF. Greek letters are often used as variables for the measure of an angle. The table below has a list of Greek letters commonly used to represent angles, and a sample angle is shown below. θ φ α β γ theta phi alpha beta gamma Angle creation is a dynamic process. We start with two rays lying on top of one another. We leave one fixed in place, and rotate the other. The fixed ray is the initial side, and the rotated ray is the terminal side. In order to identify the different sides, we indicate the rotation with a small arrow close to the vertex as shown below.

11 4.1 Angles You can play around with dynamically creating an angle here: As we discussed at the beginning of the section, there are many applications for angles, but in order to use them correctly, we must be able to measure them. The measure of an angle is the amount of rotation from the initial side to the terminal side. Probably the most familiar unit of angle measurement is the degree. One degree is 1 of a circular rotation, so a complete circular rotation contains 60 degrees. An angle 60 measured in degrees should always include the unit degrees after the number, or include the degree symbol. For example, 90 degrees = 90. To formalize our work, we will begin by drawing angles on an x-y coordinate plane. Angles can occur in any position on the coordinate plane, but for the purpose of comparison, the convention is to illustrate them in the same position whenever possible. An angle is in standard position if its vertex is located at the origin, and its initial side extends along the positive x-axis. See the image below. If the angle is measured in a counterclockwise direction from the initial side to the terminal side, the angle is said to be a positive angle. If the angle is measured in a clockwise direction, the angle is said to be a negative angle.

12 4 Foundations of Trigonometry Drawing an angle in standard position always starts the same way draw the initial side along the positive x-axis. To place the terminal side of the angle, we must calculate the fraction of a full rotation the angle represents. We do that by dividing the angle measure in degrees by 60. For example, to draw a 90 angle, we calculate that = 1 4. So, the terminal side will be one-fourth of the way around the circle, moving counterclockwise from the positive x-axis. To draw a 60 angle, we calculate that = 1. So the terminal side will be 1 complete rotation around the circle, moving counterclockwise from the positive x-axis. In this case, the initial side and the terminal side overlap. See below. Definition 1: Quadrantal Angles An angle is a quadrantal angle if its terminal side lies on an axis, including 0, 90, 180, 70, or 60. Since we define an angle in standard position by its terminal side, we have a special type of angle whose terminal side lies on an axis, a quadrantal angle. This type of angle can have a measure of 0, 90, 180, 70, or 60. See below.

13 4.1 Angles 5 Definition : Given an angle measured in degrees, draw the angle in standard position. Express the angle measure as a fraction of 60. Reduce the fraction to simplest form. Draw an angle that contains that same fraction of the circle, beginning on the positive x-axis and moving counterclockwise for positive angles and clockwise for negative angles. Example 1: Drawing an Angle in Standard Position Measured in Degrees A. Sketch an angle of 0 in standard position B. Sketch an angle of 15 in standard position. A. Sketch an angle of 0 in standard position Divide the angle measure by 60 : 0 60 = 1 1 To rewrite the fraction in a more familiar fraction, we can recognize that ( ) = 1 One-twelfth equals one-third of a quarter, so by dividing a quarter rotation into thirds, we can sketch a line at 0, as in the picture below: 4 B. Sketch an angle of 15 in standard position. Divide the angle measure by 60 : = 8

14 6 Foundations of Trigonometry In this case, we can recognize that ( ) 1 8 = 4 Three-eighths is one and one-half times a quarter, so we place a line by moving clockwise one full quarter and one-half of another quarter, as shown below: Converting Between Degrees and Radians Dividing a circle into 60 parts is an arbitrary choice, although it creates the familiar degree measurement. We may choose other ways to divide a circle. To find another unit, think of the process of drawing a circle. Imagine that you stop before the circle is completed. The portion that you drew is referred to as an arc. An arc may be a portion of a full circle, a full circle, or more than a full circle, represented by more than one full rotation. The length of the arc around an entire circle is called the circumference of that circle. The circumference of a circle is C = πr. If we divide both sides of this equation by r, we create the ratio of the circumference, which is always π, to the radius, regardless of the length of the radius. So the circumference of any circle is π 6.8 times the length of the radius. That means that if we took a string as long as the radius and used it to measure consecutive lengths around the circumference, there would be room for six full string-lengths and a little more than a quarter of a seventh, as shown below

15 4.1 Angles 7 This brings us to our new angle measure. One radian is the measure of a central angle of a circle that intercepts an arc equal in length to the radius of that circle. A central angle is an angle formed at the center of a circle by two radii. So this is how we form a radian: take the radius of the circle, lay it along the circumference. Connect the endpoints of that radius-length curve to the center of the circle. The angle formed is one radian. Definition : Radian A radian is the measure of a central angle of a circle that intercepts an arc equal in length to the radius of that circle. Note that when an angle is described without a specific unit, it refers to radian measure. For example, an angle measure of indicates radians. In fact, radian measure is dimensionless, since it is the quotient of a length (circumference) divided by a length (radius) and the length units cancel.

16 8 Foundations of Trigonometry Because degrees and radians both measure angles, we need to be able to convert between them. We can easily do so using a proportion where θ is the measure of the angle in degrees and θ R is the measure of the angle in radians. Since we know that a whole rotation of a circle is π radians, it follows that half a circle is π radians. In degrees, half a circle is 180. So we create a proportion where we compare radians to degrees: θ 180 = θ R π We can solve this proportion for either θ or θ R to see how to convert between degrees and radians: Definition 4: Converting Between Degrees and Radians To change degrees into radians, use the formula: θ π 180 = θ R To change radians into degrees, use the formula: θ R 180 π = θ Example Convert each radian measure to degrees. A. π 6 B. A. π 6 To change radians into degrees, multiply by 180 π : π π Now, typically we want to cancel and simplify as much as possible: = π π = 0 B. Now, this one is unusual, because there s no π, which is usually the giveaway that we re dealing with radians. However, there s also no label, so we assume the label is radians and proceed with

17 4.1 Angles 9 multiplying by 180 π to convert to degrees: 180 π 17 Check Point 1 Convert the angle 7π 4 from radians to degrees. Example : Convert the degree measure to radians 15 To convert degrees to radians, multiply by analysis!) π 180 (so the degrees label cancels, like in dimensional 15 π 180 Now, we usually will write our answer in terms of π, and simplify the numerical part of the fraction as much as we can: = 15 = π 180 cancel 1 15π = π 1 Check Point What is the radian measure of an angle of 90? Finding Coterminal Angles Converting between degrees and radians can make working with angles easier in some applications. For other applications, we may need another type of conversion. Negative angles and angles greater than a

18 10 Foundations of Trigonometry full revolution are more awkward to work with than those in the range of 0 to 60, or 0 to π. It would be convenient to replace those out-of-range angles with a corresponding angle within the range of a single revolution. It is possible for more than one angle to have the same terminal side. Look at the figure below. The angle of 140 is a positive angle, measured counterclockwise. The angle of 0 is a negative angle, measured clockwise. But both angles have the same terminal side. If two angles in standard position have the same terminal side, they are coterminal angles. Every angle greater than 60 or less than 0 is coterminal with an angle between 0 and 60, and it is often more convenient to find the coterminal angle within the range of 0 to 60 than to work with an angle that is outside that range. Any angle has infinitely many coterminal angles because each time we add 60 to that angle or subtract 60 from it the resulting value has a terminal side in the same location. For example, 100 and 460 are coterminal for this reason, as is 60. You can experiment with coterminal angles using the interactive applet found through this link by Math Open Reference: Definition 5: Coterminal Angles Coterminal angles are two angles in standard position that have the same terminal side. If an angle is larger than 60 or π radians, you can find a smaller coterminal angle by i subtracting /i a full revolution of a circle; that is, subtract 60 or π radians. If an angle is negative, you can find a positive coterminal angle by i adding /i a full revolution of a circle; that is, add 60 or π radians.

19 4.1 Angles 11 Example 4 Find the smallest positive angle that is coterminal with each of the following: A. 159 B. 41π 18 A. 159 Since this angle is larger than 60, we subtract a full revolution of a circle: = 1169 That s still bigger than 60, so we subtract another revolution: and another and one more revolution should get us there = = = 89 Now our angle is less than 60, so we know we have the smallest possible positive coterminal angle B. 41π 18 This angle is negative, and in radians, so we ll want to add a full revolution in radians: π 41π 18 + π Now, to add, we want a common denominator. So, we can rewrite π as π = 6π 18 : 41π 18 + π = 41π π 18 = 5π 18

20 1 Foundations of Trigonometry This is still a negative coterminal angle, so we need to add another revolution, 6π 18 : That s our smallest positive coterminal angle. 5π π 18 = 1π 18 Check Point A. What is the angle between 0 and 60 that is coterminal with 19? B. What is the angle between 0 and 60 that is coterminal with 1551? C. What is the angle between 0 and π that is coterminal with 55π? D. What is the angle between 8 0 and π that is coterminal with 1π 10?

21 4.1 Angles 1 Determining the Length of an Arc An arc of a circle is a small chunk of its circumference, as pictured below: In this figure, we have an arc of length s subtended by an angle measure of θ in a circle of radius r. We can compare this small portion of the circle to the whole circle by setting up a proportion. That proportion will look slightly different depending on whether we have the central angle, θ, given in degrees or in radians. In general, though, we want to compare the part of the circle to the whole. In the following proportion, we put the partial arc information on the left side of the proportion, and the whole circle information on the right: arc length central angle subtending the arc = circumference total angle measure of circle Symbolically, that is... s θ = πr total angle measure of circle What we use for that total angle measure will depend on whether we are given θ in degrees or radians.

22 14 Foundations of Trigonometry Definition 6: Calculating length of an arc: Degrees:To find the length of arc s subtended by angle θ in a circle of radius r, use the proportion. s θ = πr 60 Radians: To find the length of arc s subtended by angle θ R in a circle of radius r, use the proportion s = πr θ R π that is, or even more simply... s θ R = r s = r θ R Example 5: Finding the Length of an Arc A. Find the arc length along a circle of radius 10 units subtended by an angle of 15. B. Find the arc length along a circle of radius units subtended by an angle of π 4. A. Find the arc length along a circle of radius 10 units subtended by an angle of 15. Let s gather up our variables: r = 10, θ = 15, s =unknown. Since the angle is given in degrees, we ll use the degrees version of our proportions: s θ = πr 60 s 15 = π(10) 60 To solve for s, we multiply both sides of the equation by 15 and evaluate from there. s = 15 0π

23 4.1 Angles 15 B. Find the arc length along a circle of radius units subtended by an angle of π 4. Here we have r =, θ = π. Since our angle is given in radians, we can use the simpler formula 4 s = rθ R : s = π 4 = 9π 4 Example 6 Assume the orbit of Mercury around the sun is a perfect circle. Mercury is approximately 6 million miles from the sun. In one Earth day, Mercury completes of its total revolution. How many miles does it travel in one day? Figuring out our initial variables is a little trickier here. We know r = 6 (thinking of the sun as the center of our big circle for simplicity s sake). We also know Mercury will complete of its revolution. So, in terms of angles, that s = So θ = Now we create our degrees proportion: s = π(6) 60 s = π So Mercury has travelled approximately.58 million miles. Check Point 4 Find the length of an arc if the radius of the arc is 4 ft and the measure is 15π 4. Example 7 The length of an arc is π centimeters on a circle of radius 10 cm. What is the measure (in degrees) of the central angle that subtends this arc?

24 16 Foundations of Trigonometry Here, we are given a slightly different set of information. We know r = 10 and s = π and we want to solve for θ in degrees. We can still use our degrees proportion: s θ = πr 60 π θ = π(10 60 So, to solve for θ, we can first cross multiply to get: π 60 = 0π θ then divide by 0π to isolate θ: π 60 = θ 0π θ = 54

25 4.1 Angles 17 Finding the area of a sector A sector is a chunk of a circle, like a pizza slice. In the image below, a sector subtended by an angle of measure θ is shaded in yellow: The good news here is that finding the area of that pizza slice is basically the same process as finding an arc length, but now we ll use the area formula in our proportion instead of the circumference. sector area central angle subtending the sector = circle area total angle measure of circle Symbolically, that is... a θ = πr total angle measure of circle What we use for that total angle measure will depend on whether we are given θ in degrees or radians.

26 18 Foundations of Trigonometry Definition 7: Calculating area of an sector: Degrees: To find the area of a sector a subtended by angle θ in a circle of radius r, use the proportion a θ = πr 60. Radians: To find the area of a sector a subtended by angle θ R in a circle of radius r, use the proportion a = πr θ R π that is, or even more simply... a θ R = r a = r θ R Example 8 An automatic lawn sprinkler sprays a distance of 0 feet while rotating 0 degrees, as shown below. What is the area of the sector of grass the sprinkler waters? The angle is given in degrees, and we want to find a sector area, so we use the proportion a θ = πr 60

27 4.1 Angles 19 Here, θ = 0, r = 0 and a is unknown. a 0 = π(0 ) 60 Multiply both sides by 0 to isolate a and evaluate. a = 400π The area is about ft. Example 9 In central pivot irrigation, a large irrigation pipe on wheels rotates around a center point. A farmer has a central pivot system with a radius of 400 meters. If water restrictions only allow her to water 150 thousand square meters a day, what angle should she set the system to cover? Write the answer in radian measure to two decimal places. Here, we already know the sector area, and we re looking for an angle in radians, so we can use the proportion: a θ R = r where a = , r = 400 and θ R is unknown. Multiply by θ R : Then isolate θ R : So the angle is approximately radians = 400 θ R = θ R = θ R

28 0 Foundations of Trigonometry Check Point 5 A sector of a circle has a central angle of 10. Find the area of the sector if the radius of the circle is 10 cm. Use Linear and Angular Speed to Describe Motion on a Circular Path In addition to finding the area of a sector, we can use angles to describe the speed of a moving object. An object traveling in a circular path has two types of speed. Linear speed is speed along a straight path and can be determined by the distance it moves along (its displacement) in a given time interval. For instance, if a wheel with radius 5 inches rotates once a second, a point on the edge of the wheel moves a distance equal to the circumference, or 10π inches, every second. So the linear speed of the point is 10π in./s. Really here, we re talking about the relationship between distance, rate, and time, which you may recognize as the equation d = rt We want to reframe this a bit, so we ll think of the rate as v and the distance as s (like arc length): If we solve for the speed, v, we have s = vt v = s t Angular speed results from circular motion and can be determined by the angle through which a point rotates in a given time interval. In other words, angular speed is angular rotation per unit time. So, for instance, if a gear makes a full rotation every 4 seconds, we can calculate its angular speed as 60degrees 4 seconds = 90 degrees per second. Angular speed can be given in radians per second, rotations per minute, or degrees per hour, for example. The equation for angular speed is as follows, where ω (read as omega) is angular speed, θ is the angle traversed, and t is time: ω = θ t Combining the definition of angular speed with the arc length equation, s = rθ, we can find a relationship between angular and linear speeds. The angular speed equation can be solved for θ, giving θ = ωt. Substituting this into the arc length equation gives: s = rθ s = rωt

29 4.1 Angles 1 Substituting this into the linear speed equation gives: v = s t. v = rωt t v = rω Definition 8: Angular and Linear Speed As a point moves along a circle of radius r, its angular speed, ω, is the angular rotation θ per unit time, t: ω = θ t The linear speed, v, of the point can be found as the distance traveled, arc length s, per unit time, t: v = s t When the angular speed is measured in radians per unit time, linear speed and angular speed are related by the equation v = rω This equation states that the angular speed in radians, ω, representing the amount of rotation occurring in a unit of time, can be multiplied by the radius r to calculate the total arc length traveled in a unit of time, which is the definition of linear speed.

30 Foundations of Trigonometry Example 10: Finding Angular Speed A water wheel, shown below, completes 1 rotation every 5 seconds. Find the angular speed in radians per second. The wheel completes 1 rotation, or passes through an angle of π radians in 5 seconds, so the angular speed would be That s about 1.57 radians per second. ω = π Example 11: Finding a linear speed A bicycle has wheels 8 inches in diameter. A tachometer determines the wheels are rotating at 180 RPM (revolutions per minute). Find the speed the bicycle is traveling down the road. Here, we have an angular speed and need to find the corresponding linear speed, since the linear speed of the outside of the tires is the speed at which the bicycle travels down the road. We begin by converting from rotations per minute to radians per minute. It can be helpful to utilize the units to make this conversion: 180 rotations π radians radians minute = 60π rotation minute Using the formula from above along with the radius of the wheels, we can find the linear speed: v = (14 inches)(60π radians minute ) = 5040π inches minute

31 4.1 Angles Remember that radians are a unitless measure, so it is not necessary to include them. Finally, we may wish to convert this linear speed into a more familiar measurement, like miles per hour. 5040π inches minute 1feet 1inches miles per hour 1mile 580feet 60minutes 1hour Check Point 6 A bicycle with 0-in.-diameter wheels has its gears set so that the chain has a 6-in. radius on the front sprocket and 4-in. radius on the rear sprocket. The cyclist pedals at 00 rpm. Find the linear speed of the bicycle in in/min (correct to at least two decimal places) Check Point 7 A truck with 6-in.-diameter wheels is traveling at 45 mi/h. Find the angular speed of the wheels in rad/min. How many revolutions per minute do the wheels make?

32 4 Foundations of Trigonometry Exercises In the following exercises, convert the angle from degree measure into radian measure, giving the exact value in terms of π In the following exercises, convert the angle from radian measure into degree measure. 9. π 10. π 11. 7π π 6 1. π 14. 5π 15. π π For the following exercises, find the angle between 0 and 60 that is coterminal to the given angle For the following exercises, find the angle between 0 and π in radians that is coterminal to the given angle. 1. π 9. 10π. 1π π 9 5. A yo-yo which is.5 inches in diameter spins at a rate of 4500 revolutions per minute. How fast is the edge of the yo-yo spinning in miles per hour? Round your answer to two decimal places. 6. How many revolutions per minute would the yo-yo in the previous exercise have to complete if the edge of the yo-yo is to be spinning at a rate of 4 miles per hour? Round your answer to two decimal places. 7. In the yo-yo trick Around the World, the performer throws the yo-yo so it sweeps out a vertical circle whose radius is the yo-yo string. If the yo-yo string is 8 inches long and the yo-yo takes seconds to complete one revolution of the circle, compute the speed of the yo-yo in miles per hour. Round your answer

33 4.1 Angles 5 to two decimal places. 8. A computer hard drive contains a circular disk with diameter.5 inches and spins at a rate of 700 revolutions per minute. Find the linear speed of a point on the edge of the disk in miles per hour. 9. A rock got stuck in the tread of my tire and when I was driving 70 miles per hour, the rock came loose and hit the inside of the wheel well of the car. How fast, in miles per hour, was the rock traveling when it came out of the tread? (The tire has a diameter of inches.) 0. The Giant Wheel at Cedar Point is a circle with diameter 18 feet which sits on an 8 foot tall platform making its overall height is 16 feet. It completes two revolutions in minutes and 7 seconds. Assuming the riders are at the edge of the circle, how fast are they traveling in miles per hour?

34 6 Foundations of Trigonometry In the following exercises, compute the areas of the circular sectors with the given central angles and radii. 1. θ = π 6, r = 1. θ = 5π 4, r = 100. θ = 0, r = θ = π, r = 1 5. θ = 40, r = 5 6. θ = 1, r = Imagine a rope tied around the Earth at the equator. Show that you need to add only π feet of length to the rope in order to lift it one foot above the ground around the entire equator. (You do NOT need to know the radius of the Earth to show this.)

35 4. Right Triangle Trig 7 4. Right Triangle Trig Learning Objectives In this section you will: Write proportions using similar triangles Use right triangles to evaluate trigonometric functions Use trigonometric functions to solve triangles Find function values for common angles Use right-triangle trigonometry to solve applied problems. Similar Triangles Two triangles are said to be similar if their corresponding angles have the same measure. In the interactive activity from Geogebra found at this link ( triangle ABC is similar to triangle DEF because angles A and D have the same measure, angles B and E have the same measure, and angles C and F have the same measure. On the blue triangle on the left, drag around points A, B, and/or C. Observe the ratios described beneath the triangles (like c/b and f/e). What do you notice about the ratios between the two similar triangles? This is the other important property of similar triangles: corresponding sides are proportional. So, as long as we have a pair of similar triangles, ABC and DEF, their corresponding sides are proportional: c b = f e, c a = f d, and b a = e, to name a few such proportions. d Similar Right Triangles Now, we re building up to something magical. Recall (from somewhere in the depths of your geometry education) that the angles in a triangle should always add up to 180. You can check with the triangles in the applet above. We re going to use this fact to create classes of right triangles. In other words, if we know we have a right triangle (a triangle with a 90 angle), we can use one of the other angles in as a reference and create a whole class of similar triangles. Let s start with a right triangle with another angle that is 8. This is enough to describe a class of similar triangles, since the third angle in the triangle must be = 6 ; therefore, we have defined a group of similar triangles with angle measures of 90, 8, and 6. Now, if we create ratios using the sides of those triangles, we know that those ratios will always produce the same value. Check it out in the

36 8 Foundations of Trigonometry demo applet found here: If we resize the triangle, but don t change the angles, the ratios of the sides stay the same. This is such a special feature of right triangles that mathematicians decided to name those ratios!

37 4. Right Triangle Trig 9 Right Triangle Trigonometry We can define the trigonometric functions in terms a reference angle t and the lengths of the sides of the triangle. The adjacent side is the leg closest to the reference angle. (Adjacent means next to. ) The opposite side is the leg across from the angle. The hypotenuse is the side of the triangle opposite the right angle. These sides are labeled in the figure below: Given a class of right triangles with reference angle t, we know that the ratio of, say, the opposite side and the hypotenuse will always be a particular number given a particular value of t. So, we re going to name each of those ratios. Definition 9: The Trigonometric Ratios Given a right triangle with an acute angle of t, we define the following trigonometric ratios: Sine: sin(t) = opposite hypotenuse Cosine: cos(t) = adjacent hypotenuse Tangent: tan(t) = opposite adjacent A common mnemonic for remembering these relationships is SohCahToa (or as some students like to write it S o h C a h T o ), formed from the first letters of Sine is opposite over hypotenuse, Cosine is adjacent a over hypotenuse, Tangent is opposite over adjacent. Example 1: Evaluating the trigonometric functions of a right triangle Given the triangle shown below, find the value of the three main trigonometric ratios:

38 0 Foundations of Trigonometry First, we want to label the sides. Since α is our reference, the side adjacent to the angle is 15, and the hypotenuse of the triangle is 17. The opposite side currently has an unknown length. Good news, though, we can use our old friend the Pythagorean Theorem to find the length of the opposite side: 15 + x = 17 x = x = 64 x = 8 Now, we find the values of the trigonometric ratios: sin(α) = opposite hypotenuse = 8 17 cos(α) = adjacent hypotenuse = tan(α) = opposite adjacent = 8 15 Check Point 8 Find the values of the sine, cosine, and tangent of the angle A in the following right triangle:

39 4. Right Triangle Trig 1 Check Point 9 Find the values of the sine, cosine, and tangent of the angle A in the following right triangle: The Reciprocal Trigonometric Functions In addition to sine, cosine, and tangent, there are three more functions. They can be defined either in terms of the sides of the right triangle, or by their relationship to the three main trigonometric functions. Definition 10: The Reciprocal Trigonometric Ratios Given a right triangle with an acute angle of t, we define the following trigonometric ratios: Cosecant: csc(t) = hypotenuse opposite = 1 sin(t) Secant: sec(t) = hypotenuse adjacent = 1 cos(t) Cotangent: cot(t) = adjacent opposite = 1 tan(t) Note the confusing terminology: cosecant is the reciprocal of sine, while secant is the reciprocal of cosine. This is one of those inconvenient math definitions that we have to accept and make our piece with. Many problems ask for all six trigonometric functions for a given angle in a triangle. A possible strategy to use is to find the sine, cosine, and tangent of the angles first. Then, find the other trigonometric functions easily using the reciprocals. Example 1 Using the triangle shown below, evaluate sin(α), cos(α), tan(α), csc(α), sec(α), and cot(α).

40 Foundations of Trigonometry Again, we don t know the length of one of the legs, but we can figure that out using the Pythagorean Theorem: + x = 5 x = 5 9 x = 16 x = 4 Now, name the sides of the triangle according to the reference angle: Next, evaluate the main trig ratios: sin(α) = opposite hypotenuse = 4 5 cos(α) = adjacent hypotenuse = 5 tan(α) = opposite adjacent = 4 Now, to find the rest, we use the fact that cosecant is the reciprocal of sine: 1 csc(α) = sin(α) = 5 4

41 4. Right Triangle Trig secant is the reciprocal of cosine: and cotangent is the reciprocal of tangent: sec(α) = 1 cos(α) = 5. cot(α) = 1 tan(α) = 4 Check Point 10 Find the values of the cosecant, secant, and cotangent of the angle A in the following right triangle: Using Trigonometric Functions to Solve Triangles In previous examples, we evaluated the sine and cosine in triangles where we knew all three sides. But the real power of right-triangle trigonometry emerges when we look at triangles in which we know an angle but do not know all the sides. Definition 11: Given a right triangle, the length of one side, and the measure of one acute angle, find the remaining sides. For each side, select the trigonometric function that has the unknown side as either the numerator or the denominator. The known side will in turn be the denominator or the numerator. Write an equation setting the function value of the known angle equal to the ratio of the corresponding sides. Using the value of the trigonometric function and the known side length, solve for the missing side length.

42 4 Foundations of Trigonometry Example 14: Finding Missing Side Lengths Using Trigonometric Ratios Find the unknown sides of the triangle: Let s start by naming all of the sides of the triangle. Using the 0 degree angle as a reference, the side labelled a is adjacent, the side labelled 7 is opposite, and the side labelled c is the hypotenuse. If we want to solve for side a, then we are looking for the adjacent side, and we know the lenght of the opposite side. The common trigonometric ratio involving the opposite and adjacent sides is tangent. Now, we use the tangent ratio to create an equation: tan(α) = opposite adjacent tan(0 ) = 7 a We want to solve for a, but it s in the denominator. So, we can multiply on both sides of the equation by a, then divide to isolate a: a tan(0 ) = 7 7 a = tan(0 ) Because of the magical properties of similar right triangles, we know tan(0 ) is always a particular value, and we can use our calculator to get an approximation of this value. Be careful not to round 7 too soon though; it s more accurate to enter tan(0 all at once in your calculator. If you do so, the ) approximate answer rounded to four decimal places is a

43 4. Right Triangle Trig 5 Next, let s solve for c. To get the most accurate answer, we want to use as much of the original, given information as possible, and avoid using our rounded answer for a if at all possible. So, c is the hypotenuse, and we were given the length of the opposite side. The basic trig ratio involving the opposite side and the hypotenuse is sine, so we create the equation: sin(α) = opposite hypotenuse sin(0 ) = 7 c 7 c = sin(0 ) Luckily, this time our answer happens to be exact (no need to round): c = 14 Check Point 11 For the right triangle below, find the length of x.

44 6 Foundations of Trigonometry Check Point 1 For the right triangle below, find the length of x. We can also use trigonometric ratios to solve for missing angles in a right triangle. First, we must gently introduce inverse trigonometric functions. We will study these inverse functions more deeply in a later section. For now, we only need to know the following relationship: Definition 1: Inverse Trigonometric Functions If sin(a) = b and 90 < a < 90, then a = sin 1 (b) If cos(a) = b and 0 < a < 180, then a = cos 1 (b) If tan(a) = b and 90 < a < 90, then a = tan 1 (b) There are a couple of extremely important caveats to notice and understand. First of all: sin 1 (x) is not the same as cos 1 (x) is not the same as tan 1 (x) is not the same as 1 sin(x) 1 cos(x) In other words, that exponent is not really an exponent. It s rather a tag that means we want to use the inverse function here. Confusing, we know. If someone would make us monarchs of math, we d get rid of this confusing notation for good. Alas. If we did want, for example,, we d simply use csc(x). Also notice that the inverse trigonometric functions swap the inputs and outputs of the trig functions. For example, with sin(a) = b, the input (a) is an angle and the output, b, is the ratio of the opposite side over the 1 tan(x) 1 sin(x)

45 4. Right Triangle Trig 7 hypotenuse of a triangle with reference angle a. With sin 1 (b) = a, the input is that ratio of opposite side over hypotenuse, and the output tells us the reference angle we d need to produce that ratio. Therefore, we ll be able to use these inverse functions to solve for angles. Finally, notice that these inverse trigonometric functions are defined only for particular intervals of angles. That will be nothing to worry about in this section, since we re dealing in right triangles. Later on, we ll have to be a little bit more careful when interpreting the results after applying an inverse trigonometric function. Definition 1: Given two sides of a right triangle, solve for an angle Determine which sides of the triangle are involved and the corresponding main trigonometric ratio (i.e. sine, cosine, or tangent). Write an equation using that trigonometric ratio and the sides involved. Rewrite the equation using the corresponding inverse trigonometric function. Use a calculator to evaluate approximately. Example 15 Solve for angle θ In the picture, the side labeled 9 is adjacent to the reference angle, and the side labeled 1 is the hypotenuse. The common ratio involving adjacent and hypotenuse is cosine, so cos(θ) = 9 1 Rewrite as an inverse function: ) θ = cos 1 ( 9 1

46 8 Foundations of Trigonometry Use a calculator to evaluate approximately: θ Example 16: Solve for angle θ In the picture, the side labeled 6 is opposite to the reference angle, and the side labeled 10 is the hypotenuse. The common ratio involving opposite and hypotenuse is sine, so sin(θ) = 6 10 Rewrite as an inverse function: ) Use a calculator to evaluate approximately: θ = sin 1 ( 6 10 θ

47 4. Right Triangle Trig 9 Check Point 1 For the right triangle below, find the measure of the indicated angle. Check Point 14 For the right triangle below, find the measure of the indicated angle.

48 40 Foundations of Trigonometry Special Right Triangles Triangles There are some triangles that are so common, and so special, that it s worth knowing the values of sine and cosine that they produce. To build the first type of special right triangle, we ll start with an equilateral triangle. In an equilateral triangle, all the angles are the same measure (60 ) and all the sides are the same length (we ll call that length x for now): To create a right triangle that we can work with, we ll draw an altitude in this triangle. That will split the base exactly in half, and it will split the top angle exactly in half:

49 4. Right Triangle Trig 41 Now we can use the Pythagorean Theorem to solve for the height of this triangle: x + h = (x) x + h = 4x h = x h = x That gives us this reference triangle: Now we can figure out the sine and cosine of 0 : sin(0 ) = x x = 1 cos(0 ) = x x = and the sine and cosine of 60 : sin(60 ) = x x = cos(60 ) = x x = 1

50 4 Foundations of Trigonometry Triangles There s another set of special sine and cosine values that we want to explore. Our door into these will be an isoscoles right triangle: Again, we ll use the Pythagorean Theorem to solve, but this time for the hypotenuse: x + x = c x = c c = x And then we can find the sine and cosine of 45 : sin(45 ) = cos(45 ) = x x = 1 x x = 1 Commonly, we ll rationalize the denominator of this fraction: 1 = sin(45 ) = cos(45 ) =

51 4. Right Triangle Trig 4 The sines and cosines of these angles are going to become our particular friends in this chapter. We ll also want to get to know them well when we re thinking in terms of radians. We recommend that you commit to memorizing the following (like back in the days when you learned your times tables!). Definition 14: Sines and Cosines of Special Angles ( π ) sin(0 ) = sin = 1 6 ( π ) cos(0 ) = cos = 6 ( π ) sin(60 ) = sin = ( π ) cos(60 ) = cos = 1 ( π ) sin(45 ) = sin = 4 ( π ) cos(45 ) = cos = 4 Check Point 15 Evaluate exactly: cos(60 ) = ( π ) sin = 4

52 44 Foundations of Trigonometry Using Right Triangle Trigonometry to Solve Applied Problems Right-triangle trigonometry has many practical applications. For example, the ability to compute the lengths of sides of a triangle makes it possible to find the height of a tall object without climbing to the top or having to extend a tape measure along its height. We do so by measuring a distance from the base of the object to a point on the ground some distance away, where we can look up to the top of the tall object at an angle. The angle of elevation of an object above an observer relative to the observer is the angle between the horizontal and the line from the object to the observer s eye. The right triangle this position creates has sides that represent the unknown height, the measured distance from the base, and the angled line of sight from the ground to the top of the object. Knowing the measured distance to the base of the object and the angle of the line of sight, we can use trigonometric functions to calculate the unknown height. Similarly, we can form a triangle from the top of a tall object by looking downward. The angle of depression of an object below an observer relative to the observer is the angle between the horizontal and the line from the object to the observer s eye. Example 17 How long a ladder is needed to reach a windowsill 50 feet above the ground if the ladder rests against the building making an angle of 75 with the ground? Round to the nearest foot. First, draw the situation:

53 4. Right Triangle Trig 45 We know the side opposite the reference angle is 50 feet, and we want to solve for the hypotenuse (which represents the length of the ladder). Therefore, we ll use a sine function to solve the problem: sin(75 ) = 50 x x = 50 sin(75 x ft Example 18 A radio tower is located 5 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is 4, and that the angle of depression to the bottom of the tower is 1. How tall is the tower? Please forgive the author s hand-drawn illustration:

54 46 Foundations of Trigonometry As pictured above, this situation creates two right triangles. We want to solve in each of the opposite side to our reference angles, and we know that the adjacent side is of length 5 ft. Hence, we can use the tangent function: tan(4 ) = x 1 5 x 1 = 5 tan(4 ) and tan(1 ) = x 5 x = 5 tan(1 ) If we add up these values, we have the total height of the tower (remember not to round until the very end!): x 1 + x = 5 tan(4 ) + 5 tan(1 ) ft Check Point 16 The angle of elevation to the top of a Building in New York is found to be 7 degrees from the ground at a distance of 1 mile from the base of the building. Using this information, find the height of the building.

55 4. Right Triangle Trig 47 Check Point 17 A radio tower is located 75 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is 7 and that the angle of depression to the bottom of the tower is 0. How tall is the tower? 4..1 Exercises In the following exercises, find the requested values. 1. Find θ, a, and c.. Find α, b, and c.. Find θ, a, and c. 4. Find β, b, and c.

56 48 Foundations of Trigonometry In the following exercises, answer assuming θ is an angle in a right triangle. 5. If θ = 0 and the side opposite θ has length 4, how long is the side adjacent to θ? 6. If θ = 15 and the hypotenuse has length 10, how long is the side opposite θ? 7. If θ = 87 and the side adjacent to θ has length, how long is the side opposite θ? 8. If θ = 8. and the side opposite θ has lengh 14, how long is the hypoteneuse? 9. If θ =.05 and the hypotenuse has length.98, how long is the side adjacent to θ? 10. If θ = 4 and the side adjacent to θ has length 1, how long is the side opposite θ? In the following exercises, find the two acute angles in the right triangle whose sides have the given lengths. Express your answers using degree measure rounded to two decimal places. 11., 4, and , 1, and , 57, and A tree standing vertically on level ground casts a 10 foot long shadow. The angle of elevation from the end of the shadow to the top of the tree is 1.4. Find the height of the tree to the nearest foot. With the help of your classmates, research the term umbra versa and see what it has to do with the shadow in this problem. 15. The broadcast tower for radio station WSAZ (Home of Algebra in the Morning with Carl and Jeff ) has two enormous flashing red lights on it: one at the very top and one a few feet below the top. From a point 5000 feet away from the base of the tower on level ground the angle of elevation to the top light is and to the second light is Find the distance between the lights to the nearest foot. 16. From a firetower 00 feet above level ground in the Sasquatch National Forest, a ranger spots a fire off in the distance. The angle of depression to the fire is.5. How far away from the base of the tower is the

57 4. Right Triangle Trig 49 fire? 17. The ranger in question 16. sees a Sasquatch running directly from the fire towards the firetower. The ranger takes two sightings. At the first sighting, the angle of depression from the tower to the Sasquatch is 6. The second sighting, taken just 10 seconds later, gives the the angle of depression as 6.5. How far did the Saquatch travel in those 10 seconds? Round your answer to the nearest foot. How fast is it running in miles per hour? Round your answer to the nearest mile per hour. If the Sasquatch keeps up this pace, how long will it take for the Sasquatch to reach the firetower from his location at the second sighting? Round your answer to the nearest minute. 18. When I stand 0 feet away from a tree at home, the angle of elevation to the top of the tree is 50 and the angle of depression to the base of the tree is 10. What is the height of the tree? Round your answer to the nearest foot. 19. From the observation deck of the lighthouse at Sasquatch Point 50 feet above the surface of Lake Ippizuti, a lifeguard spots a boat out on the lake sailing directly toward the lighthouse. The first sighting had an angle of depression of 8. and the second sighting had an angle of depression of 5.9. How far had the boat traveled between the sightings? 0. A guy wire 1000 feet long is attached to the top of a tower. When pulled taut it makes a 4 angle with the ground. How tall is the tower? How far away from the base of the tower does the wire hit the ground? 1. A guy wire 1000 feet long is attached to the top of a tower. When pulled taut it touches level ground 60 feet from the base of the tower. What angle does the wire make with the ground? Express your answer using degree measure rounded to one decimal place.. At Cliffs of Insanity Point, The Great Sasquatch Canyon is 7117 feet deep. From that point, a fire is seen at a location known to be 10 miles away from the base of the sheer canyon wall. What angle of depression is made by the line of sight from the canyon edge to the fire? Express your answer using degree measure rounded to one decimal place.. Shelving is being built at the Utility Muffin Research Library which is to be 14 inches deep. An 18-inch rod will be attached to the wall and the underside of the shelf at its edge away from the wall, forming a right triangle under the shelf to support it. What angle, to the nearest degree, will the rod make with the wall? 4. A parasailor is being pulled by a boat on Lake Ippizuti. The cable is 00 feet long and the parasailor is 100 feet above the surface of the water. What is the angle of elevation from the boat to the parasailor? Express your answer using degree measure rounded to one decimal place. 5. A tag-and-release program to study the Sasquatch population of the eponymous Sasquatch National Park is begun. From a 00 foot tall tower, a ranger spots a Sasquatch lumbering through the wilderness directly towards the tower. Let θ denote the angle of depression from the top of the tower to a point on the

58 50 Foundations of Trigonometry ground. If the range of the rifle with a tranquilizer dart is 00 feet, find the smallest value of θ for which the corresponding point on the ground is in range of the rifle. Round your answer to the nearest hundreth of a degree. 6. The rule of thumb for safe ladder use states that the length of the ladder should be at least four times as long as the distance from the base of the ladder to the wall. Assuming the ladder is resting against a wall which is plumb (that is, makes a 90 angle with the ground), determine the acute angle the ladder makes with the ground, rounded to the nearest tenth of a degree.

59 4. Trig in the Plane Trig in the Plane Learning Objectives In this section you will: Find the quadrant in which the terminal side of an angle in standard position falls Find reference angles in the coordinate plane Use reference triangles to calculate trigonometric ratios in the coordinate plane Use reference triangles and special right triangles to calculate trigonometric ratios exactly for common angles. Reference Angles So far, we ve only defined trigonometric ratios based on what s going on inside of a right triangle. We want to break free of those limitations; for instance, can we define a trigonometric value for an angle that s too big to fit inside a right triangle, like say 190? For this, we ll return to our angles in standard position in the coordinate plane and define their reference angles. To formalize our work, we will begin by drawing angles on an x-y coordinate plane. Angles can occur in any position on the coordinate plane, but for the purpose of comparison, the convention is to illustrate them in the same position whenever possible. An angle is in standard position if its vertex is located at the origin, and its initial side extends along the positive x-axis.

60 5 Foundations of Trigonometry Within this framework, we want to be able to identify in which quadrant a particular angle s terminal side will fall. As we see in the illustration above, the angles with terminal sides in quadrant I will be between 0 and 90, that is, between 0 and π radians. The angles with terminal sides in quadrant II will be between 90 and 180, or π and π. The angles with terminal sides in quadrant III will be between 180 and 70, or π and π. Finally, angles with terminal sides in quadrant IV will be between 70 and 60 or π and π. Note that there are also many coterminal angles larger or smaller than this first full revolution that may fall in these quadrants, but for our purposes now we will focus only on positive angles within one revolution.

61 4. Trig in the Plane 5 Definition 15: Reference Angles Let a be an angle in standard position in the coordinate plane. The reference angle, r, is the smallest angle formed between the terminal side of a and the x axis. Reference Angles in each quadrant, along with relevant formulas, are pictured below. Q1 QII QIII QIV We ll practice finding reference angles next. Rather than strictly memorizing the formulas above, it is more useful to think of the geometry of the situation; we re using the fact that a straight line is an angle of 180 or π radians to help calculate QII and QIII references. Similarly, we re using the fact that a full revolution is an angle of 60 or π radians in our QIV calculations.

62 54 Foundations of Trigonometry Example 19 Given each angle, find the reference angle A. 156 B. 5π 4 A. 156 First, we must identify which quadrant the terminal side of this angle lies in. 156 is between 90 and 180, so its terminal side lies in QII. To find a reference angle in QII, we subtract the actual angle from the measure of a straight line, 180 : r = = 4 B. 5π 4 Part of our challenge here is to think in radians, not convert to degrees. To identify the quadrant in which the terminal side of this angle lies, consider that 5π 4 is between π and π = 6π. This tells 4 us that the terminal side of this angle lies in QIII. To find a reference angle in QIII, we subtract the measure of a straight line (π radians) from the actual angle: r = 5π 4 π = 5π 4 4π 4 = π 4 Check Point 18 Find the reference angles for 11,, 11, π, π 4, 11π 6.

63 4. Trig in the Plane 55 Reference Triangles Now that we have a notion of what a reference angle is, we can start building reference right triangles in the coordinate plane. From there, we can start to think about the trigonometric ratios formed by these reference triangles. The most important thing to note as we build these triangles is that, by convention, the hypotenuse is always a positive (distance) value, whereas the signs of the legs will depend on the quadrant. Definition 16: Reference Triangles To build a reference triangle for angle a in standard position, we connect the terminal side of angle a to the x-axis with a right angle. Q1 QII QIII QIV Because we are working in the coordinate plane, in the first quadrant, both of the legs of the triangle will have positive values. Hence, the sin(r) and the cos(r) will both be positive. In quadrant II, the side adjacent

64 56 Foundations of Trigonometry to the reference angle will have a negative value, meaning the cos(r) will be negative. In this way, we can determine the signs of the trig ratios in each quadrant. Definition 17: Signs of the Trigonometric Ratios in Each Quadrant This brings us to how we will define trigonometric ratios on the coordinate plane. Using these reference triangles, we will say, for instance, that the sin(a) = ± sin(r). How that ± is determined will depend on the quadrant in which our reference triangle lies.

65 4. Trig in the Plane 57 Definition 18: Reference Angle Theorem Suppose r is the reference angle for a. Then cos(a) = ± cos(r) and sin(a) = ± sin(r), where the choice of the (±) depends on the quadrant in which the terminal side of a lies. Basically this means we can use the reference triangle to find the sine, cosine, and tangent of any angle a in the coordinate plane, as long as we pay attention to the quadrant that we re working in so that we can figure out the signs! Let s try it out. Example 0 The terminal side of angle α in standard position passes through the point (, 4). values of the six trigonometric functions of α. Find the It helps immensely to draw a picture of the situation: The coordinates we are given help us to determine that the side adjacent to the reference angle is labelled -, and the side opposite the reference angle is labelled -4. To figure out sine and cosine, we also need to find the length of the hypotenuse. We can use the Pythagorean Theorem to figure

66 58 Foundations of Trigonometry this out: ( ) + ( 4) = c = c c = 0 Since the hypotenuse is always positive, we do not have to worry about considering the 0 for the hypotenuse label. Now that we have all of the lengths, and we know the terminal side is located in QIII, we can find all of the trigonometric ratios for angle α. sin(α) = sin(r) = 4 = 0 4 = csc(α) = cos(α) = cos(r) = 0 = 5 = 5 sec(α) = 1 sin(α) = 5 1 cos(α) = 1 = tan(α) = tan(r) = 4 = cot(α) = 1 tan(α) = 1 Check Point 19 The terminal side of angle θ in standard position goes through the point (14, -15). Find the values of the six trigonometric functions of θ. Example 1 If sin(θ) = 1 and the terminal side of angle θ is in quadrant II, find cos(θ). 1 Again, it helps us to draw a picture. Remember, trigonometry operates using the properties of similar triangles; we can draw any triangle in QII that has a sine of 1, so let s draw the simplest 1 version of that: a triangle with opposite side 1 and hypotenuse 1.

67 4. Trig in the Plane 59 All we need in order to find the cosine is the length of the adjacent side. Pythagorean Theorem to solve for the length of that side. Again, let s use the b + 1 = 1 b = 169 b = 5 b = ±5 To decide whether b = 5 or 5, consider the quadrant; in QII, the adjacent side is negative, so b = 5. Therefore cos(θ) = cos(r) = 5 1 Check Point 0 If sin(θ) = 4 and the terminal side of θ falls in QII, find the exact value of cos(θ). 5

68 60 Foundations of Trigonometry Calculating exact values of special angles For our next task, it pays to know the cosine and sine values for certain common angles. In the table below, we summarize the values which we consider essential and must be memorized. θ (degrees) θ (radians) cos(θ) sin(θ) 0 π π 4 60 π 1 Example Find the cosine and sine of the following angles. A. θ = 5 B. θ = 11π 6 C. θ = 5π 4 D. θ = 7π A. θ = 5 We begin by plotting θ = 5 in standard position and find its terminal side overshoots the negative x-axis to land in Quadrant III. Hence, we obtain θ s reference angle r by subtracting: r = θ 180 = = 45. Since θ is a Quadrant III angle, both cos(θ) < 0 and sin(θ) < 0. The Reference Angle Theorem yields: cos (5 ) = cos (45 ) = sin (5 ) = sin (45 ) =. and B. θ = 11π 6 The terminal side of θ = 11π, when plotted in standard position, lies in Quadrant IV, just shy of the 6

69 4. Trig in the Plane 61 positive x-axis. To find θ s reference angle r, we subtract: r = π θ = π 11π = π 6 6. Since θ is a Quadrant IV angle, cos(θ) > 0 and sin(θ) < 0, so the Reference Angle Theorem gives: cos ( 11π 6 ) = cos ( π 6 ) = and sin ( 11π 6 ) = sin ( π 6 ) = 1. C. θ = 5π 4 To plot θ = 5π 4, we rotate i clockwise /i an angle of 5π from the positive x-axis. The terminal 4 side of θ, therefore, lies in Quadrant II making an angle of r = 5π π 4 = π radians with respect 4 to the negative x-axis. Since θ is a Quadrant II angle, the Reference Angle Theorem gives: cos ( 5π ) ( 4 = cos π ) 4 = and sin ( 5π ) ( 4 = sin π ) 4 =. D. θ = 7π Since the angle θ = 7π measures more than π = 6π, we find the terminal side of θ by rotating one full revolution followed by an additional r = 7π π = π radians. Since θ and r are coterminal, cos ( ) ( 7π = cos π ) = 1 and sin ( ) ( 7π = sin π ) =. Check Point 1 Without using a calculator, compute the sine and cosine of 15. Check Point Without using a calculator, compute the sine and cosine of 15π 4.

70 6 Foundations of Trigonometry 4..1 Exercises In the following exercises, let θ be the angle in standard position whose terminal side contains the given point. Compute cos(θ) and sin(θ). 1. (-7, 4). (, 4). (5, -9) 4. (-, -11) In the following exercises, solve for the requested value. 5. If sin(θ) = 7 with θ in Quadrant IV, what is cos(θ)? 5 6. If cos(θ) = 4 with θ in Quadrant I, what is sin(θ)? 9 7. If sin(θ) = 5 with θ in Quadrant II, what is cos(θ)? 1 8. If cos(θ) = with θ in Quadrant III, what is sin(θ)? If sin(θ) = with θ in Quadrant III, what is cos(θ)? 10. If cos(θ) = 8 with θ in Quadrant IV, what is sin(θ)? If sin(θ) = 5 and π < θ < π, what is cos(θ)? If cos(θ) = 10 and π < θ < 5π, what is sin(θ)? Find the exact value of the cosine and sine of the given angle (without using a calculator). 1. θ = π 14. θ = π θ = 7π θ = 5π θ = 4π 18. θ = 5π 19. θ = 7π 4 0. θ = π 6 1. θ = 4π 6

71 4. Trig in the Plane 6. θ = π 4. θ = π 6 4. θ = 10π

72 64 Foundations of Trigonometry

73 Chapter 5 The Circular Functions 5.1 The Unit Circle Learning Objectives In this section you will: Build the unit circle using special right reference triangles Create basic trigonometric identities using the geometry of the unit circle Looking for a thrill? Then consider a ride on the Singapore Flyer, the world s tallest Ferris wheel. Located in Singapore, the Ferris wheel soars to a height of 541 feet: a little more than a tenth of a mile! Described as an observation wheel, riders enjoy spectacular views as they travel from the ground to the peak and down again in a repeating pattern. In this section, we will examine this type of revolving motion around a circle. To do so, we need to define the type of circle first, and then place that circle on a coordinate system. Then we can discuss circular motion in terms of the coordinate pairs.

74 66 The Circular Functions The Unit Circle We have already defined the trigonometric functions in terms of right triangles. In this section, we will redefine them in terms of the unit circle. The Unit Circle is a circle centered at the origin with radius 1. Therefore, the equation of the Unit Circle is (x 0) + (y 0) = 1. x + y = 1 For a point on the unit circle, we can create a reference triangle by connecting the point to the origin and dropping a perpendicular side down to the x-axis: Using that reference triangle, we can figure out the sine and cosine of θ: sin(θ) = y 1 y = sin(θ) cos(θ) = x 1 x = cos(θ)

75 5.1 The Unit Circle 67 This gives us a new way of defining and thinking about sine and cosine: we can say that if a point on the unit circle, (x, y) corresponds to a central angle of θ, then cos(θ) = x and sin(θ) = y. By associating the point P = (x, y) with the angle θ, we are assigning a position on the Unit Circle to the angle θ. Since for each angle θ, the terminal side of θ, when graphed in standard position, intersects The Unit Circle only once, the mapping of θ to P is a function. Since there is only one way to describe a point using rectangular coordinates, the mappings of θ to each of the x and y coordinates of P are also functions. When we break the sine and cosine free from triangles this way and instead define them in regards to the unit circle, we call them Circular Functions. This new definition works for angles in the other quadrants as well, and it corresponds to what we learned in Section 4.. For a point in quadrant II, for example, we could draw a reference triangle like this: Now, because this triangle is in quadrant II, the value of the x-coordinate will be negative, meaning the cosine of our reference angle will take on a negative value, as we know it should in quadrant II.

76 68 The Circular Functions Definition 19: The Unit Circle and The Circular Functions The Unit Circle is a circle centered at the origin with radius 1 and equation x + y = 1. Suppose an angle θ is graphed in standard position. Let P(x, y) be the point of intersection of the terminal side of P and the Unit Circle. The x-coordinate of P is called the cosine of θ, written cos(θ). The y-coordinate of P is called the sine of θ, written sin(θ). Using this new definition of sine and cosine, we can break out of a right triangle and find sine and cosine values of big angles, negative angles, even angles like 90 and 180! First, we ll work with angles on the axes, which we will call Quadrantal Angles. Example : Finding the sine and cosine of a quadrantal angle Using the unit circle, find the sine and cosine of 90 If we were stuck inside a triangle, this wouldn t make much sense. Using SOHCAHTOA, we can only find sine and cosine of reference angles, not the 90 angle. If we think about the unit circle, though, we can think about the coordinates which correspond to a 90 reference angle:

77 5.1 The Unit Circle 69 Since the length of the radius is 1, the point that corresponds to our central angle of 90 is (0,1). Therefore cos(θ) = x cos(90 ) = 0 and sin(θ) = y sin(90 ) = 1

78 70 The Circular Functions Check Point Find the sine and cosine of 180. Let s jump right into our picture on the unit circle: So, the coordinates that correspond with our 180 central angle are ( 1, 0), meaning cos(180 ) = 1 and sin(180 ) = 0 To help with visualizing the relationships between the sine and cosine of reference triangles and sine and cosine defined as circular functions, you can play around with this interactive demonstration: In particular, look at what happens with a central angle of 0. Drag the point on the unit circle down until it rests on the positive x-axis, watching the reference triangle as you go. Think about the cosine of θ in that triangle, the ratio of the adjacent side and the hypotenuse. As we drag the point down, the hypotenuse and adjacent sides get closer and closer to the same length until they are the same length. Therefore, we end up with a cosine of 1!

79 5.1 The Unit Circle 71 Similarly, if we think about the sine of that reference angle, we look at the opposite side and the hypotenuse. As we drag that point towards the positive x-axis, the opposite side gets smaller and smaller until it vanishes to 0. Therefore the sine ends up being 0! Check Point 4 Evaluate each of the following: cos(0 ) sin(90 ) sin(70 ) Special Right Triangles Now we turn towards a large memorization task. In Section 4., we learned about the sines and cosines of some special angles (0, 60, and 45 ). Definition 0: Sines and Cosines of Special Angles ( π ) sin(0 ) = sin = 1 6 ( π ) cos(0 ) = cos = 6 ( π ) sin(60 ) = sin = ( π ) cos(60 ) = cos = 1 ( π ) sin(45 ) = sin = 4 ( π ) cos(45 ) = cos = 4

80 7 The Circular Functions We re going to use these special values and the properties of the unit circle to construct a cheat sheet of sorts with all of the most common angles and their sine and cosine values. Let s begin with the 0 (aka ) angle values: π 6 From the properties of special triangles, we know that the cos(0 ) = label the point of intersection of the terminal side and the unit circle as length of the horizontal leg is and the length of the vertical leg is 1. ( and sin(0 ) = 1, so we can ). That also tells us that the, 1 Now, let s reflect that triangle across the y-axis. We ll then have a reference angle of 0 in QII, which corresponds to an angle of = 150 in standard position. Because ( of the reflection, we know that ) the point of intersection of the terminal side and the unit circle will be, 1, since our point is to the left of the origin and above the x-axis! So, then, we know cos(150 ) = and sin(150 ) = 1. Convert to radians, and we have cos ( ) 5π 6 = and sin ( ) 5π 6 = 1.

81 5.1 The Unit Circle 7 We continue reflecting this triangle around to quadrants III and IV, and we have the beginnings of our special values cheat sheet! In Quadrant III, our reference of 0 gives us a standard angle of = 10, which is 7π 6 radians. In Quadrant IV, we have 60 0 = 0 or 11π radians. Using the fact that the x- 6 and y-coordinates are both negative in QIII, and only the y s are negative in QIV, we then have ( ) 7π cos(10 ) = cos = 6 ( ) 7π sin(10 ) = sin = 1 6 ( ) 11π cos(0 ) = cos = 6 ( ) 11π sin(0 ) = sin = 1 6

82 74 The Circular Functions As we continue building up our key unit circle values, the unit circle itself will become a shortcut way of remembering these important cosine and sine values. Eventually, we want to commit all of these important values to memory for quickier and easier calculations in our math futures.

83 5.1 The Unit Circle 75 Check Point 5 Using the properties of the unit circle, find the cosine and sine values of the angles with a 60 reference angle in each quadrant.

84 76 The Circular Functions Check Point 6 Using the properties of the unit circle, find the cosine and sine values of the angles with a 45 reference angle in each quadrant.

85 5.1 The Unit Circle 77 Now we can put this information all together, along with what we ve learned about quadrantal angles, to create our Unit Circle with common trigonometric values! Definition 1: The Unit Circle with Trigonometric Values for Special Angles Note that in the coming sections, we will be shifting our focus to thinking in terms of radians rather than in degrees, so all of the angles in this reference unit circle are labelled with radians, while only the quadrantal and QI angles also are labelled with their degree measures. Once we have these special angles memorized on the unit circle, we can quickly figure out their cosine and sine values. For instance, consider 7π. Find that spoke on the unit circle; the corresonding coordinates are ( ) 4,, so cos ( ) 7π 4 = and sin ( ) 7π 4 =.

86 78 The Circular Functions Moving Beyond the Unit Circle Now that we have these basic special angles memorized (or on the way to being memorized), we can work on finding the values of trigonometric functions of coterminal angles. Example 4 Find the exact value of sin ( ) 14π 14π is simply too big for us to deal with right now, so let s find a smaller coterminal angle: Still too big! Let s go smaller... 14π π = 14π 6π = 11π 11π 6π = 5π That s an angle we recognize from our common unit circle values! Since this is coterminal with our original angle, ( ) ( ) 14π 5π sin = sin = Example 5 Find the exact value of cos ( 5π 6 ) Again, we start with finding a positive coterminal angle: 5π 6 + π

87 5.1 The Unit Circle 79 = 5π 6 + 1π 6 = 7π 6 So we have cos ( 5π 6 ) ( ) 7π = cos = 6 Check Point 7 Compute the exact value of sin ( 17π 6 ). Check Point 8 Compute the exact value of cos( 4π). So far, we define the sine and cosine functions using the Unit Circle, x + y = 1. It turns out that we can use any circle centered at the origin to determine the sine and cosine values of angles. To show this, we essentially recycle the same similarity arguments to show the trigonometric ratios described so far are independent of the choice of right triangle used. Consider for the moment the acute angle θ drawn below in standard position. Let Q(x, y) be the point on the terminal side of θ which lies on the circle x + y = r, and let P(x, y ) be the point on the terminal side of θ which lies on the Unit Circle. Now consider dropping perpendiculars from P and Q to create two right triangles, OPA and OQB. These triangles are similar, thus it follows that = r = r, so x = rx 1 and, similarly, we find y = ry. Since, by definition, x = cos(θ) and y = sin(θ), we get the coordinates of Q to be x = r cos(θ) and y = r sin(θ). By reflecting these points through the x-axis, y-axis and origin, we obtain the result for all non-quadrantal angles θ, and we leave it to the reader to verify these formulas hold for the quadrantal angles as well. x x

88 80 The Circular Functions Not only can we describe the coordinates of Q in terms of cos(θ) and sin(θ) but since the radius of the circle is r = x + y, we can also express cos(θ) and sin(θ) in terms of the coordinates of Q. These results are summarized in the following theorem. Definition If Q(x, y) is the point on the terminal side of an angle θ, plotted in standard position, which lies on the circle x + y = r then x = r cos(θ) and y = r sin(θ). Moreover, cos(θ) = x r = x and sin(θ) = y x + y r = y x + y Example 6 Suppose that the terminal side of an angle θ, when plotted in standard position, contains the point Q(4, ). Find sin(θ) and cos(θ). Since we are given both the x and y coordinates of a point on the terminal side of this angle, we can use our theorem directly. First, we find r = x + y = ( ) + 4 = 0 = 5. This means the point Q lies on a circle of radius 5 units as seen below on the left. Hence,

89 5.1 The Unit Circle 81 cos(θ) = x r = 4 = and sin(θ) = y r = 5 = 5 5. Example 7 Suppose π 8 < θ < π with sin(θ) =. Find cos(θ). 17 We are told π < θ < π, so, in particular, θ is a Quadrant II angle. Per our theorem, sin(θ) = 8 17 = y r where y is the y-coordinate of the intersection point of the circle x + y = r and the terminal side of θ (when plotted in standard position, of course!) For convenience, we choose r = 17 so that y = 8, and we get the diagram below on the right. Since x + y = r, we get x + 8 = 17. We find x = ±15, and since θ is a Quadrant II angle, we get x = 15. Hence, cos(θ) = Fundamental Trigonometric Identities Now with the Unit Circle in hand, we can go about establishing some fundamental identities. Recall that the equation for the unit circle is x + y = 1. Because x = cos(θ) and y = sin(θ), we can substitute these trigonometric functions into the equation of the unit circle for x and y: x + y = 1 (cos(θ)) + (sin(θ)) = 1 It is common practice to write structures like (cos(θ)) as cos (θ)), so we finish by rewriting this equation as cos (θ) + sin (θ) = 1 The nifty, revolutionary, thing about this equation is that it s true for any real value of θ! So, we know that ( π ) ( π ) cos + sin = and cos ( 87 ) + sin ( 87 ) = 1 and so on.

90 8 The Circular Functions Because it s so revolutionary, we name this identity; since we can connect it to the Pythagorean Theorem when we re thinking about right triangles, we call it the Pythagorean Identity. Definition : The Pythagorean Identity For any real number θ, cos (θ) + sin (θ) = 1 Example 8 If sin(θ) =, and the terminal side of θ is in the second quadrant, find cos(θ). 7 According to the Pythagorean Identity, cos (θ) + sin (θ) = 1 So, we can substitute in in place of sin(θ): 7 ( ) cos (θ) + = 1 7 cos (θ) = 1 cos (θ) = cos (θ) = Now things get a little interesting. If we apply the square root to both sides of this equation, we have: 40 cos (θ) = cos(θ) = cos(θ) = or 7 7

91 5.1 The Unit Circle 8 Well which one is it? We re talking about a specific angle θ here, so the cosine can t be both positive and negative 40 ; cosine is a function afterall. We need to look at the other piece of given 7 information to make a decision: the terminal side of θ is in the second quadrant. In the second quadrant, the cosine of an angle is always negative (think about the fact that x-coordinates are negative in the second quadrant), so our final answer is 40 cos(θ) = 7 Definition 4: Using the Pythagorean Identity to solve for a sin(θ) given cos(θ) or vice versa Substitute the known value into the Pythagorean Identity. Solve for the unknown value. Choose the appropriate sign for the solution based on the quadrant in which the terminal side of the angle is located. Example 9 If cos(t) = 4 and the terminal side of t is located in the fourth quadrant, find sin(t) Substitute the known value into the Pythagorean Identity. ( ) 4 + sin (t) = 1. Solve for the unknown value sin (t) = 1 sin (t) = 9 5 sin(t) = ± 5

92 84 The Circular Functions. Choose the appropriate sign for the solution based on the quadrant in which the terminal side of the angle is located. Since the terminal side of t is located in the fourth quadrant, the sine (y-values) should be negative, so our final answer is sin(t) = 5 Check Point 9 If cos(θ) = 1 and θ is in the 1st quadrant, find sin(θ). 6 Check Point 0 If sin(θ) = 1 and θ is in the nd quadrant, find cos(θ) Exercises 1. Describe the unit circle.. What do the x- and y-coordinates of the points on the unit circle represent? For the following exercises, use the given sign of the sine and cosine functions to find the quadrant in which the terminal point determined by θ lies.. sin(θ) < 0 and cos(θ) < 0 4. sin(θ) > 0 and cos(θ) > 0 5. sin(θ) > 0 and cos(θ) < 0 6. sin(θ) < 0 and cos(θ) > 0 In the following exercises, find the exact value of the cosine and sine of the given angle. 7. θ = 0 8. θ = π 4 9. θ = π 10. θ = π 11. θ = π 1. θ = π 4 1. θ = π 14. θ = 7π θ = 5π θ = 4π 17. θ = π 18. θ = 5π 19. θ = 7π 4 0. θ = π 6 1. θ = 1π. θ = 4π 6. θ = π 4 4. θ = π 6 5. θ = 10π 6. θ = 117π

93 5.1 The Unit Circle 85 In the following exercises, let θ be the angle in standard position whose terminal side contains the given point then compute cos(θ) and sin(θ). 7. P( 7, 4) 8. Q(, 4) 9. R(5, 9) 0. T (, 11) 1. If sin(θ) = 7 with θ in Quadrant IV, what is cos(θ)? 5. If cos(θ) = 4 with θ in Quadrant I, what is sin(θ)? 9. If sin(θ) = 5 with θ in Quadrant II, what is cos(θ)? 1 4. If cos(θ) = with θ in Quadrant III, what is sin(θ)? If sin(θ) = with θ in Quadrant III, what is cos(θ)? 6. If cos(θ) = 8 with θ in Quadrant IV, what is sin(θ)? 5 7. If sin(θ) = 5 and π < θ < π, what is cos(θ)? If cos(θ) = 10 and π < θ < 5π, what is sin(θ)? 9. If sin(θ) = 0.4 and π < θ < π, what is cos(θ)? 40. If cos(θ) = 0.98 and π < θ < π, what is sin(θ)?

94 86 The Circular Functions 5. Other Circular Functions Learning Objectives In this section you will: Discover the connection between sine, cosine, and the other trigonometric ratios. Evaluate the values of all trig ratios for special unit circle angles. Evaluate the values of trig ratios beyond the unit circle. A wheelchair ramp that meets the standards of the Americans with Disabilities Act must make an angle with the ground whose tangent is 1 or less, regardless of its length. A tangent represents a ratio, so this 1 means that for every 1 inch of rise, the ramp must have 1 inches of run. Trigonometric functions allow us to specify the shapes and proportions of objects independent of exact dimensions. We have already defined the sine and cosine functions of an angle. Though sine and cosine are the trigonometric functions most often used, there are four others. Together they make up the set of six trigonometric functions. In this section, we will investigate the remaining functions. Connecting Sine and Cosine to the other Trigonometric Functions We can also define the remaining functions in terms of the unit circle with a point (x,y) corresponding to an angle of θ, as shown below:

95 5. Other Circular Functions 87 From right triangle trigonometry, we know that which in the case of our triangle here, means tan(θ) = opposite adjacent tan(θ) = y x. In a similar way, we can connect all of our trigonometric functions to the coordinates of the unit circle. Definition 5: The Circular Functions Suppose an angle θ is graphed in standard position. Let P(x, y) be the point of intersection of the terminal side of P and the Unit Circle. The sine of θ, denoted sin(θ), is defined by sin(θ) = y. The cosine of θ, denoted cos(θ), is defined by cos(θ) = x. The tangent of θ, denoted tan(θ), is defined by tan(θ) = y, provided x 0. x The secant of θ, denoted sec(θ), is defined by sec(θ) = 1, provided x 0. x The cosecant of θ, denoted csc(θ), is defined by csc(θ) = 1, provided y 0. y The cotangent of θ, denoted cot(θ), is defined by cot(θ) = x, provided y 0. y In Section 5.1, we saw that we can define the x coordinate of that point on the unit circle as the cos(θ), and the y coordinate as sin(θ). Now, consider the tangent ratio. It then follows that tan(θ) = sin(θ) cos(θ). Similarly we can show that cot(θ) = cos(θ) sin(θ). These facts are true for any value of θ; therefore, we call them identities.

96 88 The Circular Functions Definition 6: The Quotient Identities tan(θ) = sin(θ), provided cos(θ) 0; if cos(θ) = 0, tan(θ) is undefined. cos(θ) cot(θ) = cos(θ), provided sin(θ) 0; if sin(θ) = 0, cot(θ) is undefined. sin(θ) We had also defined the cosecant and secant functions by their relationship to sine and cosine. We ll now call those definitions the reciprocal identities. Definition 7: The Reciprocal Identities sec(θ) = csc(θ) = 1, provided cos(θ) 0; if cos(θ) = 0, sec(θ) is undefined. cos(θ) 1, provided sin(θ) 0; if sin(θ) = 0, csc(θ) is undefined. sin(θ) Example 0 Find tan ( π 6 ) and sec ( π 6 ). Since π 6 is a special angle, we know (or should know soon after spending some time practicing) that ( π ) cos = 6 and ( π ) sin = 1 6 We also know from the quotient identities that ( π ) tan 6 = sin ( ) π 6 cos ( π 6 ).

97 5. Other Circular Functions 89 So substituting in our values for the sine and cosine, we have: ( π ) tan = 1/ 6 / = 1 = 1 = Now, since the reciprocal identities tell us that ( π ) sec = 6 we can substitute in our value for cosine to get 1 cos ( π 6 ( π ) sec = 1 6 ) = = Hence our final answer is tan ( ) π 6 = and sec ( ) π 6 =. Example 1 Find the exact value of csc ( 14π ) ( and cot 14π ) 14π is simply too big for us to deal with right now, so let s find a smaller coterminal angle: 14π π = 14π 6π = 11π

98 90 The Circular Functions Still too big! Let s go smaller... 11π 6π = 5π That s an angle we recognize from our common unit circle values! Since this is coterminal with our original angle, ( ) ( ) 14π 5π cos = cos = 1 ( ) ( ) 14π 5π sin = sin = Now we turn to the question at hand. Using reciprocal identities, we have ( ) 14π 1 csc = sin ( ) 14π = 1 = = Then, using quotient identities, we have ( ) 14π cot = cos ( ) 14π sin ( ) 14π = 1/ / = 1 = 1 = Therefore, our final answers are csc ( ) 14π = and cot ( ) 14π =. Check Point 1: F nd the exact values of sec ( π 4 ), csc ( π 4 ), tan ( π 4 ), and cot ( π 4 ).

99 5. Other Circular Functions 91 Check Point : F nd the exact value of csc ( π 6 ). Beyond the Unit Circle...Again Once again, we can extend our definitions of these trigonometric functions beyond the unit circle using similar triangles, like we did in Section 5.1. Definition 8: Circular Functions Theorem Suppose Q(x, y) is the point on the terminal side of an angle θ (plotted in standard position) which lies on the circle of radius r, x + y = r. Then: sin(θ) = y r = cos(θ) = x r = y x + y x x + y tan(θ) = y, provided x 0. x sec(θ) = r x x = + y, provided x 0. x csc(θ) = r x y = + y, provided y 0. y cot(θ) = x, provided y 0. y Example Suppose the terminal side of θ, when plotted in standard position, contains the point Q(, 4). Find the values of the six circular functions of θ.

100 9 The Circular Functions Since x = and y = 4, from x + y = r, () + (4) = r so r = 5, or r = 5. Our Circular Functions Theorem tells us sin(θ) = 4 5, cos(θ) = 5, tan(θ) = 4, sec(θ) = 5, csc(θ) = 5 4, and cot(θ) = 4. Example Suppose θ is a Quadrant IV angle with cot(θ) = 4. Find the values of the five remaining circular functions of θ. In order to use Circular Functions Theorem, we need to find a point Q(x, y) which lies on the terminal side of θ, when θ is plotted in standard position. We have that cot(θ) = 4 = x. Since θ is a Quadrant IV angle, we also know x > 0 and y < 0. y Rewriting 4 = 4 1, we choosea x = 4 and y = 1 so that r = x + y = (4) + ( 1) = 17. Applying Circular Functions Theorem, we find sin(θ) = 1 17 = 17 17, cos(θ) = 4 17 = , tan(θ) = 1 4, sec(θ) = 17 4, and csc(θ) = 17. a We could have just as easily chosen x = 8 and y = - just so long as x > 0, y < 0 and x y = 4. Example 4 Find cos (θ), where tan(θ) = and π < θ < π. We are told tan(θ) = and π < θ < π, so we know θ is a Quadrant III angle. To find cos(θ) using Circular Functions Theorem, we need to find the x-coordinate of a point Q(x, y) on the terminal side of θ, when θ is plotted in standard position, and the corresponding radius, r.

101 5. Other Circular Functions 9 Since tan(θ) = y and θ is a Quadrant III angle, we rewrite tan(θ) = = x 1 = y and choose x = 1 x and y =. From x + y = r, we get r = 10. Hence, cos(θ) = x r = 1 10 = Exercises In the following exercises, find the exact value of the cosine and sine of the given angle. ( π ) 1. tan 4 ( π ). sec ( 6 ) 5π. csc ( 6 ) 4π 4. cot ( 5. tan 11π ) ( 6 6. sec π ) ( 7. csc π ) ( ) 1π 8. cot 9. tan (117π) ( 10. sec 5π ) 11. csc (π) 1. cot ( 5π) ( ) 1π 1. tan ( π ) 14. sec 4 ( 15. csc 7π ) ( ) 4 7π 16. cot ( 6 ) π 17. tan 18. sec (( 7π) π ) 19. csc ( ) π 0. cot 4 In the following exercises, use the given the information to find the exact values of the circular functions of θ. 1. sin(θ) = with θ in Quadrant II 5. tan(θ) = 1 with θ in Quadrant III 5. csc(θ) = 5 with θ in Quadrant I 4 4. sec(θ) = 7 with θ in Quadrant IV 5. csc(θ) = with θ in Quadrant III cot(θ) = with θ in Quadrant II 7. tan(θ) = with θ in Quadrant IV. 8. sec(θ) = 4 with θ in Quadrant II. 9. cot(θ) = 5 with θ in Quadrant III. 0. cos(θ) = 1 with θ in Quadrant I.

102 94 The Circular Functions 1. cot(θ) = with 0 < θ < π.. csc(θ) = 5 with π < θ < π.. tan(θ) = 10 with π < θ < π. 4. sec(θ) = 5 with π < θ < π.

103 5. Graphs of Sine and Cosine Graphs of Sine and Cosine Learning Objectives In this section you will: Describe how transformations affect the graphs of functions. Create Parent Tables and Graphs for Sine and Cosine functions. Graph Sine and Cosine Functions using transformations. Describe how transformations affect the graphs of functions. Before we concern ourselves with the graphs of the circular functions, let s refresh our skills of graphical transformations. Our motivational example for the results in this section is the graph of y = f (x) below. While we could formulate an expression for f (x) as a piecewise-defined function consisting of linear and constant parts, we wish to focus more on the geometry here. That being said, we do record some of the function values - the key points if you will - to track through each transformation.

104 96 The Circular Functions x (x, f (x)) f (x) 0 (0, 1) 1 (, ) 4 (4, ) 5 (5, 5) 5 Vertical and Horizontal Shifts Suppose we wished to graph g(x) = f (x) +. From a procedural point of view, we start with an input x to the function f and we obtain the output f (x). The function g takes the output f (x) and adds to it. Using the sample values for f from the table above we can create a table of values for g below, hence generating points on the graph of g. x (x, f (x)) f (x) g(x) = f (x) + (x, g(x)) 0 (0, 1) = (0, ) (, ) + = 5 (, 5) 4 (4, ) + = 5 (4, 5) 5 (5, 5) = 7 (5, 7) In general, if (a, b) is on the graph of y = f (x), then f (a) = b. Hence, g(a) = f (a) + = b +, so the point (a, b + ) is on the graph of g. In other words, to obtain the graph of g, we add to the y-coordinate of each point on the graph of f. To get a feel for what s happening here, go to and scroll down to the interactive example pictured here.

105 5. Graphs of Sine and Cosine 97 You can adjust the slider labelled d in the following interactive demonstration. Consider the following: as d becomes larger, how does the graph change? What about as d becomes smaller (more negative)? Slide slider d to, and you ll see that geometrically, adding to the y-coordinate of a point moves the point units above its previous location. Adding to every y-coordinate on a graph en masse is moves or shifts the entire graph of f up units. Notice that the graph retains the same basic shape as before, it is just units above its original location. In other words, we connect the four key points we moved in the same manner in which they were connected before. You ll note that the domain of f and the domain of g are the same, namely [0, 5], but that the range of f is [1, 5] while the range of g is [, 7]. In general, shifting a function vertically like this will leave the domain unchanged, but could very well affect the range. You can easily imagine what would happen if we wanted to graph the function j(x) = f (x). Instead of adding to each of the y-coordinates on the graph of f, we d be subtracting. Geometrically, we would be moving the graph down units. We leave it to the reader to verify that the domain of j is the same as f, but the range of j is [ 1, ]. In general, we have:

106 98 The Circular Functions Definition 9: Vertical Shifts Suppose f is a function and d is a real number. To graph F (x) = f (x) + d, add d to each of the y-coordinates of the points on the graph of y = f (x). NOTE: This results in a vertical shift up d units if d > 0 or down d units if d < 0. Keeping with the graph of y = f (x) above, suppose we wanted to graph g(x) = f (x + ). In other words, we are looking to see what happens when we add to the input of the function. Let s try to generate a table of values of g based on those we know for f. We quickly find that we run into some difficulties. For instance, when we substitute x = 4 into the formula g(x) = f (x + ), we are asked to find f (4 + ) = f (6) which doesn t exist because the domain of f is only [0, 5]. The same thing happens when we attempt to find g(5). x (x, f (x)) f (x) g(x) = f (x + ) (x, g(x)) 0 (0, 1) 1 g(0) = f (0 + ) = f () = (0, ) (, ) g() = f ( + ) = f (4) = (, ) 4 (4, ) g(4) = f (4 + ) = f (6) =? 5 (5, 5) 5 g(5) = f (5 + ) = f (7) =? What we need here is a new strategy. We know, for instance, f (0) = 1. To determine the corresponding point on the graph of g, we need to figure out what value of x we must substitute into g(x) = f (x + ) so that the quantity x +, works out to be 0. Solving x + = 0 gives x =, and g( ) = f (( ) + ) = f (0) = 1 so (, 1) on the graph of g. To use the fact f () =, we set x + = to get x = 0. Substituting gives g(0) = f (0 + ) = f () =. Continuing in this fashion, we produce the table below. x x + g(x) = f (x + ) (x, g(x)) 0 g( ) = f ( + ) = f (0) = 1 (, 1) 0 g(0) = f (0 + ) = f () = (0, ) 4 g() = f ( + ) = f (4) = (, ) 5 g() = f ( + ) = f (5) = 5 (, 5) In summary, the points (0, 1), (, ), (4, ) and (5, 5) on the graph of y = f (x) give rise to the points (, 1), (0, ), (, ) and (, 5) on the graph of y = g(x), respectively. In general, if (a, b) is on the graph of y = f (x), then f (a) = b. Solving x + = a gives x = a so that g(a ) = f ((a ) + ) = f (a) = b. As

107 5. Graphs of Sine and Cosine 99 such, (a, b) is on the graph of y = g(x). The point (a, b) is exactly units to the left of the point (a, b) so the graph of y = g(x) is obtained by shifting the graph y = f (x) to the left units. Note that while the ranges of f and g are the same, the domain of g is [, ] whereas the domain of f is [0, 5]. In general, when we shift the graph horizontally, the range will remain the same, but the domain could change. If we set out to graph j(x) = f (x ), we would find ourselves adding to all of the x values of the points on the graph of y = f (x) to effect a shift to the right units. Generalizing these notions produces the following result. Definition 0: Horizontal Shifts Suppose f is a function and c is a real number. To graph F (x) = f (x + c), subtract c from each of the x-coordinates of the points on the graph of y = f (x). Note: This results in a horizontal shift right c units if c < 0 or left c units if c > 0. These observations about vertical and horizontal shifts present a theme which will run common throughout the section: changes to the outputs from a function result in some kind of vertical change; changes to the inputs to a function result in some kind of horizontal change. Reflections about the Coordinate Axes We now turn our attention to reflections. You may know from algebra courses that to reflect a point (x, y) across the x-axis, we replace y with y. If (x, y) is on the graph of f, then y = f (x), so replacing y with y is the same as replacing f (x) with f (x). Hence, the graph of y = f (x) is the graph of f reflected across the x-axis. Similarly, the graph of y = f ( x) is the graph of y = f (x) reflected across the y-axis.

108 100 The Circular Functions Definition 1: Reflections Suppose f is a function. To graph F (x) = f (x), multiply each of the y-coordinates of the points on the graph of y = f (x) by 1. NOTE: This results in a reflection across the x-axis. To graph F (x) = f ( x), multiply each of the x-coordinates of the points on the graph of y = f (x) by 1. NOTE: This results in a reflection across the y-axis. Using the language of inputs and outputs, we are saying that multiplying the outputs from a function by 1 reflects its graph across the horizontal axis, while multiplying the inputs to a function by 1 reflects the graph across the vertical axis. Applying these ideas to the graph of y = f (x) given at the beginning of the section, we can graph y = f (x) by reflecting the graph of f about the x-axis. x (x, f (x)) f (x) g(x) = f (x) (x, g(x)) 0 (0, 1) 1 1 (0, 1) (, ) (, ) 4 (4, ) (4, ) 5 (5, 5) 5 5 (5, 5)

109 5. Graphs of Sine and Cosine 101 By reflecting the graph of f across the y-axis, we obtain the graph of y = f ( x). x x g(x) = f ( x) (x, g(x)) 0 0 g(0) = f ( ( 0)) = f (0) = 1 (0, 1) g( ) = f ( ( )) = f () = (, ) 4 4 g( 4) = f ( ( 4)) = f (4) = ( 4, ) 5 5 g( 5) = f ( ( 5)) = f (5) = 5 ( 5, 5)

110 10 The Circular Functions Scalings We now turn our attention to our last class of transformations: scalings. A thorough discussion of scalings can get complicated because they are not as straight-forward as the previous transformations. A quick review of what we ve covered so far, namely vertical shifts, horizontal shifts and reflections, will show you why those transformations are known as rigid transformations. Simply put, rigid transformations preserve the distances between points on the graph - only their position and orientation in the plane change. If, however, we wanted to make a new graph twice as tall as a given graph, or one-third as wide, we would be affecting the distance between points. These sorts of transformations are hence called non-rigid. As always, we motivate the general theory with an example. Suppose we wish to graph the function g(x) = f (x) where f (x) is the function whose graph is given at the beginning of the section. From its graph, we can build a table of values for g as before. x (x, f (x)) f (x) g(x) = f (x) (x, g(x)) 0 (0, 1) 1 (0, ) (, ) 6 (, 6) 4 (4, ) 6 (4, 6) 5 (5, 5) 5 10 (5, 10) Graphing, we get:

111 5. Graphs of Sine and Cosine 10 In general, if (a, b) is on the graph of f, then f (a) = b so that g(a) = f (a) = b puts (a, b) on the graph of g. In other words, to obtain the graph of g, we multiply all of the y-coordinates of the points on the graph of f by. Multiplying all of the y-coordinates of all of the points on the graph of f by causes what is known as a vertical scaling by a factor of. If we wish to graph y = 1 f (x), we multiply the all of the y-coordinates of the points on the graph of f by 1. This creates a vertical scaling by a factor of 1 as seen below. We generalize these results below. Definition : Vertical Scalings. Suppose f is a function and a > 0 is a real number. To graph F (x) = af (x), multiply each of the y-coordinates of the points on the graph of y = f (x) by a. If a > 1, we say the graph of f has undergone a vertical stretch by a factor of a. If 0 < a < 1, we say the graph of f has undergone a vertical shrink by a factor of 1 a. Referring to the graph of f given at the beginning of this section, suppose we want to graph g(x) = f (x). In other words, we are looking to see what effect multiplying the inputs to f by has on its graph. If we attempt to build a table directly, we quickly run into the same problem we had in our discussion leading up to our horizontal shifts, as seen in the table on the left below.

112 104 The Circular Functions We solve this problem in the same way we solved this problem before. For example, if we want to determine the point on g which corresponds to the point (, ) on the graph of f, we set x = so that x = 1. Substituting x = 1 into g(x), we obtain g(1) = f ( 1) = f () =, so that (1, ) is on the graph of g. Continuing in this fashion, we obtain the table on the lower right. x (x, f (x)) f (x) g(x) = f (x) (x, g(x)) 0 (0, 1) 1 f ( 0) = f (0) = 1 (0, 1) (, ) f ( ) = f (4) = (, ) 4 (4, ) f ( 4) = f (8) =? 5 (5, 5) 5 f ( 5) = f (10) =? x x g(x) = f (x) (x, g(x)) 0 0 g(0) = f ( 0) = f (0) = 1 (0, 0) 1 g(1) = f ( 1) = f () = (1, ) 4 g() = f ( ) = f (4) = (, ) 5 5 g ( ) ( ) ( 5 = f 5 = f (5) = 5 5, 5) In general, if (a, b) is on the graph of f, then f (a) = b. Hence g ( a ) = f ( a ) = f (a) = b so that ( a, b) is on the graph of g. In other words, to graph g we divide the x-coordinates of the points on the graph of f by. This results in a horizontal scaling by a factor of 1.

113 5. Graphs of Sine and Cosine 105 If, on the other hand, we wish to graph y = f ( 1 x), we end up multiplying the x-coordinates of the points on the graph of f by which results in a horizontal scaling by a factor of, as demonstrated below. We generalize these results below. Definition : Horizontal Scalings. Suppose f is a function and b > 0 is a real number. To graph F (x) = f (bx), divide each of the x-coordinates of the points on the graph of y = f (x) by b. If 0 < b < 1, we say the graph of f has undergone a horizontal stretch by a factor of 1 b. If b > 1, we say the graph of f has undergone a horizontal shrink by a factor of b.

114 106 The Circular Functions Transformations in Sequence Now that we have studied three basic classes of transformations: shifts, reflections, and scalings, we present a result below which provides one algorithm to follow to transform the graph of y = f (x) into the graph of y = af (bx + c) + d. Definition 4: Transformations in Sequence Suppose f is a function. If a, b 0, then to graph g(x) = af (bx + c) + d start with the graph of y = f (x) and follow the steps below. Subtract c from each of the x-coordinates of the points on the graph of f. NOTE: This results in a horizontal shift to the left if h < 0 or right if h > 0. Divide the x-coordinates of the points on the graph obtained in Step 1 by b. NOTE: This results in a horizontal scaling, but includes a reflection about the y-axis if b < 0. Multiply the y-coordinates of the points on the graph obtained in Step by a. NOTE: This results in a vertical scaling, but includes a reflection about the x-axis if a < 0. Add k to each of the y-coordinates of the points on the graph obtained in Step. NOTE: This results in a vertical shift up if k > 0 or down if k < 0. A convenient way to remember this order is C-BAD. That s a good amount of review to be sure. We ll get plenty of practice using these techniques once we develop the parent graphs and tables for the sine and cosine functions.

115 5. Graphs of Sine and Cosine 107 Graphing the Sine and Cosine Functions In this chapter, we discussed how to interpret the sine and cosine of real numbers. To review, we identify a real number t with an oriented angle θ measuring t radians and define sin(t) = sin(θ) and cos(t) = cos(θ). Since every real number can be identified with one and only one angle θ this way, the domains of the functions f (t) = sin(t) and g(t) = cos(t) are all real numbers, (, ). When it comes to range, recall that the sine and cosine of angles are coordinates of points on the Unit Circle and hence, each fall between 1 and 1 inclusive. Since the real number line, when wrapped around the Unit Circle completely covers the circle, we can be assured that every point on the Unit Circle corresponds to at least one real number. Putting these two facts together, we conclude the range of f (t) = sin(t) and g(t) = cos(t) are both [ 1, 1]. We summarize these two important facts below. Definition 5: Domain and Range of the Cosine and Sine Functions: The functions f (t) = sin(t) and f (t) = cos(t) have domain (, ) and range [ 1, 1]. Our aim in this section is to become familiar with the graphs of f (t) = sin(t) and g(t) = cos(t). To that end, we begin by making a table and plotting points. We ll start by graphing f (t) = sin(t) by making a table of values and plotting the corresponding points. We ll keep the independent variable t for now and use the default y as our dependent variable. Keep in mind that we re using y here to denote the output from the sine function. It is a coincidence that the y-values on the graph of y = sin(t) correspond to the y-values on the Unit Circle. Note in the graph below, the scale of the horizontal and vertical axis is far from 1:1. (We will present a more accurately scaled graph shortly.)

116 108 The Circular Functions t sin(t) (t, sin(t)) 0 0 ( (0, 0) ) π 4, π 4 π π 4 1 ( π (, 1) ) π 4, π 0 (π, 0) ( ) 5π 5π 4 4, ( π 1 π, 1) ( ) 7π 7π 4 4, π 0 (π, 0) If we plot additional points, we soon find that the graph repeats itself. In fact, we expect the function to repeat itself every π units. Below is a more accurately scaled graph highlighting the portion we had already graphed above. The graph is often described as having a wavelike nature and is sometimes called a sine wave or, more technically, a sinusoid.

117 5. Graphs of Sine and Cosine 109 Note that by copying the highlighted portion of the graph and pasting it end-to-end, we obtain the entire graph of f (t) = sin(t). We give this repeating property a name. Definition 6: Periodic Functions A function f is said to be periodic if there is a real number c so that f (t + c) = f (t) for all real numbers t in the domain of f. The smallest positive number p for which f (t + p) = f (t) for all real numbers t in the domain of f, if it exists, is called the period of f. We know that sin(t + π) = sin(t) for all real numbers t but the question remains if any smaller real number will do the trick. Suppose p > 0 and sin(t + p) = sin(t) for all real numbers t. Then, in particular, sin(0+p) = sin(0) = 0 so that sin(p) = 0. From this we know p is a multiple of π. Since sin ( ) ( π sin π + π), we know p π. Hence, p = π so the period of f (t) = sin(t) is π. Having period π essentially means that we can completely understand everything about the function f (t) = sin(t) by studying one interval of length π, say [0, π]. For this reason, when graphing sine (and cosine) functions, we typically restrict our attention to graphing these functions over the course of one period to produce one cycle of the graph. Not surprisingly, the graph of g(t) = cos(t) exhibits similar behavior as f (t) = sin(t) as seen below. Here note that the dependent variable y represents the outputs from g(t) = cos(t) which are x-coordinates on the Unit Circle.

118 110 The Circular Functions t cos(t) (t, cos(t)) 0 1 ( (0, 1) ) π 4, π 4 π 0 π 4 ( π (, 0) ) π 4, π 1 (π, 1) ( ) 5π 5π 4 4, ( π 0 π, 0) ( ) 7π 7π 4 4, π 1 (π, 1) Like f (t) = sin(t), g(t) = cos(t) is a wavelike curve with period π. Moreover, the graphs of the sine and cosine functions have the same shape - differing only in what appears to be a horizontal shift. As we ll prove in a later section, sin ( t + π ) = cos(t), which means we can obtain the graph of y = cos(t) by shifting the graph of y = sin(t) to the left π units. Hence, we can obtain the graph of y = sin(t) by shifting the graph of y = cos(t) to the right π units: cos ( t π ) = sin(t).

119 5. Graphs of Sine and Cosine 111 While arguably the most important property shared by f (t) = sin(t) and g(t) = cos(t) is their periodic wavelike nature (this is the reason they are so useful in the Sciences and Engineering) their graphs suggest these functions are both continuous and smooth. Like polynomial functions, the graphs of the sine and cosine functions have no jumps, gaps, holes in the graph, vertical asymptotes, corners or cusps. Moreover, the graphs of both f (t) = sin(t) and g(t) = cos(t) meander and never settle down as t ± to any one real number. So even though these functions are trapped (or bounded) between 1 and 1, neither graph has any horizontal asymototes. Lastly, the graphs of f (t) = sin(t) and g(t) = cos(t) suggest each enjoy one of the symmetries of functions. The graph of y = sin(t) appears to be symmetric about the origin while the graph of y = cos(t) appears to be symmetric about the y-axis. Indeed, as we ll prove in a later section, f (t) = sin(t) is, in fact, an odd function: that is, sin( t) = sin(t) and g(t) = cos(t) is an even function, so cos( t) = cos(t). We summarize all of these properties in the following result.

120 11 The Circular Functions Definition 7: Properties of the Cosine and Sine Functions The function f (t) = sin(t)... has domain (, ) has range [ 1, 1] is continuous and smooth is odd (so sin( t) = sin(t)) has period π hr The function f (t) = cos(t)... has domain (, ) has range [ 1, 1] is continuous and smooth is even (so cos( t) = cos(t)) has period π

121 5. Graphs of Sine and Cosine 11 Graph Sine and Cosine Functions using transformations Now that we know the basic shapes of the graphs of y = sin(t) and y = cos(t), we can use our knowledge of graphical transformations to graph more complicated functions using transformations. As mentioned already, the fact that both of these functions are periodic means we only have to know what happens over the course of one period of the function in order to determine what happens to all points on the graph. To that end, we graph the fundamental cycle - the portion of each graph generated over the interval [0, π] - for each of the sine and cosine functions below. In working through transformations, is very helpful to track key points through the transformations. The key points we ve indicated on the graphs below correspond to the quadrantal angles and generate the zeros and the extrema of functions. Since the quadrantal angles divide the interval [0, π] into four equal pieces, we shall refer to these angles henceforth as the quarter marks. We include a table of the quarter marks below the graphs. t sin(t) 0 0 π 1 π 0 π 1 π 0 t cos(t) 0 1 π 0 π 1 π 0 π 1

122 114 The Circular Functions Example 5 Graph one cycle of the function. State the period. f (t) = sin(t) One way to proceed is to use our CBAD procedure. Here, we have a = and b =. Starting with the fundamental cycle of y = sin(t), we divide each t-coordinate by and multiply each y-coordinate by to obtain one cycle of y = sin(t). Fundamental Cycle t y = sin(t) 0 0 π 1 π 0 π 1 π 0 b =, so divide the t-coordinates by (or, equivalently, multiply them by 1 ): 1 t y = sin(t) 1 0 = π = π = π 0 1 = π = π 0 Now, keeping those t-coordinates, we apply a = by multiplying the y-coordinates by : 1 t y 0 0 = 0 π 4 1 = π 0 = 0 π 4 1 = π 0 = 0

123 5. Graphs of Sine and Cosine 115 To be sure that we have our numbering scaled correctly on the t-axis, it can be useful to rewrite all those t-coordinates with a common denominator (in this case, 4). When we do that, we have our final table. When we plot the points in the table, we create one cycle of the transformed sine function, graphed below: t y 0 0 π 4 π 4 0 π 4 4π 4 0 When we compare our transformed function with the fundamental cycle of sin(t), we see that the function has undergone a horizontal compression and a vertical stretch. Since one cycle of y = f (t) is completed over the interval [0, π], the period of f is π. Example 6 Graph one cycle of the function. State the period. f (t) = cos ( t + π ) Again, from a transformations perspective, we have a =, c = π, and d = 1. Going in CBAD order,

124 116 The Circular Functions we will first subtract π from our t-coordinates, then multiply the y s by, and finally add 1 to all of the y s. We will condense this work into a single table below, starting with the fudamental cycle in the middle, and working outwards to apply the transformations: apply c parent t parent cos(t) apply a apply d 0 π = π = + 1 = π π = 0 π 0 0 = = 1 π π = π π 1 1 = + 1 = 1 π π = π π 0 0 = = 1 π π = π π 1 1 = + 1 = Now, our final transformed t-values are in the first column, and the final transformed y-values are in the last column. If you like, you can condense these into a final, transformed table, then use that to graph the coordinates of the transformed cosine function: t y π 0 1 π 1 π 1 π We can see that the transformed function has been shifted left, vertically stretched, and shifted up.

125 5. Graphs of Sine and Cosine 117 We find one cycle of y = g(t) is completed over the interval [ π, π ], the period is π ( π ) = π. Check Point Sketch a graph of the function f (x) = sin ( 1 x). Check Point 4 Sketch a graph of the function f (x) = cos ( x + 5π 4 ). Features of Sinusoidal Functions As previously mentioned, the curves graphed in the examples above are examples of sinusoids. A sinusoid is the result of taking the graph of y = sin(t) or y = cos(t) and performing any of the transformations mentioned above. We graph one cycle of a generic sinusoid below. Sinusoids can be characterized by four properties: period, phase shift, vertical shift (or baseline ), and amplitude. We have already discussed the period of a sinusoid. If we think of t as measuring time, the period is how

126 118 The Circular Functions long it takes for the sinusoid to complete one cycle and is usually represented by the letter T. The standard period of both sin(t) and cos(t) is π, but horizontal scalings will change this. In the example above, for instance, the function f (t) = sin(t) has period π instead of π because the graph is horizontally compressed by a factor of as compared to the graph of y = sin(t). However, the period of g(t) = cos ( t + π ) +1 is the same as the period of cos(t), π, since there are no horizontal scalings. The phase shift of the sinusoid is the overall horizontal shift. Again, thinking of t as time, the phase shift of a sinusoid can be thought of as when the sinusoid starts as compared to t = 0. Assuming there are no reflections across the y-axis, we can determine the phase shift of a sinusoid by finding where the value t = 0 on the graph of y = sin(t) or y = cos(t) is mapped to under the transformations. Consider, then the transformed function g(t) = a f (bt + c) + d. If we want to know where t = 0 from the parent ends up after these transformations, we consider where bt + c = 0. Subtract c and divide by b, and we have t = b c, which is the phase shift in general for a sinusoidal function. For f (t) = sin(t), the phase shift is 0 since the value t = 0 on the graph of y = sin(t) remains stationary under the transformations. Loosely speaking, this means both y = sin(t) and y = sin(t) start at the same time. The phase shift of g(t) = cos ( t + π ) + 1 is π or π to the left since the value t = 0 on the graph of y = cos(t) is mapped to t = π on the graph of y = cos ( t + π ) + 1. Again, loosely speaking, this means y = cos ( t + π ) + 1 starts π time units earlier than y = cos(t). The vertical shift of a sinusoid is exactly the same as the vertical shifts in generic transformations, and determines the new baseline of the sinusoid. Thanks to symmetry, the vertical shift can always be found by averaging the maximum and minimum values of the sinusoid. For f (t) = sin(t), the vertical shift is 0 whereas the vertical shift of g(t) = cos ( t + π ) + 1 is 1 or 1 up. The amplitude of the sinusoid is a measure of how tall the wave is, as indicated in the figure below. Said differently, the amplitude measures how much the curve gets displaced from its baseline. The amplitude of the standard cosine and sine functions is 1, but vertical scalings can alter this. In the example above, the amplitude of f (t) = sin(t) is, owing to the vertical stretch by a factor of as

127 5. Graphs of Sine and Cosine 119 compared with the graph of y = sin(t). In the case of g(t) = cos ( t + π ) + 1, the amplitude is due to its vertical stretch as compared with the graph of y = cos(t). Note that the +1 here does not affect the amplitude of the curve; it merely changes the baseline from y = 0 to y = 1. The following theorem shows how these four fundamental quantities relate to the parameters which describe a generic sinusoid. Definition 8: Features of sinusoidal functions The graphs of F (t) = A sin(bt + C) + D and G(T ) = A cos(bt + C) + D have the following features: period T = π B amplitude A phase shift C B vertical shift or baseline D Example 7: Find the period, phase shift, amplitude, and vertical shift of the following function: f (t) = cos ( ) πt π + 1 First, we need to rewrite f (t) in the form prescribed by our CBAD methodology. To that end, we rewrite: f (t) = cos ( ) ( πt π + 1 = cos π t + ( π )) + 1.

128 10 The Circular Functions From this, we identify A =, B = π, C = π π and D = 1. Therefore, the period is T = B = π π/ = 4, the phase shift is C B = π/ = 1 (indicating a shift to the right 1 unit), the amplitude is π/ A = =, and the vertical shift is D = 1 (indicating a shift up 1 unit. If we graph this function using our transformation strategies, we will see all of these features. The cycle starts at t = 1 and ends at t = 5, giving a period of 5 1 = 4. The start of the cycle has been shifted from t = 0 to t = 1, hence we have a phase shift of 1. The baseline has been shifted up from y = 0 to y = 1, showcasing the vertical shift. Finally, the distance from that baseline to the top of the cycle (from 1 to 4) is, the amplitude. Example 8: Find the period, phase shift, amplitude, and vertical shift of the following function: f (t) = 1 sin (π t) + We first note that the t comes second inside the function, which will confuse things from our CBAD perspective. We can reorder that inside so that we have f (t) = 1 sin( t + π) +, and now identify A = 1, B =, C = π, and D =. Now we find that the period is T = π B = π = π, the phase shift is C B = π = π (indicating a shift to the right π unit), the amplitude is A = 1 = 1, and the vertical shift is D = (indicating a

129 5. Graphs of Sine and Cosine 11 shift up unit. If we graph this function using our transformation strategies, we will see all of these features. The cycle starts at t = π and ends at t = π, giving a period of π = π. The start of the cycle has been shifted from t = 0 to t = π, hence we have a phase shift of π. The baseline has been shifted up from y = 1 to y = /, showcasing the vertical shift. Finally, the distance from that baseline to the top of the cycle (from / to ) is 1/, the amplitude. π In the next example, we use the properties of sinusoidal functions to determine the formula of a sinusoid given the graph of one cycle. Note that in some disciplines, sinusoids are written in terms of sines whereas in others, cosines functions are preferred. To cover all bases, we ask for both. Example 9 Below is the graph of one complete cycle of a sinusoid y = f (t).

130 1 The Circular Functions a. Write f (t) in the form G(T ) = A cos(bt + C) + D. A useful strategy is to think about this in the reverse of CBAD: first let s look for D, a vertical shift. We notice the middle points are at a height of 1, implying a vertical shift, so D = 1. Next, we look for the amplitude. From the baseline to the top of the curve, there is a distance of 5 1 =, indicating the amplitude is, and A =. For simplicity, let s make A =. Now we consider our horizontal transformations. We know that the cycle starts at t = 1 and ends at t = 5, so the period is 5 ( 1) = 6. Since the period is also π B, we can solve for B: π B = 6 π = 6 B B = π Again, for simplicity, let s say B = π. Finally, observe that the fundamental cosine function starts at the top of the cycle, then travels down; here, that top of the cycle starts at t = 1, implying that the phase shift is 1, so C B = 1. We can substitute in π for B and solve for C: C π/ = 1 C = π

131 5. Graphs of Sine and Cosine 1 C = π Therefore, we conclude that the function is G(T ) = cos( π t + π ) + 1. Example 40 Write f (t) from the previous example in the form F(T ) = A sin(bt + C) + D. Most of our thought process here will likely be the same. We know D = 1, and A =. This time, though, we ll need to consider how to treat those absolute value bars. Recall that the fundamental cycle of sine starts at 0, then goes up to the top of the cycle. If we look at our first point on the baseline of this graph, ( 1, ) 1, the curve proceeds down to the bottom of the cycle instead. This indicates we have a vertical reflection, so we should let A =. We can still use B = π (based off of the same period of 6), but now we ll tweak our treatment of the phase shift and C. If we consider ( 1, ) 1 to be the start of our cycle, then the phase shift is 1. Hence 1 = C B 1 = C π/ C = π 6 C = π 6 This time, then, we find that the function is F (T ) = sin( π t π 6 ) + 1. ( Now, we could instead choose to treat 7, ) 1 as the start of the cycle. In this case, the function goes up as it should, so we can keep our A-value positive. This time, the phase shift becomes 7, and when we solve for C, we get 7π 6. Therefore, another valid answer is F (T ) = sin( π t 7π 6 ) + 1!

132 14 The Circular Functions 5..1 Exercises In the following exercises, graph one cycle of the given function. State the period, amplitude, phase shift and vertical shift of the function. 1. f (t) = sin(t). g(t) = sin(t). h(t) = cos(t) 4. f (t) = cos ( t π ) 5. g(t) = sin ( t + π ) 6. h(t) = sin(t π) 7. f (t) = 1 cos ( 1 t + π ) 8. g(t) = cos(t π) h(t) = sin ( t π 4 ) 10. f (t) = cos ( π 4t) g(t) = cos ( t + π ) 1 1. h(t) = 4 sin( πt + π) In the following exercises, a sinusoid is graphed. Find a formula for the sinusoid in the form S(t) = A sin(bt + C) + D and C(t) = A cos(bt + C) + D. Check your answer by graphing.

133 5. Graphs of Sine and Cosine 15

134 16 The Circular Functions In the following exercises, use a graphing utility to graph each function and discuss the related questions with your classmates. 17. f (t) = cos(t) + sin(t). Is this function periodic? If so, what is the period? 18. f (t) = sin(t). What appears to be the horizontal asymptote of the graph? t 19. f (t) = t sin(t). Graph y = ±t. What do you notice? 0. f (t) = sin ( ) 1 t. What s happening as t 0? 1. f (x) = x tan(x). Graph y = x on the same set of axes and describe the behavior of f.. f (t) = e 0.1t (cos(t) + sin(t)). Graph y = ±e 0.1t on the same set of axes. What do you notice?. f (t) = e 0.1t (cos(t) + sin(t)). Graph y = ±e 0.1t on the same set of axes. What do you notice?

135 5.4 Graphs of Other Circular Functions Graphs of Other Circular Functions Learning Objectives In this section you will: Create Parent Tables and Graphs for Tangent, Cotangent, Secant and Cosecant functions. Graph Trigonometric Functions Using Transformations Describe the Features of Trigonometric Functions We know the tangent function can be used to find distances, such as the height of a building, mountain, or flagpole. But what if we want to measure repeated occurrences of distance? Imagine, for example, a police car parked next to a warehouse. The rotating light from the police car would travel across the wall of the warehouse in regular intervals. If the input is time, the output would be the distance the beam of light travels. The beam of light would repeat the distance at regular intervals. The tangent function can be used to approximate this distance. Asymptotes would be needed to illustrate the repeated cycles when the beam runs parallel to the wall because, seemingly, the beam of light could appear to extend forever. The graph of the tangent function would clearly illustrate the repeated intervals. In this section, we will explore the graphs of the tangent and other trigonometric functions. /section

136 18 The Circular Functions Parent graphs and tables for Tangent and Cotangent In Section 5. we learned that tan(t) = sin(t). We ll use this fact to help to develop our parent graph (or cos(t) fundamental cycle) for tan(t). For reasons that we ll discuss momentarily, we re going to look at common unit circle values from π to π to develop our table. t π π π 4 π 6 tan ( π tan ( π 6 tan(t) = sin(t) (t, tan(t)) cos(t) tan ( ) π sin( π/) = cos( π/) = sin(π/) = 1 = DNE DNE cos(π/) 0 ) sin( π/) = cos( π/) = sin(π/) = = ( cos(π/) 1/ π, ) ( π, 1.7) tan ( ) π sin( π/4) 4 = cos( π/4) = sin(π/4) = / ( = 1 cos(pi/4) / π 4, 1) ) ( ) sin( π/6) = cos( π/6) = sin(π/6) = 1/ = cos(pi/6) / π 6, 1 ( π 6, ) 0 tan(0) = sin(0) π tan ( ) π sin(π/6) 6 6 = cos(π/6) = sin(π/6) π tan ( ) π sin(π/4) 4 4 = π tan ( ) π sin(π/) = π tan ( π cos(0) = 0 1 cos(pi/6) = 1/ = 1 / cos(π/4) = sin(π/4) cos(pi/4) = / cos(π/) = sin(π/) cos(π/) = = 0 (0, 0) ( ) π 6, 1 ( π ( = 1 π / 4, 1) 1/ = ) = sin(π/) cos(π/) = sin(π/) cos(π/) = 1 0 = DNE 6, ) ( π, ) ( π, 1.7) DNE This table contains some curiosities for us. We see that tan(t) does not always exist, so what does that mean for its graph? Investigating the behavior near the values excluded from the domain, we find as t π, sin(t) 1 and cos(t) 0 +. Hence, tan(t) = sin(t) producing a vertical asymptote to the cos(t) graph at t = π. Similarly, we get that as t π +, tan(t). Therefore, rather than simply marking DNE in our table, we can specify that there are vertical asymptotes at these locations. Putting all of this information together, and streamlining the table a bit, we can create the fundamental cycle of tan(t).

137 5.4 Graphs of Other Circular Functions 19 Definition 9: The Fundamental Cycle of tan(t) t π π 4 tan(t) Vertical Asymptote (VA) π 4 1 π VA After the usual copy and paste procedure, we create the graph of y = tan(t) below:

138 10 The Circular Functions The graph of y = tan(t) suggests symmetry through the origin. Indeed, tangent is odd since sine is odd and cosine is even: tan( t) = sin( t) cos( t) = sin(t) = tan(t). cos(t) Moreover, the period of the tangent function is π, and we see that reflected in the graph. This means we can choose any interval of length π to serve as our fundamental cycle. Looking at the full graph, we can see that if we chose (0, π) for our interval, we would be splitting up the shape of the tangent curve in an odd way, with an asymptote in the middle. To create a nice continuous chunk of the tangent shape, then, we choose our fundamental cycle interval to be ( π, π ). Viewing tan(t) = sin(t), we find the domain of J excludes all values where cos(t) = 0. Hence, the domain cos(t) of tan(t) is {t t π + πk, for integers k}. We also see the graph suggests the range of tan(t) is all real numbers, (, ). We will summarize all of this in a moment, but first let s look at cot(t), which is very similar to tan(t). Like tangent, the graph of cotangent will have vertical asymptotes, but this time, they will occur where sin(t) = 0 since cot(t) = cos(t). Therefore, to create our fundamental cycle, we can again start at 0 (which sin(t) will be an asymptote) and proceed to the next asymptote, in this case π (since sin(π) = 0). t cot(t) (t, cot(t)) 0 cot(0) = cos(0) = 1 sin(0) 0 = DNE Vertical asymptote (VA) π cot ( ) π cos(π/4) / ( 4 4 = = = 1 π sin(π/4) / 4, 1) π cot ( ) π cos(π/) = = 0 sin(π/) 1 = 0 ( π, 0) π cot ( ) π cos(π/4) 4 4 = = / ( = 1 π sin(π/4) / 4, 1) π cot(π) = cos(π) sin(π) = 1 0 = DNE Vertical asymptote (VA) We summarize below.

139 5.4 Graphs of Other Circular Functions 11 Definition 40: The Fundamental Cycle of cot(t) t cot(t) 0 Vertical Asymptote (VA) π 1 4 π 0 π 1 4 π VA As usual, pasting copies end to end produces the graph of cot(t) below.

140 1 The Circular Functions As with tan(t), the graph of cot(t) suggests cot(t) is odd, a fact we leave to the reader to prove in the exercises. Also, we see that the period of cotangent (like tangent) is π and the range is (, ). We take as one fundamental cycle the graph as traced out over the interval (0, π), highlighted above, with quarter marks: t = 0, t = π 4, t = π, t = π and t = π. 4 The features of the tan(t) and cot(t) functions The properties of the tangent and cotangent functions are summarized below. Each of the results below can be traced back to properties of the cosine and sine functions and the definition of the tangent and cotangent functions as quotients thereof.

141 5.4 Graphs of Other Circular Functions 1 Definition 41: The Properties of tan(t) and cot(t) The function tan(t) has domain { t t π + πk, k is an integer} as range (, ) is continuous and smooth on its domain is odd has period π The function cot(t) has domain {t t πk, k is an integer} has range (, ) is continuous and smooth on its domain is odd has period π Note that the tangent and cotangent functions have different periods than sine and cosine. Moreover, in the case of the tangent function, the fundamental cycle we ve chosen starts at π instead of 0. Nevertheless, we can use the same notions graphical transformations to graph transformed tangent and cotangent functions, as long as we have the fundamental cycle and our CBAD strategy. Graphing Tangent and Cotangent Functions using Transformations Techniques Now, once again, we ll practice applying our CBAD transformation technique on these new functions, and see how those graphs relate to the properties described above. Example 41 Graph one cycle of the following function. State the period. f (t) = 1 tan( t π)

142 14 The Circular Functions Let s do a quick rewrite so that we recognize our A, B, C, and D: f (t) = tan( 1 t π) + 1 Therefore, we know that A = 1, B = 1, C = π, and D = 1. Once again, we can create a single table with the fundamental cycle in the middle, and work our way to the outsides in CBAD order. t+pi) 1 apply B apply C parent parent apply A apply D = (t + π)) t ( π) = t + π t y = tan(t) 1 y 1 y + 1 π = π π + π = π π VA VA VA π 4 = π π 4 + π = π 4 π = = π = π 0 + π = π = = 1 5π 4 = 5π π 4 + π = 5π 4 π = π π + π = π π = = 0 π VA VA VA Notice that the vertical transformations (A and D) have no effect on the vertical asymptote. If we try to slide up or verticaly stretch a vertical line, nothing really happens! Now, we can take our transformed table and sketch the graph: The period is π π = π, which is confirmed by the formula T = π B = π 1/ = π.

143 5.4 Graphs of Other Circular Functions 15 Example 4 Graph one cycle of the following function. State the period. f (t) = cot(π πt) 1 Once again, do a quick rewrite so that we recognize our A, B, C, and D: f (t) = cot( πt + π) 1 Therefore, we know that A =, B = π, C = π, and D = -1. We can create a single table with the fundamental cycle in the middle, and work our way to the outsides in CBAD order. apply B apply C parent parent apply A apply D (t+π) π t π t y = cot(t) y y 1 π = 0 π = π 0 VA VA VA π 7π/4 π = 7 4 π/ π = 5π/4 π = 5 4 π π π 4 π = 7π 4 π π = 1 = 1 π π = π 0 0 = = 1 π 4 = 5π π = 1 = = 1 π π = π π VA VA VA Because of our negative B value, our t coordinates have been flipped around. To sort things out, let s write the final table with the t s ordered from least to greatest: t y 1 = 4 4 VA 5 4 = = 8 4 VA When we graph only this cycle, we have:

144 16 The Circular Functions Note that this pattern repeats! The period seen on the graph is 1 = 1. If we use the formula, we still find the period is π π = 1. Creating Parent Tables and Graphs for Cosecant and Secant We ll continue to use sine and cosine functions to analyze secant and cosecant. For instance, Rewriting F(t) = sec(t) =, we first note that F(t) is undefined whenever cos(t) = 0. That happens when 1 cos(t) t = π, π, and if we continue to sweep around full circles and find coterminal angles, we ll that when t = 5π, 7π... cos(t) also equals zero. These are odd multiples of π, so cos(t) = 0 whenever t = π + πk for integers k. This is a hint that the graph of secant will also have asymptotes all across its domain. To investigate what s happening around those asymptotes, let s once more create a table of values, this time for secant. We ll use the fact that secant is the reciprocal of cosine to help us along the way.

145 5.4 Graphs of Other Circular Functions 17 t cos(t) sec(t) (t, sec(t)) (0, 1) ( π 4, ) π 4 π π 4 0 undefined VA ( π 4, ) π 1 1 (π, 1) 5π 4 π ( 5π 4, ) 0 undefined VA ( 7π 4, ) 7π 4 π 1 1 (π, 1) We can more closely analyze the behavior of sec(t) near the values excluded from its domain. We find as t π, cos(t) 0 +, so sec(t). Similarly, we get as t π +, sec(t) ; as t π, sec(t) ; and as t π +, sec(t). This helps us to properly graph the behavior of the graph at the asymptotes, as seen below: To get a graph of the entire secant function, we paste copies of the fundamental cycle end to end to produce the graph below. The graph suggests that F(t) = sec(t) is even. Indeed, since cos(t) is even, that 1 1 is, cos( t) = cos(t), we have sec( t) = = = sec(t). Hence, along with its period, the secant cos( t) cos(t)

146 18 The Circular Functions function inherits its symmetry from the cosine function. As one would expect, to graph G(t) = csc(t) we begin with y = sin(t) and take reciprocals of the corresponding y-values. Here, we encounter issues at t = 0, t = π, t = π, and, in general, at all whole number multiples of π, so the domain of G is {t t πk, for integers k}. Not surprisingly, these values produce vertical asymptotes. Proceeding as above, we graph produce the graph of the fundamental cycle of y = csc(t) below along with the dotted graph of y = sin(t) for reference. x sin(x) csc(x) (x, csc(x)) 0 0 undefined ( π 4, ) π 4 π ( 1 1 π (, 1) π 4, ) π 4 π 0 undefined 5π ( 5π 4 4, ) ( π 1 1 π, 1) 7π ( 7π 4 4, ) π 0 undefined

147 5.4 Graphs of Other Circular Functions 19 Pasting copies of the fundamental period of y = csc(t) end to end produces the graph below. Since the graphs of y = sin(t) and y = cos(t) are merely phase shifts of each other, it is not too surprising to find the graphs of y = csc(t) and y = sec(t) are as well. As with the graph of secant, the graph below suggests symmetry. Indeed, since the sine function is odd, 1 that is sin( t) = sin(t), so too is the cosecant function: csc( t) = sin( t) = 1 = csc(t). Hence, the sin(t) graph of csc(t) is symmetric about the origin. Note that, on the intervals between the vertical asymptotes, both F(t) = sec(t) and G(t) = csc(t) are continuous and smooth. In other words, they are continuous and smooth on their domains.

148 140 The Circular Functions Definition 4: Properties of the Secant and Cosecant Functions The function F(t) = sec(t)... has domain { t t π + πk, k is an integer} has range (, 1] [1, ) is continuous and smooth on its domain is even has period π The function G(t) = csc(t)... has domain {t t πk, k is an integer} has range (, 1] [1, ) is continuous and smooth on its domain is odd has period π Graphing Secant and Cosecant Functions Using Transformations Techniques In the next example, we discuss graphing more general secant and cosecant curves. We make heavy use of the fact they are reciprocals of sine and cosine functions and apply what we learned about the CBAD technique for transformations with one very important caveat: we have to stall applying D until the very last moment!! Example 4 Graph one cycle of the function. State the period. f (t) = 1 sec(t) One way of graphing this is to think about the related cosine curve. Let s rewrite it so that we can identify our A, B, C, and D values: g(t) = cos(t) + 1 Now we can see that A =, B = and D = 1. For the time being we re going to ignore D and only graph the rest of the structure. More about that in a minute. Below, we create the tranformation

149 5.4 Graphs of Other Circular Functions 141 table for g 1 (t) = cos(t). apply B parent parent apply A t = t 1 t cos(t) y 0 1 = = π 1 = π 4 π 1 π 1 = π 4 π 1 π 0 0 = 0 = π π 1 1 = π 0 0 = 0 = π π 1 1 = Now, before we turn our attention to D, we need to consider how the secant function works. We 1 know that sec(t) =, so we need to be concerned about where cos(t) = 0. According to our cos(t) table currently, we know that an input of π into our transformed cosine function will produce an 4 output of 0. So, if we try to input π into our function f (t) = 1 sec(t), we will have a division by 4 zero problem: ( π ) ( f = 1 sec π ) 4 4 ( π ) = 1 sec 1 = 1 cos(π/) = = DNE This means that on our graph, we will have vertical asymptotes at t = π 4 and t = π. In other words, 4 we must graph our cosine function before applying the vertical shift transformation, D in order to identify the asymptotes of the corresponding secant function.

150 14 The Circular Functions Example continued Now, let s take our cosine table, and finally, carefully apply that +D transformation. Note that we are going to disregard any place where we know that there will be an asymptote in the final secant graph. t cos(t) apply D y = 1 π 4 0 (VA on secant graph) π + 1 = π 4 0 (VA on secant graph) π + 1 = 1 Finally, we think back to the structure of the secant graph. We know that there are parabola-like curves extending off the tops of the peaks of the underlying cosine graph, and off of the bottom of the valleys of the underlying cosine graph. We say parabola-like, because these curves are reminiscent of parabolas, but they are constrained by the asymptotes of the secant function. Here, then, we ll want to draw such structures extending up from ( π, ), and extending down from (0, 1) and (π, 1), and all contained by the asymptotes we detected at t = π 4 and π. Therefore, our final secant function looks like this: 4 From the graph, we can see that the cyclical pattern repeats from 0 to π, so the period is π. We can find this same period using the formula T = π B = π = π. Example 44 Graph one cycle of the function. State the period. g(t) = csc( πt π) 5

151 5.4 Graphs of Other Circular Functions 14 Admittedly, this function is a mess. It could definitely do with a rewrite so that we can identify A, B, C, and D: g(t) = 1 csc( πt π) 5 Now we can tell that A = 1, B = π, C = π, and D = 5. We ll want to apply these transformations to the underlying sine graph, but we must remember to be careful when applying D! After you apply A, note that any point that at that moment has a y-coordinate of 0 will be a vertical asymptote on the corresponding cosecant graph, regardless of the application of D. Apply B Apply C parent parent Apply A Apply D t+π 1 π t + π t sin(t) y 1 y 5 π π = π = π = 0 VA on cosecant graph π/ π = π π + π = π π 1 π = π + π = π π 0 1 5π/ π = 5 π + π = 5π π 1 1 π π = π + π = π π = = 4 0 = 0 VA on cosecant graph 1 = = 6 = 0 = 0 VA on cosecant graph Using this sine information as a guide, we graph the cosecant graph as shown below. The period is π B = π π =, which we see on the graph if we look from the start of our cycle at t = 5/ to the end of our cycle at t = 1/: 1 ( 5 ) = 4 =. All that s left is to practice graphing these functions!

152 144 The Circular Functions Exercises In the following ( exercises, graph one cycle of the given function. State the period of the function. 1. y = tan t π ) ( ) 1. y = tan 4 t. y = 1 tan( t π) + 1 ( 4. y = sec t π ) ( 5. y = csc t + π ) ) 6. y = 1 ( 1 sec t + π 7. y = csc(t π) 8. y = sec(t ( π) y = csc t π ) ( y = cot t + π ) ( 6 ) y = 11 cot 5 t 1. y = 1 ( cot t + π ) + 1

153 5.5 Inverse Trig Functions Inverse Trig Functions Learning Objectives In this section you will: Review the conditions for a function to be invertible Determine the properties of the inverses of the circular functions Evaluate inverse circular functions Find exact values of composite functions with inverse trigonometric functions. Inverse is the fancy mathematical term for undo. In this section, we re looking for ways to undo our trigonometric (circular) functions, but we have to be careful about how we approach this question. We ll start by reviewing conditions for creating inverse functions, then figure out how to apply these concepts to our circular functions. A Quick Review of Inverse Functions Before we worry about our inverse trigonometric functions, we must review what an inverse function is, and what conditions we must have in order to create an inverse function. Definition 4: Inverse functions f and g are inverses of one another if (f g)(x) = x for all x in the domain of g and (g f )(x) = x for all x in the domain of f That is, f, and g undo one another for inputs in their domains. If g is the inverse of f, we notate this as g = f 1. The notation f 1 is an unfortunate choice since you ve been programmed since Elementary Algebra to think of this as 1. This is most definitely not the case since, for instance, f (x) = x + 4 has as its inverse f f 1 (x) = x 4 1, which is certainly different than f (x) = 1 x+4. Notice that f 1 (x) undoes any action of f. For instance, if x = 5, then f (5) = (5) + 4 = = 19.

154 146 The Circular Functions Taking the output 19 from f, we substitute it into f 1 to get f 1 (19) = 19 4 = 15 = 5, which is our original input to f. To check that f 1 does the job for all x in the domain of f, we take the generic output from f, f (x) = x + 4, and substitute that into f 1. That is, we simplify f 1 (f (x)) = f 1 (x + 4) = (x+4) 4 = x = x, which is our original input to f. If we carefully examine the arithmetic as we simplify f 1 (f (x)), we actually see f 1 first undoing the addition of 4, and then undoing the multiplication by. Not only does f 1 undo f, but f also undoes f 1. That is, if we take the output from f 1, f 1 (x) = x 4, and substitute that into f, we get f (f 1 (x)) = f ( ) ( x 4 = x 4 ) + 4 = (x 4) + 4 = x. Definition 44: Conditions for invertibility In order to find the inverse of a function, that function must be one-to-one. That is, each input must correspond to exactly one output and each output must correspond to exactly one input. Graphically, to check that a function is invertible, we see if it passes both the Vertical Line Test (to make sure it s a function) and the Horizontal Line Test (to make sure the function is one-to-one). That is, if we draw a vertical or a horizontal line anywhere on the graph, it should only intersect the graph in one location. Certain functions do not naturally meet these requirements. For instance, a quadratic function, which has a parabola for a graph, does not pass the horizontal line test. The natural candidate for an inverse of f (x) = x is g(x) = x. However, if we examine g(f ( )) we notice a problem. In this case, we have f ( ) = ( ) = 4, but g(4) = 4 =, not the - that we started with. The common practice, then, is to limit the domain of a non-one-to-one function before finding its inverse. In the case of f (x) = x, we limit the domain of f to x 0, and now g(x) = x is an inverse that works for all of the x values on our limited domain. Finally, before we see how all of this applies to trigonometric functions, let s review the propoerties of inverse functions.

155 5.5 Inverse Trig Functions 147 Definition 45: Properties of Inverse Functions Suppose f is an invertible function. There is exactly one inverse function for f, denoted f 1 (read f -inverse ) The range of f is the domain of f 1 and the domain of f is the range of f 1 f (a) = c if and only if a = f 1 (c) NOTE: In particular, for all y in the range of f, the solution to f (x) = y is x = f 1 (y). (a, c) is on the graph of f if and only if (c, a) is on the graph of f 1 NOTE: This means graph of y = f 1 (x) is the reflection of the graph of y = f (x) across y = x. f 1 is an invertible function and (f 1 ) 1 = f. The Inverse Sine and Inverse Cosine Functions Now we are ready to turn our attention to the inverses of the circular (trigonometric) functions. Our immediate problem is that, owing to their periodic nature, none of the six circular functions is one-to-one. To remedy this, we restrict the domains of the circular functions in the same way we restricted the domain of the quadratic function in the example above to obtain a one-to-one function. We start with f (t) = sin(t) and restrict our domain to [ π, π ] in order to keep the range as [ 1, 1] as well as the properties of being smooth and continuous. As we saw in our quick review, the inverse of a function f is typically denoted f 1. For this reason, some textbooks use the notation f 1 (t) = sin 1 (t) for the inverse of f (t) = sin(t). The obvious pitfall here is our convention of writing (sin(t)) as sin (t), (sin(t)) as sin (t) and so on. It is far too easy to confuse sin 1 (t)

156 148 The Circular Functions with 1 = csc(t) so we will not use this notation in our text. But be aware that many books do! As always, sin(t) be sure to check the context! Instead, we use the notation f 1 (t) = arcsin(t), read arc-sine of t. We ll explain the arc in arcsine shortly. For now, we graph f (t) = sin(t) and f 1 (t) = arcsin(t), where we obtain the latter from the former by reflecting it across the line y = t There are a few things to notice here. First, the limited, one-to-one domain of our sine function is [ π, π ] ; that becomes the range of the arcsine function. Similarly, the range of sine, [ 1, 1], is the domain of our new arcsine function. This domain and range will have important implications when it comes time for us to calculate special common values using the arcsine function. Next, we consider g(t) = cos(t). If we try the same domain restriction as we used for sine, we have a problem; that will only capture y-values from 0 to 1, and that chunk of the cosine graph fails the Horizontal Line Test and hence is not one-to-one. Instead, we select the interval [0, π] for our restriction. Reflecting the across the line y = t produces the graph y = g 1 (t) = arccos(t).

157 5.5 Inverse Trig Functions 149

158 150 The Circular Functions Let s summarize some important properties of these functions. Definition 46: Properties of the Arccosine and Arcsine Functions Properties of F (x) = arcsin(x) Domain: [ 1, 1] Range: [ π, π ] arcsin(x) = t if and only if sin(t) = x and π t π sin(arcsin(x)) = x provided 1 x 1 arcsin(sin(t)) = t provided π t π F (x) = arcsin(x) is odd Properties of G(x) = arccos(x) Domain: [ 1, 1] Range: [0, π] arccos(x) = t if and only if cos(t) = x and 0 t π cos(arccos(x)) = x provided 1 x 1 arccos(cos(t)) = t provided 0 t π Before moving to an example, we take a moment( to understand ) the arc in arcsine. Consider the figure below which illustrates the specific case of arcsin.

159 5.5 Inverse Trig Functions 151 By definition, the real number t = arcsin ( ) satisfies sin(t) = with π t π. In other words, we ) = π. are looking for angle measuring t radians between π and π ( with a sine of. Hence, arcsin In general, the function f (t) = sin(t) takes a real number input t, associates it with the angle θ = t radians, and returns the value sin(θ). The value sin(θ) = sin(t) is the y-coordinate of the terminal point on the Unit Circle of an oriented arc of length t whose initial point is (1, 0). Hence, we may view the inputs to f (t) = sin(t) as oriented arcs and the outputs as y-coordinates on the Unit Circle. Therefore, the function f 1 reverses this process and takes y-coordinates on the Unit Circle and return oriented arcs, hence the arc in arcsine. It is high time for some examples, don t you think? Example 45 Find the exact ( ) values of the following: a. arcsin The best way to approach these problems is to remember that arcsin(x) and arccos(x) are real numbers which correspond to the radian measure of angles that fall within a certain prescribed range. ( ) To find arcsin, we need the angle measuring t radians which lies between π and π

160 15 The Circular Functions with sin(t) = ( ). Hence, arcsin = π 4. b. arccos ( ) 1 To find arccos ( 1 ), we are looking for the angle measuring t radians which lies between 0 and π that has cos(t) = 1. Our answer is arccos ( ) 1 = π. With positive inputs, arcsine and arccosine are fairly easy to evaluate. What happens, though, when negative inputs get involved? Tread carefully, especially with arcsine! Example 46 Find the exact values of the following: a. arcsin ( ) 1 For arcsin ( ) 1, we are looking for an angle measuring t radians which lies between π and π with sin(t) = 1. We want to look in QIV of the unit circle, but we need to be sure to give our answer a negative angle name, so that our final answer is between π and π we see that the angle 11π has a sine of 1 6 arcsin ( 1 ) = π 6. as required. In QIV, 11π. The negative angle coterminal with is π 6 6. Hence, Alternatively, we could use the fact that the arcsine function is odd, so arcsin ( ( ) 1 = arcsin 1 ). We find arcsin ( ) 1 = π 6, so arcsin ( ( ) 1 = arcsin 1 ) = π 6. ( b. arccos ( For arccos ) ). Hence, arccos (, we need the angle measuring t radians which lies between 0 and π with cos(t) = ) = π 4.

161 5.5 Inverse Trig Functions 15 Check Point 5 Evaluate cos 1 ( ). Next, we consider how to evaluate compositions of inverse functions and trig functions. It s tempting to take shortcuts when approaching these problems, but we must always carefully consider our (restricted) domains when evaluating these compositions! Example 47 Find the exact values of the following: a. arccos ( cos ( )) π 6 Since 0 π π, we could simply invoke the properties of inverse trig functions to get 6 arccos ( cos ( )) π 6 = π 6. However, in order to make sure we understand why this is the case, we choose to work the example through using the definition of arccosine. Working from the inside out, arccos ( cos ( )) ( ) π 6 = arccos. To find arccos angle measuring t radians which lies between 0 and π that has cos(t) = arccos ( cos ( )) ( ) π 6 = arccos = π 6. b. arccos ( cos ( )) 11π 6 ( ), we need an. We get t = π, so that 6 Since 11π does not fall between 0 and π, we can t look at the outer arccosine as simply 6 undoing the inner cosine. We are forced to work through from the inside out starting with arccos ( cos ( )) ( ) ( ) 11π 6 = arccos. From the previous problem, we know arccos = π 6. Hence, arccos ( cos ( )) 11π 6 = π 6.

162 154 The Circular Functions Check Point 6 Evaluate arcsin ( sin ( 4π )). Another interesting application of inverse trig functions is figuring out how we can use them to solve for non-special values. To see what we mean, check out the next example. Example 48 Find the exact value of sin ( arccos ( 5 )) First, let s think about what this jumble of information means. When we write arccos ( 5), we mean that there is an angle between 0 and π that has an x-coordinate of on the unit circle. 5 Let s temporarily name that angle t, so that t = arccos ( 5). Well then, by definition of arccosine, cos(t) =. We also know that the terminal side of t must fall in QII, since t must be between 0 5 and π and its cosine is negative. Now, we can use our friend the Pythagorean Identity to figure out the sine of t. cos (t) + sin (t) = 1 ( 5) + sin (t) = sin (t) = 1 sin (t) = 16 5 sin(t) = ± 4 5 Because we know that t is a QII angle, its sine must be positive, so sin(t) = 4 5. Therefore, sin ( arccos ( 5)) = 4 5. Note that if you prefer to draw a reference triangle in Quadrant II, you can arrive at the same final answer.

163 5.5 Inverse Trig Functions 155 Example 49 Find the exact value of tan ( arcsin ( 5 7)). Again, we let t = arcsin ( ) 5 7. Then we can start by saying that there is an angle, t, between π and π that has a y-coordinate of 5 on the Unit Circle. That places our t in QI, and we know that 7 sin(t) = 5. Using the Pythagorean Identity, then, we have 7 ( ) 5 cos (t) + = 1 7 cos (t) = 1 cos (t) = cos(t) = 7 We have chosen the positive solution, since cosine is positive in QI. Now, since t = arcsin ( ( ( 5 7), our expression tan arcsin 5 7)) is equivalent to tan(t). Now, we know the sine and cosine of our angle, but in the end, we want to know its tangent. Recall that tan(t) = sin(t) cos(t). So to calculate our tangent, we use the sine and cosine values from above: tan(t) = 5/7 4/7 = 5 4 = = Check Point 7 Evaluate cos ( arcsin ( 1 0)).

164 156 The Circular Functions Example 50 Rewrite f (x) = tan(arccos(x)) as an algebraic functions of x and state the domain. We begin this problem in the same manner we began the previous two problems. t = arccos(x), so our goal is to find a way to express tan (arccos (x)) = tan(t) in terms of x. We let Since t = arccos(x), we know cos(t) = x where 0 t π. One approach to finding tan(t) is to use the quotient identity tan(t) = sin(t). Since we know cos(t), we just need to find sin(t). cos(t) Using the Pythagorean Identity, we get sin (t) = 1 cos (t) = 1 x so that sin(t) = ± 1 x. Since 0 t π, sin(t) 0, so we choose sin(t) = 1 x. Thus, tan(t) = sin(t) cos(t) = 1 x, so f (x) = tan(arccos(x)) = 1 x. x x To determine the domain, we consider that the function f (x) = tan (arccos (x)) can be thought of as a two step process: first, take the arccosine of a number, and second, take the tangent of whatever comes out of the arccosine. Since the domain of arccos(x) is 1 x 1, the domain of f will be some subset of [ 1, 1]. The range of arccos(x) is [0, π], and of these values, only π will cause a problem for the tangent function. Since arccos(x) = π happens when x = cos ( ) π = 0, we exclude x = 0 from our domain. Hence, the domain of f (x) = tan (arccos (x)) = 1 x is [ 1, 0) (0, 1]. x Note that in this particular case, we could have obtained the correct domain of f using its algebraic description: f (x) = tan(arccos(x)) = caution! 1 x x. This is not always true, however, so proceed with

165 5.5 Inverse Trig Functions 157 Example 51 Rewrite g(x) = cos(arcsin(x)) as an algebraic functions of x and state the domain. We proceed as in the previous problem by writing t = arcsin(x) so that t lies in the interval [ π, π with sin(t) = x. We aim to express cos (arcsin(x)) = cos(t) in terms of x. ] Since we know x = sin(t), then by the Pythagorean Identity we have cos (t) + x = 1 cos (t) = 1 x cos(t) = ± 1 x Again, because t lies in the interval [ π, π ], the cosine of t must be positive, so cos(t) = 1 x. To find the domain of g(x) = cos (arcsin(x)), we once again consider the inner function first. The domain of arcsin(x) is [ 1, 1], and since there are no domain restrictions on cosine, the domain of g is [ 1, 1]. Check Point 8 Simplify cot ( sin 1 (y) ).

166 158 The Circular Functions Inverse Tangent and Inverse Cotangent The next pair of functions we wish to discuss are the inverses of tangent and cotangent. First, we restrict f (t) = tan(t) to its fundamental cycle on ( π, π ) to obtain the arctangent function, f 1 (t) = arctan(t). Among other things, note that the vertical asymptotes t = π and t = π of the graph of f (t) = tan(t) become the horizontal asymptotes y = π and y = π of the graph of f 1 (t) = arctan(t). Next, we restrict g(t) = cot(t) to its fundamental cycle on (0, π) to obtain g 1 (t) = arccot(t), the arccotangent function. Once again, the vertical asymptotes t = 0 and t = π of the graph of g(t) = cot(t) become the horizontal asymptotes y = 0 and y = π of the graph of g 1 (t) = arccot(t). Below we summarize the important properties of the arctangent and arccotangent functions.

167 5.5 Inverse Trig Functions 159 Definition 47: Properties of the Arctangent and Arcotangent Functions Properties of F (x) = arctan(x) Domain: (, ) Range: ( π, π ) as x, arctan(x) π + ; as x, arctan(x) π arctan(x) = t if and only if tan(t) = x and π < t < π arctan(x) = arccot ( 1 x ) for x > 0 tan (arctan(x)) = x for all real numbers x arctan(tan(t)) = t provided π < t < π F (x) = arctan(x) is odd Properties of G(x) = arccot(t) Domain: (, ) Range: (0, π) as x, arccot(x) π ; as x, arccot(x) 0 + arccot(x) = t if and only if cot(t) = x and 0 < t < π arccot(x) = arctan ( 1 x ) for x > 0 cot (arccot(x)) = x for all real numbers x arccot(cot(t)) = t provided 0 < t < π

168 160 The Circular Functions When it comes time to evaluate inverse trigonometric functions, it s useful to have a list of the common tangent and cotangent values for special angles. You may not be as comfortable with these the corresponding sines and cosines, so we list them for your convenience. t π π π 4 π 6 0 π 6 π 4 π π t 0 π 6 π 4 π π π π 4 5π 6 π sin(t) cos(t) 1 tan(t) 0 DNE / 1/ / / 1 1/ / 1 = / 1 = / / 1 / / 1/ 1 0 DNE cos(t) sin(t) 1 / 1/ / 1 / cot(t) DNE 0 1/ / 1 = / / 1 = / / / 1 1/ 1 0 DNE Example 5 Find the exact values of the following. a. arctan( ) To find arctan( ), we need the angle measuring t radians which lies between π and π with tan(t) =. We find arctan( ) = π.

169 5.5 Inverse Trig Functions 161 b. arccot( ) To find arccot( ), we need the angle measuring t radians which lies between 0 and π with cot(t) =. Hence, arccot( ) = 5π 6. Check Point 9 Find the exact value of cot 1 ( ). Example 5 Find the exact values of the following. a. arccot ( cot ( )) 7π 6 This is another case where we are better off taking things a step at a time, so that we don t make assumptions that result in the wrong answer. In other words, the arccotangent does not directly undo the cotangent here, because the 7π input to the cotangent is not a valid output for the 6 arccotangent function; arccotangent only gives answers between 0 and π! First, cot ( ) 7π cos(7π/6) 6 = = / =. That means we can write arccot ( cot ( )) 7π sin(7π/6) 1/ 6 as arccot ( ). The angle between 0 and π with a cotangent of is π, so our final answer is 6 arccot ( cot ( 7π 6 )) = π 6. Check Point 40 Evaluate arctan ( tan ( 4π )) exactly. Example 54 Find the exact value of the following. sin ( arctan ( )) 4

170 16 The Circular Functions We start simplifying sin ( arctan ( ( ) )) 4 by letting t = arctan 4. By definition, tan(t) = 4 for some angle measuring t radians which lies between π and π. Since tan(t) < 0, we know, in fact, t corresponds to a Quadrant IV angle. We are given tan(t) but wish to know sin(t). Since there is no direct identity to marry the two, we make a quick sketch of the situation below. Since tan(t) = 4 = 4, we take P(, 4) as a point on the terminal side of θ = t = arctan ( ) 4 radians. We find r = x + y = () + ( 4) = 5, so sin(t) = 4 5. Hence, sin ( arctan ( 4 )) = 4 5. Check Point 41 Find the exact value of sin ( arctan ( 8 9)). Example 55 Rewrite the following composite function as and algebraic function of x and state the domain. cos(arccot(x)) To get started, we let t = arccot(x) so that cot(t) = x where 0 < t < π. In terms of t, cos(arccot(x)) = cos(t), and our goal is to express the latter in terms of x.

171 5.5 Inverse Trig Functions 16 Note that if t = arccot(x), then cot(t) = x. If we think way back to right triangles, this means we could draw a right triangle with reference angle t, where the ratio of the adjacent side to the opposite side is x. The easiest way to make such a triangle is to let the adjacent side be x and the opposite side be 1. Since we want to find cos(t), we will need to find an expression for the hypotenuse of this triangle. We can once more lean on the Pythagorean Theorem: (x) + 1 = c 4x + 1 = c c = 4x + 1 Note: It may be tempting to try to simplify this expression, but the addition inside of the root makes that impossible! Because cos(t) is adjacent over hypotenuse, our answer is cos(t) = x 4x + 1 To find the domain, viewing g(x) = cos(arccot(x)) as a sequence of steps, we see we first double the input x, then take the arccotangent, and, finally, take the cosine. Since each of these processes are valid for all real numbers, the domain of g is (, ).

172 164 The Circular Functions Check Point 4 Simplify sec(tan 1 (B)). Inverse Cosecant and Inverse Secant The last two functions to invert are secant and cosecant. A portion of each of their graphs are given below with the fundamental cycles highlighted. It is clear from the graph of secant that we cannot find one single continuous piece of its graph which covers its entire range of (, 1] [1, ) and restricts the domain of the function so that it is one-to-one. The same is true for cosecant. Thus in order to define the arcsecant and arccosecant functions, we must settle for a piecewise approach wherein we choose one piece to cover the top of the range, namely [1, ), and another piece to cover the bottom, namely (, 1]. There are two generally accepted ways make these choices which restrict the domains of these functions so that they are one-to-one. In this subsection, we restrict the secant and cosecant functions to coincide with the restrictions on cosine and sine, respectively. For f (t) = sec(t), we restrict the domain to [ 0, π ) ( π, π] and we restrict g(t) = csc(t) to [ π, 0) ( 0, π ]. These choices are arbitrary; in this text, we ll use these domain restrictions because they are, in a sense, the prettiest. Calculus texts will often choose different domains, but once you ve chosen a domain, the rest of the principles we are learning will remain the same.

173 5.5 Inverse Trig Functions 165

174 166 The Circular Functions Definition 48: Properties of the Arcsecant and Arccosecant Functions Properties of F (x) = arcsec(x) Domain: {x x 1} = (, 1] [1, ) Range: [ 0, π ) ( π, π] as x, arcsec(x) π + ; as x, arcsec(x) π arcsec(x) = t if and only if sec(t) = x and 0 t < π or π < t π arcsec(x) = arccos ( 1 x ) provided x 1 sec (arcsec(x)) = x provided x 1 arcsec(sec(t)) = t provided 0 t < π or π < t π Properties of G(x) = arccsc(x) Domain: {x x 1} = (, 1] [1, ) Range: [ π, 0) ( 0, π ] as x, arccsc(x) 0 ; as x, arccsc(x) 0 + arccsc(x) = t if and only if csc(t) = x and π t < 0 or 0 < t π arccsc(x) = arcsin ( 1 x ) provided x 1 csc (arccsc(x)) = x provided x 1 arccsc(csc(t)) = t provided π t < 0 or 0 < t π G(x) = arccsc(x) is odd

175 5.5 Inverse Trig Functions 167 The particular reason we stick with the ranges here is specifically because of two properties listed in above: arcsec(x) = arccos ( 1 x ) and arccsc(x) = arcsin ( 1 x ). These formulas essentially allow us to always convert arcsecants and arccosecants back to arccosines and arcsines, respectively. We see this play out in our next example. Example 56 Find the exact values of the following. a. arcsec() Using the property highlighted above, we have arcsec() = arccos ( ) 1 = π. b. arccsc( ) Once again, we can rewrite in terms of sine, giving arccsc( ) = arcsin ( 1 ) = π 6. Check Point 4 Find the exact value of sec 1 (). Check Point 44 ( Evaluate csc 1 ). Example 57 Find the exact values of the following. a. arcsec ( sec ( )) 5π 4

176 168 The Circular Functions Since 5π 4 doesn t fall between 0 and π or π and π, we cannot use the inverse property directly. Hence, we work from the inside out. We get: arcsec ( sec ( )) ( ) ( ) 5π 4 = arcsec( ) = arccos 1 = arccos = π 4. Check Point 45 Evaluate sec(sec 1 ( )). Check Point 46 Evaluate sec 1 ( sec ( 4π )). Example 58 Evaluate exactly. cot (arccsc ( )) We begin simplifying cot (arccsc ( )) by letting t = arccsc( ). Then, csc(t) =. Since csc(t) < 0, t lies in the interval [ π, 0), so t corresponds to a Quadrant IV angle. All right, so we know that our angle lies in QIV, with csc(t) =. Therefore, sin(t) = 1. Again, let s draw a simple triangle with these properties:

177 5.5 Inverse Trig Functions 169 In the end we want to know cot(t), which is the ratio of the adjacent side over the opposite side, so we ll need to solve for the length of the missing leg using the Pythagorean Theorem. x + ( 1) = x + 1 = 9 x = 8 x = ± 8 Since our missing side lies on the positive x-axis, we have x = 8 = for this side. Therefore cot(t) = 1 cot(t) = Check Point 47 Evaluate sin ( sec ( ))

178 170 The Circular Functions Example 59 Rewrite the following composite function as an algebraic function of x and state the domain. f (x) = tan(arcsec(x)) Proceeding as above, we let t = arcsec(x). Then, sec(t) = x for t in [ 0, π ) ( π, π]. We seek a formula for tan(arcsec(x)) = tan(t) in terms of x. [ Since ) sec(t) ( = x, then we know that cos(t) = 1. Here we have to be careful; t can be in x 0, π π, π], so 1 might be negative or positive depending on the quadrant we find ourselves x in. The pictures here can get unneccessarily confusing, so we re going to try a different strategy: rewriting the Pythagorean Identity. If then we can divide every term by cos (t) to get cos (t) + sin (t) = 1 cos (t) cos (t) + sin (t) cos (t) = 1 cos (t) 1 + tan (t) = sec (t) Now, since sec(t) = x, we have 1 + tan (t) = x tan (t) = x 1 tan(t) = ± x 1 If t belongs to [ 0, π ) then tan(t) 0. On the the other hand, if t belongs to ( π, π] then tan(t) 0. As a result, we get a piecewise defined function for tan(t): { x 1, if 0 t < π tan(t) = x 1, if π < t π Now we need to determine what these conditions on t mean for x. Since x = sec(t), when 0 t < π, x 1, and when π < t π, x 1. Hence, { x 1, if x 1 f (x) = tan(arcsec(x)) = x 1, if x 1

179 5.5 Inverse Trig Functions 171 To find the domain of f, we consider f (x) = tan(arcsec(x)) as a two step process. First, we have the arcsecant function, whose domain is (, 1] [1, ). Since the range of arcsec(x) is [ 0, π ) ( π, π], taking the tangent of any output from arcsec(x) is defined. Hence, the domain of f is (, 1] [1, ). Example 60 Rewrite the following composite function as an algebraic function of x and state the domain. f (x) = cos(arccsc(4x)) Taking a cue from the previous problem, we start by letting t = arccsc(4x). Then csc(t) = 4x for t in [ π, 0) ( 0, π ]. Our goal is to rewrite cos(arccsc(4x)) = cos(t) in terms of x. From csc(t) = 4x, we get sin(t) = 1 4x cos (t) = 1 sin (t). Substituting sin(t) = 1 4x gives cos (t) = 1 ( 1 4x denominator and extracting square roots, we obtain: cos(t) = ±, so to find cos(t), we can make use of the Pythagorean Identity: ) = 1 1. Getting a common 16x 16x 1 16x = ± 16x 1. 4 x Since t belongs to [ π, 0) ( 0, π ], we know cos(t) 0, so we choose cos(t) = 16 x 4 x. (The absolute values here are necessary, since x could be negative.) Hence, g(x) = cos(arccsc(4x)) = 16 x. 4 x To find the domain of g(x) = cos(arccsc(4x)), as usual, we think of g as a series of processes. First, we take the input, x, and multiply it by 4. Since this can be done to any real number, we have

180 17 The Circular Functions no restrictions here. Next, we take the arccosecant of 4x. Using interval notation, the domain of the arccosecant function is written as: (, 1] [1, ). Hence to take the arccosecant of 4x, the quantity 4x must lie in one of these two intervals. a That is, 4x 1 or 4x 1, so x 1 4 or x 1 4. The third and final process coded in g(x) = cos(arccsc(4x)) is to take the cosine of arccsc(4x). Since the cosine accepts any real number, we have no additional restrictions. Hence, the domain of g is (, 1 4] [ 1 4, ). a Alternatively, we can write the domain of arccsc(x) as x 1, so the domain of arccsc(4x) is 4x 1.

181 5.5 Inverse Trig Functions Exercises In the following exercises, find the exact value. 1. arcsin ( 1) ( ). arcsin ( ). arcsin ( 4. arcsin 1 ) 5. arcsin (0) ( ) 1 6. arcsin ( ) 7. arcsin ( ) 8. arcsin 9. arcsin (1) 10. arccos ( 1) ( ) 11. arccos ( ) 1. arccos ( 1. arccos 1 ) 14. arccos (0) ( ) arccos ( ) 16. arccos ( ) 17. arccos 18. arccos (1) 19. arctan ( ) 0. arctan ( 1) ( ) 1. arctan. arctan ( (0) ). arctan 4. arctan (1) 5. arctan ( ) 6. arccot ( ) 7. arccot ( 1) ( ) 8. arccot 9.arccot (0) ( ) 0. arccot 1. arccot (1). arccot ( ). arcsec () 4. arccsc () 5. arcsec ( ) 6. arccsc ( ) ( ) 7.arcsec ( ) 8. arccsc 9. arcsec (1) 40. arccsc (1) In the following exercises, assume that the range of arcsecant is [ 0, π ) ( π, π] and that the range of arccosecant is [ π, 0) ( 0, π ] when finding the exact value. 41. arcsec ( ) 4. arcsec ( ) ( 4. arcsec ) 44. arcsec ( 1) 45. arccsc ( ) 46. arccsc ( ) ( 47. arccsc ) 48. arccsc ( 1) In the following exercises, find the exact value or state that it is undefined.

182 174 The Circular Functions ( ( )) sin arcsin ( ( )) 50. sin arcsin ( ( )) 51. sin arcsin 5 5. sin (arcsin ( ( 0.4)) ( 5 5. sin arcsin ( 4 ( )) 54. cos arccos ( ( 55. cos arccos 1 )) ( ( )) cos arccos cos (arccos ( 0.998)) 58. cos (arccos (π)) 59. tan (arctan ( 1)) 60. tan ( arctan ( )) ( ( )) tan arctan 1 6. tan (arctan (0.965)) 6. tan (arctan (π)) 64. cot (arccot (1)) 65. cot ( arccot ( )) ( ( 66. cot arccot 7 )) cot ((arccot ( 0.001)) ( )) 17π 68. cot arccot sec (arcsec ()) 70. sec (arcsec ( ( 1)) ( )) sec arcsec 7. sec (arcsec (0.75)) 7. sec (arcsec (117π)) 74. csc ( arccsc ( )) ( ( 75. csc arccsc )) ( ( )) 76. csc arccsc 77. csc ((arccsc (1.0001)) ( π 78. csc arccsc 4 In the following exercises, find the exact value or state that it is undefined. ( ( π )) 79. arcsin sin ( ( arcsin sin π )) ( ( )) π 81. arcsin sin ( ( 4 )) 11π 8. arcsin sin ( ( 6 )) 4π 8. arcsin sin ( ( π )) 84. arccos cos ( ( 4 )) π 85. arccos cos ( ( )) π 86. arccos cos ( ( 87. arccos cos π )) 6 ( ( )) 5π 88. arccos cos ( ( 4 π )) 89. arctan tan ( ( 90. arctan tan π )) arctan (tan ( (π)) ( π )) 9. arctan tan ( ( )) π 9. arctan tan ( ( π )) 94. arccot 95. arccot cot ( cot ( π )) 4 ( π )) 96. arccot (cot ( (π)) 97. arccot cot ( ( )) π 98. arccot cot

183 5.5 Inverse Trig Functions 175 In the following exercises, assume that the range of arcsecant is [ 0, π ) [ ) π, π and that the range of arccosecant is ( 0, π ] ( ] π, π when finding the exact value. ( ( π )) 99. arcsec sec ( ( 4 )) 4π 100. arcsec sec ( ( )) 5π 101. arcsec sec ( ( arcsec sec π )) ( ( )) 5π 10. arcsec sec ( ( π )) 104. arccsc csc 6 ( ( )) 5π 105. arccsc csc ( ( 4 )) π 106. arccsc csc ( ( 107. arccsc csc π )) ( ( )) 11π 108. arccsc csc ( ( 6 )) 11π 109. arcsec sec ( ( 1)) 9π 110. arccsc csc 8 In the following exercises, find the exact value or state that it is undefined. ( ( 111. sin arccos 1 )) ( ( )) 11. sin arccos sin (arctan ( )) 114. sin ( arccot ( 5 )) 115. sin (arccsc ( ( )) ( 116. cos arcsin 5 )) cos ( arctan ( 7 )) 118. cos (arccot ()) 119. cos ( (arcsec (5)) ( 10. tan arcsin )) 5 ( ( tan arccos 1 )) ( ( )) 5 1. tan arcsec 1. tan (arccot (1)) ( ( )) cot arcsin 1 ( ( )) 15. cot arccos 16. cot ( arccsc ( 5 )) 17. cot (arctan ( (0.5)) ( )) 18. sec arccos ( ( 19. sec arcsin 1 )) sec (arctan ( (10)) ( )) sec arccot csc ((arccot (9)) ( )) 1. csc arcsin ( ( csc arctan )) In the following exercises, rewrite each of the following composite functions as algebraic functions of x and state the domain. 15. f (x) = sin (arccos (x)) 16. f (x) = cos (arctan (x))

184 176 The Circular Functions 17. f (x) = tan (arcsin (x)) 18. f (x) = sec (arctan (x)) 19. f (x) = csc (arccos (x))

185 5.6 Trig Equations Trig Equations Learning Objectives In this section you will: Solve basic trigonometric equations. Solve trigonometric equations where the input is an angle multiple. Solve trigonometric equations involving algebra. Thales of Miletus (circa BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of similar triangles, which he developed by measuring the shadow of his staff. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the finding the dimensions of the pyramids. Solving Basic Trigonometric Equations Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions.

186 178 The Circular Functions Example 61 Find all angles that satisfy the following equations. Express your answers in radians. a. cos(θ) = 1 Since 1 is one of our special unit circle values, we start by finding solutions by i inspection /i. That is, we look at the unit circle and recall which angles have a cosine of 1. Those angles, in QI and QIV, are θ = π and θ = 5π. These are not the i only /i solutions, however. Take θ = π. Any other coterminal angle will also have a cosine of 1. That means π + π = 7π and 7π 1π + π = both have cosines 1 of and are therefore solutions. In fact, we can keep adding (or subtracting) π infinitely to come up with more solutions! Hence to describe all angles coterminal with a given angle, we add πk for integers k = 0, ±1, ±,.... Hence, we record our final answer as θ = π +πk for integers k. Proceeding similarly for the Quadrant IV case, we find the solution to cos(θ) = 1 here is 5π, so our answer in this Quadrant is θ = 5π + πk for integers k. b. sin(α) = 1 Again using inspection, we see that the common unit circle values of α = 7π have sines of 6 6. We capture all solutions by adding multiples of π to create coterminal angles: 1 and 11π α = 7π 6 + πk for integers k. α = 11π 6 + πk Check Point 48 Solve sin(x) = on all real numbers. One of the key items to take from the examples above is that, in general, solutions to trigonometric equations consist of infinitely many answers. To get a feel for these answers, the reader is encouraged to follow our mantra: When in doubt, write it out! This is especially important when checking answers to the exercises.

187 5.6 Trig Equations 179 Example 6 Find all angles that satisfy the following equations. Express your answers in radians. a.sec(θ) = To solve sec(θ) =, we convert to cosines and get θ = π + πk or θ = 5π + πk for integers k. 1 = or cos(θ) = 1. The answer is: cos(θ) b. tan(θ) = The inspection bit is a little harder on this one, but with some work we can find that the angles θ = π and 4π have tangents of. In Quadrant I, we get the solutions: θ = π + πk for integers k, and for Quadrant III, we get θ = 4π + πk for integers k. While these descriptions of the solutions are correct, they can be combined into one list as θ = π + πk for integers k. Check Point 49 Solve tan(θ) = over all real numbers. In some cases, the equation is not immediately formatted for inspection. That is, we often must first worry about isolating the trigonometric part of the equation, then proceed with inspection. Example 6 Find all angles that satisfy the following equations. Express your answers in radians. a. cos(α) = 5 First we must isolate the cos(α): cos(α) = cos(α) = 1 Now we can proceed with inspection. We see that α = π has a cosine of -1. Our final set of

188 180 The Circular Functions solutions is α = π + πk for all integers k. Check Point 50 Solve sin(x) = 1. Definition 49: Solving Trigonometric Equations Over All Real Numbers by Inspection Isolate the trigonometric function. Identify the special unit circle angles which satisfy the equation. Account for other solutions by adding multiples of π to the answer: +πk for integers k. Solving Trigonometric Equations by Inspection over Limited Domains Often times, we are only interested in the solutions to a trigonometric equation which fall in a particular domain. In other words, we don t need to describe infinite solutions, we only need to find a few, specific solutions. Example 64 a. Solve cos(α) = for 0 α < π Here, we re being asked only for the solutions that are the initial special angles on the unit circle. That is, α = π 6, 11π 6 We don t need to add any copies of π, because we ve found all the solutions between 0 and π just by inspection! b. Solve cos(α) = for 0 α < 4π We start in the same place, with α = π 6, 11π. However, we want all valid solutions all the way up to 6

189 5.6 Trig Equations 181 4π, so we need to add π to find some coterminal angles: α = π 6 + π = π 6 + 1π 6 = 1π 6 α = 11π 6 + π = π 6 We stop here, because if we were to add any more cycles of π, we would create angles larger than 4π. In other words, since we want our solutions to be between 0 and 4π, we only need to look at two rotations of the unit circle. Our final set of answers is α = π 6, 11π 6, 1π 6, π 6. c. Solve cos(α) = for π α < π Here, only our first solution, π initially falls in the relevant interval. Since we want to look at negative 6 angles, we can create coterminal angles by subtracting π: α = π 6 π = 11π 6. That solution doesn t work; it s to the left of π. Let s try our other initial special value instead; that is, even though it doesn t fall in the right interval initially, when we solve by inspection we see that α = 11π has a cosine of, so let s find a negative coterminal angle: 6 That one works! Our final solution set is α = 11π 6 π = π 6 α = π 6, π 6. Check Point 51 Solve sin(x) = 1, 0 x < π. Check Point 5 Solve cos(x) = 0, π x < π.

190 18 The Circular Functions Solving Trigonometric Equations with Complicated Inputs So far, all of our equations have had simple, single inputs in the trigonometric portion of the equation. We ve been solving sin(x), not sin(x). Here, we will outline two strategies for solving trigonometric equations with more complicated expressions inside of the trig portion. Example 65 Method 1: Solving Trig Equations with Complicated Inputs by Solving Over All Real Numbers First Solve cos(θ) = on the interval [0, π). This equation is at least in the form cos(u) = if we let u = θ. In that case, The solutions to cos(u) = are u = 5π 6 + πk or u = 7π + πk for integers k. 6 Since the argument of cosine here is θ, this means θ = 5π 6 + πk or θ = 7π 6 Solving for θ gives θ = 5π 7π + πk or θ = + πk for integers k πk for integers k. To check these answers analytically, we substitute them into the original equation. For any integer k: cos ( [ 5π 1 + πk]) = cos ( 5π 6 + πk) = cos ( ) 5π = 6 (the period of cosine is π) Similarly, we find cos ( [ 7π 1 + πk]) = cos ( 7π 6 + πk) = cos ( ) 7π 6 =. To determine which of our solutions lie in [0, π), we substitute integer values for k. The solutions we keep come from the values of k = 0 and k = 1 and are θ = 5π 1, 7π 1, 17π 19π and 1 1. Example 66 Method : Solving Trig Equations with Complicated Inputs by Modifying Intervals Solve cos(θ) = on the interval [0, π). Again, we let u = θ, but this time before we find our solutions, we consider that if 0 θ < π

191 5.6 Trig Equations 18 then 0 θ < π 0 u < 4π So, if we want to solve cos(u) =, we need to solve on the interval [0, 4π), or two rotations of the unit circle. The first set of solutions comes from u = 5π 6 and its coterminal pal u = 5π 17π + π =. The next set 6 6 come from u = 7π 6 and u 7π 19π + π = 6 6. Now we put things back in terms of the original θ: θ = 5π 6 θ = 5π 1 θ = 7π 6 θ = 7π 1 θ = 17π 6 θ = 17π 1 θ = 19π 6 θ = 19π 1 You can choose to work with whichever method you prefer. We will continue to present the next few examples using both methods. Example 67 Solve exactly on [0, π). csc( 1 θ π) = Since this equation has the form csc(u) =, we rewrite this as sin(u) = or u = π + πk for integers k. 4 and find u = π 4 + πk Since the argument of cosecant here is ( 1 θ π), 1 θ π = π 4 + πk or 1 θ π = π 4 + πk. To solve 1θ π = π 4 + πk, we first add π to both sides to get 1θ = π + πk + π. A common error 4 is to treat the πk and π terms as like terms and try to combine them when they are not.

192 184 The Circular Functions We can, however, combine the π and π 4 terms to get 1 θ = 5π 4 + πk. We now finish by multiplying both sides by to get θ = ( 5π 4 + πk) = 15π + 6πk, where k, as 4 always, runs through the integers. Solving the other equation, 1θ π = π 1π + πk produces θ = + 6πk for integers k. To check the 4 4 first family of answers, we substitute, combine line terms, and simplify. csc ( [ 1 15π + 6πk ] π ) = csc ( 5π + πk π) 4 4 = csc ( π 4 + πk) = csc ( ) π 4 = The family θ = 1π 4 + 6πk checks similarly. (the period of cosecant is π) Despite having infinitely many solutions, we find that i none /i of them lie in [0, π). First we let u = 1 θ π and find that if 0 θ < π 1 0 π 1 θ π < 1 π π π u < π That s a pretty small interval in which to search for our solution, but we ll check. We know if csc(u) = 1 then sin(u) =, so our candidates for solutions are u = π 4 and π. To see 4 if these have negative coterminal angles that fall between π and π, we subtract a full rotation, π. π 4 π = π 4 8π 4 = 7π 4 π 4 π = 5π 4 Neither of these angles falls between π and π, so our search stops here; there are no solutions

193 5.6 Trig Equations 185 in the interval. Example 68 Solve exactly on [0, π). cot(t) = 0 Since cot(t) = 0 has the form cot(u) = 0, we know u = π + πk, so, in this case, t = π integers k. + πk for Solving for t yields t = π 6 + π k. Checking our answers, we get cot ( [ π 6 + π k]) = cot ( π + πk) = cot ( ) π = 0 (the period of cotangent is π) As k runs through the integers, we obtain six answers, corresponding to k = 0 through k = 5, which lie in [0, π): x = π 6, π, 5π 6, 7π 6, π 11π and 6. If we let u = t, then 0 t < π 0 t < π 0 u < 6π So this time, we want to solve cot(u) = 0 by looking at i three /i rotations of the unit circle. Our first set of solutions comes from when u = π : u = π u = π + π = 5π u = 5π + π = 9π

194 186 The Circular Functions For the second batch, we start with u = π : u = π u = π + π = 7π u = 7π + π = 11π In each of these cases, u = t, so to finish solving, we divide each of our six answers by, producing a final solution of t = π 6, π, 5π 6, 7π 6, π 11π and 6. Check Point 5 Solve sin(x) = 1, 0 x < π. Check Point 54 Solve cos(x) = 1, 0 x < π. Solving Trigonometric Equations with Non-Special Angle Solutions So far we have limited ourselves to equations that can be solved just by investigating the special values that we have memorized in relation to the unit circle. If, on the other hand, we had been asked to find all angles with sin(θ) = 1 or solve tan(t) = for real numbers t, we would have been hard-pressed to do so. With the introduction of the inverse trigonometric functions, however, we are now in a position to solve these equations. Example 69 Find all angles θ for which sin(θ) = 1. If sin(θ) = 1, then the terminal side of θ, when plotted in standard position, intersects the Unit Circle at y = 1. Geometrically, we see that this happens at two places: in Quadrant I and Quadrant II.

195 5.6 Trig Equations 187 If we let α denote the acute solution to the equation, then all the solutions to this equation in Quadrant I are coterminal with α, and α serves as the reference angle for all of the solutions to this equation in Quadrant II as seen below. center img src = assets/img/inv/solveex.jpg alt = a reference angle in Qi and QII width = 80 /center Since 1 isn t the sine of any of the common angles we ve encountered, we use the arcsine functions to express our answers. By definition, real number t = arcsin ( ) 1 sin(t) = 1 with 0 < t < π. Hence, α = arcsin ( ) 1 radians is an acute angle with sin(α) = 1. Since all of the Quadrant I solutions θ are all coterminal with α, we get θ = α + πk = arcsin ( 1 ) + πk for integers k. Turning our attention to Quadrant II, we get one solution to be π α. Hence, the Quadrant II solutions are θ = π α + πk = π arcsin ( 1 ) + πk, for integers k. Example 70 Find all real numbers t for which tan(t) =. The real number solutions to tan(t) = correspond to angles θ with tan(θ) =. Since tangent is negative only in Quadrants II and IV, we focus our efforts there. The real number t = arctan( ) satisfies tan(t) = and π < t < 0. If we let β = arctan( ) radians, then all of the Quadrant IV solutions to tan(θ) = are coterminal with β. Moreover, the solutions from Quadrant II differ by exactly π units from the solutions in Quadrant IV (recall, the period of the tangent function is π.) Hence, all of the solutions to tan(θ) = are of the form θ = β + πk = arctan( ) + πk for some integer k. Switching back to the variable t, we record our final answer to tan(t) = as t = arctan( ) + πk for integers k.

196 188 The Circular Functions Example 71 Solve the following equation on [0, π). sin(x) = 0.87 To solve sin(x) = 0.87, we first note that it has the form sin(u) = 0.87, which has the family of solutions u = arcsin(0.87) + πk or u = π arcsin(0.87) + πk for integers k. Since the argument of sine here is x, we get x = arcsin(0.87)+πk or x = π arcsin(0.87)+πk which gives x = 1 arcsin(0.87) + πk or x = π 1 arcsin(0.87) + πk for integers k. To determine which of these solutions lie in [0, π), we first need to get an idea of the value of x = 1 arcsin(0.87). Once again, we could use the calculator, but we adopt an analytic route here. By definition, 0 < arcsin(0.87) < π so that multiplying through by 1 gives us 0 < 1 arcsin(0.87) < π 4. Starting with the family of solutions x = 1 arcsin(0.87) + πk, we use the same kind of arguments as in our solution to number?? above and find only the solutions corresponding to k = 0 and k = 1 lie in [0, π): x = 1 arcsin(0.87) and x = 1 arcsin(0.87) + π. Next, we move to the family x = π 1 arcsin(0.87) + πk for integers k. Here, we need to get a better estimate of π 1 arcsin(0.87). From the inequality 0 < 1 arcsin(0.87) < π, we first multiply 4 through by 1 and then add π to get π > π 1 arcsin(0.87) > π 4, or π < π 1 4 arcsin(0.87) < π. Proceeding with the usual arguments, we find the only solutions which lie in [0, π) correspond to k = 0 and k = 1, namely x = π 1 arcsin(0.87) and x = π 1 arcsin(0.87). All told, we have found four solutions to sin(x) = 0.87 in [0, π): x = 1 arcsin(0.87) 0.58, x = 1 arcsin(0.87)+π.669, x = π 1 arcsin(0.87) 1.04 and x = π 1 arcsin(0.87) Solving Basic Trigonometric Equations where More Algebra is Needed Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as x or u. If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric

197 5.6 Trig Equations 189 equations. Example 7 Solve exactly on [0, π) cos (θ) + cos(θ) = 0 The first thing to notice is that we have a cosine squared and a cosine in this equation. That means we can do a quick substitution to make this look like a quadratic equation. Let u = cos(θ); then u = cos (θ) and our equation becomes u + u = 0. To solve, factor and use the Zero Product Property: (u + )(u 1) = 0 u + = 0 or u 1 = 0 u = or u = 1 Now, we replace u with the original cos(θ) and finish solving for θ cos(θ) = or cos(θ) = 1 Recall that the range of cosine is [ 1, 1], so the equation cos(θ) = is extraneous; it can t have a solution!. By inspection, we know that cosine is 1 when θ = 0, so our final solution is θ = 0. Example 7 Solve exactly on [0, π) sec (t) = 4

198 190 The Circular Functions Once again, we can view this equation as a quadratic if we let u = sec(t): u = 4 To solve for u in this case, we can take the square root, but keep in mind, that means we end up with two solutions: u = or u =. Subsitution our secants back in, we have and putting this in terms of cosine we find that sec(t) = or sec(t) = cos(t) = 1 or cos(t) = 1. Now we solve by inspection; t = π, π, 4π, 5π. the common angles which have a cosine of 1 or 1 are Example 74 Solve exactly on [0, π) sin (θ) = sin (θ) One approach to solving sin (θ) = sin (θ) begins with dividing both sides by sin (θ). Doing so, however, assumes that sin (θ) 0 strong which means we risk losing solutions. /strong This brings us to a nice general principle: center h If you are tempted to divide by a variable to solve an equation, try factoring and the Zero Product Property instead! /h /center Let s once again make a substitution and try to solve this like a regular polynomial equation. Let u = sin(θ). u = u u u = 0

199 5.6 Trig Equations 191 u (u 1) = 0 Now we apply the Zero Product Property! u = 0 or u 1 = 0 u = 0, 1 Since u = sin(θ), sin(θ) = 0 or sin(θ) = 1. The solution to sin(θ) = 0 is θ = 0, π, for the two solutions which lie in [0, π). To solve sin(θ) = 1, we use the arcsine function to get θ = arcsin ( ( 1 ) +πk or θ = π arcsin 1 ) +πk for integers k. We find the two solutions here which lie in [0, π) to be θ = arcsin ( 1 ) and θ = π arcsin ( 1 ). Check Point 55 Solve tan (x) tan(x) = 0, 0 x < π. Check Point 56 Solve cos (x) 1 = 0, 0 x < π. Check Point 57 Solve sin (x) + sin(x) + 1 = 0, 0 x < π.

200 19 The Circular Functions Exercises In the following exercises, find all of the angles which satisfy the given equation. 1. sin(θ) = 1. cos(θ) =. sin(θ) = 0 4. cos(θ) = 5. sin(θ) = 6. cos(θ) = 1 7. sin(θ) = 1 8. cos(θ) = 9. cos(θ) = tan(θ) = 11. sec(θ) = 1. csc(θ) = 1 1. cot(θ) = 14. tan(θ) = sec(θ) = csc(θ) = 17. cot(θ) = tan(θ) = sec(θ) = 0 0. csc(θ) = 1 1. sec(θ) = 1. tan(θ) =. csc(θ) = 4. cot(θ) = 1 In the following exercises, solve on [0, π). 5. sin (5θ) = 0 6. cos (t) = 1 7. sin ( x) = 8. tan (6θ) = 1 9. csc (4t) = 1 0. sec (x) = 1. cot (θ) =. cos (9t) = 9 ( x ). sin = ( 4. cos θ + 5π ) = 0 6 ( 5. sin t π ) = 1 ( ) 6. cos x + 7π = 4 7. csc(θ) = 0 8. tan (t π) = 1 9. tan (x) = 40. sec (θ) = cos (t) = 1 4. sin (x) = 4 Solve the equation over all real numbers, then approximate the solutions which lie in the interval [0, π) to four decimal places.

201 5.6 Trig Equations sin(θ) = cos(θ) = sin(θ) = cos(θ) = sin(θ) = cos(θ) = tan(t) = cot(t) = sec(t) = 5. csc(t) = tan(t) = sin(t) = cos(x) = tan(x) = sin(x) = 0.50

202 194 The Circular Functions For the following exercises, find all exact solutions on [0, π). 58. sec(x) sin(x) sin(x) = tan(x) sin(x) tan(x) = cos (t) + cos(t) = sin(x) cos(x) sin(x) + cos(x) 1 = 0 6. tan (x) = 1 tan(x) 6. 8 sin (x) + 6 sin(x) + 1 = tan 5 (x) = tan(x) 65. tan (x) tan(x) = sin (x) + sin(x) = sin (x) sin(x) 4 = cos (x) + cos(x) 1 = cos (x) cos(x) = sin (x) + sin(x) 1 = tan (x) + 5 tan(x) 1 = 0

203 Chapter 6 Analytical Trigonometry 6.1 The Pythagorean Identities Learning Objectives In this section you will: Verify the fundamental trigonometric identities. Simplify trigonometric expressions using algebra and the identities. Introduction In espionage movies, we see international spies with multiple passports, each claiming a different identity. However, we know that each of those passports represents the same person. The trigonometric identities act in a similar manner to multiple passports, there are many ways to represent the same trigonometric expression. Just as a spy will choose an Italian passport when traveling to Italy, we choose the identity that applies to the given scenario when solving a trigonometric equation. In this section, we will begin an examination of the fundamental trigonometric identities, including how we can verify them and how we can use them to simplify trigonometric expressions.

204 196 Analytical Trigonometry The Reciprocal and Quotient Trigonometric Identities We ll start with the familiar: the reciprocal identities. We encountered these way back in Section 4.1 when we first learned right triangle trigonometry. Definition 50: The Reciprocal Identities csc(θ) = sin(θ) = sec(θ) = sec(θ) = cot(θ) = tan(θ) = 1 sin(θ) 1 csc(θ) 1 cos(θ) 1 cos(θ) 1 tan(θ) 1 cot(θ) Example 75 Show that sin(θ) = 1 csc(θ). We can work from the definition of csc(θ), which is csc(θ) = 1 sin(θ) : Multiply by sine: sin(θ) csc(θ) = 1 Divide by csc(θ): 1 sin(θ) = csc(θ) This is where all the identities on the right hand column of Definition 1 are coming from.

205 6.1 The Pythagorean Identities 197 When we learned about the unit circle, we also discovered a different way of defining tangent and cotangent; these are known as the quotient identities. Definition 51: The Quotient Identities tan(θ) = sin(θ) cos(θ) cot(θ) = cos(θ) sin(θ) The Pythagorean Identities Many other identities can be discovered by using the physical properties of the Unit Circle. To build up our Unit Circle, we first thought about embedded right triangles within it: When we first built the unit circle, we found that x = cos(θ) and y = sin(θ). Now, consider the implications for the Pythagorean Theorem: x + y = 1 (cos(θ)) + (sin(θ)) = 1 cos (θ) + sin (θ) = 1

206 198 Analytical Trigonometry The magic is, this is a relationship that holds true no matter the value of θ! Pythagorean Identity. Using This is known as the cos (θ) + sin (θ) = 1 as a starting point, we can then come up with a couple of versions of the Pythagorean Identity. Example 76 (cos (θ)) + (sin (θ)) = 1 cos (θ) cos (θ) cos (θ) Divide cos (θ) from both sides of the equation. 1 + sin (θ) cos (θ) = 1 cos (θ) Now we consider our reciprocal and quotient identities. Since sin(θ) cos(θ) = tan(θ) we can square both sides of this identity to see that Similarly, sin (θ) cos (θ) = tan (θ). 1 cos (θ) = sec (θ). Therefore, our new version of the Pythagorean Identity is 1 + tan (θ) = sec (θ)

207 6.1 The Pythagorean Identities 199 Check Point 58 Create another version of the Pythagorean identity by dividing cos (θ) + sin (θ) = 1 by sin (θ) (cos (θ)) + (sin (θ)) = 1 sin (θ) sin (θ) sin (θ) Divide sin (θ) from both sides of the equation. cos (θ) sin (θ) + 1 = 1 sin (θ) cot (θ) + 1 = csc (θ) In summary, here are all three of the Pythagorean Identities. They are worth memorizing, although if you know the first, you can always recreate the latter two using the methods from above. Definition 5: The Pythagorean Identities For all θ in the domain... cos (θ) + sin (θ) = tan (θ) = sec (θ) cot (θ) + 1 = csc (θ)

208 00 Analytical Trigonometry The Even/Odd Identities A function is called even if it has y-axis symmetry; a function is called odd if it has origin symmetry. In Chapter 0, we learned algebraic ways of detecting this type of symmetry: Definition 5: Testing the Graph of an Equation for Symmetry To test the graph of an equation for symmetry About the y-axis: Substitute ( x, y) into the equation and simplify. If the result is equivalent to the original equation, the graph is symmetric about the y-axis. About the origin: Substitute ( x, y) into the equation and simplify. If the result is equivalent to the original equation, the graph is symmetric about the origin. We can modify these notions to think about functions instead of merely equations. For instance, for y-axis symmetry, let y = f (x). So if we substitute x into our function, we want the result to be equivalent to the original function; that is f ( x) = f (x). If a function has this property, we will call it even. Similarly, for origin symmetry, let y = f (x). If we substitute in x, we should get back y = f (x); that is, f ( x) = f (x). If a function has this property, we will call it odd. Definition 54: Even and Odd Functions A function is even if its graph has y-axis symmetry. Then the function will also have the property that f ( x) = f (x). A function is odd if its graph has origin symmetry. Then the function will also have the property that f ( x) = f (x).

209 6.1 The Pythagorean Identities 01 Now, consider our trigonometric functions. The graph of f (x) = cos(x) is symmetric about the y-axis, so it is even. More relevant to us now, that means that cos( x) = cos(x). The graph of f (x) = sin(x) is symmetric about the origin, so it is odd and sin( x) = sin(x). From there, we can derive the rest of the even/odd identities. Definition 55: The Even/Odd Identities cos(θ) and sec(θ) are even, so for all θ in the domain... cos( θ) = cos(θ) sec( θ) = sec(θ). The rest of the trigonometric functions are odd, hence for all θ in the domain... sin( θ) = sin(θ) csc( θ) = csc(θ) tan( θ) = tan(θ) cot( θ) = cot(θ) One of the most important things that we will use these identities for is simplification. As you move on to other math class (like Calculus in particular), the goals are much less solve for x and much more manipulate this expression so it s more convenient. When we simplify trig expressions, we re generally trying to write the whole thing in terms of a single trigonometric function. There s not always a single path forward to this goal; a good way to practice your simplification skills is to try to approach a particular problem from a few different points of view. The best way to get good at this skill is through experience and practice!

210 0 Analytical Trigonometry Example 77 Simplify the expression sin(x) cos( x) sec(x) First, we want everything to be in terms of the same angle. Currently, sine and secant have a positive x input, but cosine has x. We apply the even identity for cosine: Expression Reasoning sin(x)cos( x) sec(x) = sin(x)cos(x) sec(x) cos(x) is even, so cos( x) = cos(x) Now, we have a mix of different trigonometric funtions. In such cases, it is often a good idea to convert everything to sines and cosines. Here we can use the reciprocal identities to help: Expression Reasoning = sin(x) cos(x)sec(x) Finally, we use algebra to simplify: 1 = sin(x) cos(x) Reciprocal identity: sec(x) = cos(x) 1 cos(x) Expression Reasoning = sin(x)cos(x) 1 cos(x) = sin(x) cos(x) cos(x) Algebra = sin(x) 1 Algebra = sin(x)

211 6.1 The Pythagorean Identities 0 Therefore, the expression sin(x) cos( x) sec(x) simplifies to sin(x). Example 78 Simplify the expression cot(t) + tan(t) sec( t) Expression Reasoning cot(t)+tan(t) sec( t) = cot(t)+tan(t) sec(t) Secant is even, so sec( t) = sec(t) = cos(t) sin(t) + sin(t) cos(t) sec(t) Quotient identities for tangent and cotangent = = cos(t) sin(t) + sin(t) cos(t) 1 cos(t) Reciprocal identity for sec(t) ( ) cos(t) sin(t) + sin(t) cos(t) Fraction division (multiply by the reciprocal of the denominator) cos(t) 1 = cos(t) sin(t) cos(t) + sin(t) cos(t) cos(t) Distribution = cos (t) sin(t) + sin(t) cos(t) cos(t) Simplification = cos (t) sin(t) + sin(t)

212 04 Analytical Trigonometry At this point, we may feel stuck for a moment. To get un-stuck, recall that to add fractions, we need a common denominator... Expression Reasoning = cos (t) sin(t) + sin(t) = cos (t) sin(t) + sin(t) sin(t) sin(t) Multiply by a fancy form of 1 ( sin(t) ) to create a common denominator sin(t) = cos (t) sin(t) + sin (t) sin(t) = cos (t)+sin (t) sin(t) Fraction addition = 1 sin(t) Pythagorean Identity = csc(t) Reciprocal Identity for csc(t) Hence, cot(t)+tan(t) sec( t) simplifies to csc(t).

213 6.1 The Pythagorean Identities 05 Check Point 59 Simplify the following: tan( x) cos( x) csc(x) cot(x) cot(x) cos(x) + csc(x) sin (x) sin( x) cos(x) sec(x) csc(x) tan(x) cot(x) Using Algebra and Identities to Simplify Trigonometric Expressions We have seen that algebra is very important in verifying trigonometric identities, but it is just as critical in simplifying trigonometric expressions before solving. Being familiar with the basic properties and formulas of algebra, such as the difference of squares formula, the perfect square formula, or substitution, will simplify the work involved with trigonometric expressions and equations. Conider, for example, the difference of squares formula, a b = (a + b)(a + b). We often make use of this formula when simplifying expressions like cos 4 (x) 1, which can be factored as (cos (x) + 1)(cos (x) 1). We can use many such factoring strategies when simplifying trigonometric expressions.

214 06 Analytical Trigonometry Example 79 Simplify the expression sin 4 (x) 1 sin (x) + 1 Expression Reasoning sin 4 (x) 1 sin (x)+1 = (sin (x)+1)(sin (x) 1) sin (x)+1 (sin (x)+1)(sin (x) 1) = sin (x)+1 Factoring using the difference of squares pattern Cancel common factors Now, when we see a squared trig function and a 1, that should put the Pythagorean identities in mind. Consider the first identity; since it is true for all values of θ in the domain, we can rearrange it as we please: cos (x) + sin (x) = 1 cos (x) cos (x) sin (x) = 1 cos (x) 1 1 sin (x) 1 = cos (x) Hence our expression above, sin (x) 1 can be simplified using the Pythagorean Identity to cos (x). We could have approached this more algebraically as well. Since the Pythagorean Identity

215 6.1 The Pythagorean Identities 07 states the cos (x) + sin (x) = 1, we could replace the 1 in our expression with cos (x) + sin (x): Expression = sin (x) 1 Reasoning = sin (x) (cos (x) + sin (x)) Pythagorean Identity = sin (x) cos (x) sin (x) Distribute the negative = sin (x) cos (x) sin (x) sin (x) sin (x) is 0 = cos (x) Either way, sin4 (x) 1 sin (x)+1 simplifies to cos (x). Check Point 60 Simplify the following: 1 cot 4 (x) 1 cot (x) Verifying Trigonometric Identities In verifying identities, we typically start with the more complicated side of the equation and use known identities to transform it into the other side of the equation. To be mathematically rigorous, we must commit to using only one side. Strategies such as multiplying by the same quantity on both sides only work in the context of solving an equation, and rely on the assumption that the two sides of the equation will remain balanced. Here our goal is different: we are trying to prove the two sides of the equation are exactly the same for all values of the variable in the domain. So, overall we can think of this process as simplifying towards a specific goal. div id = examplebox Verify the identity. Assume that all quantities are defined tan(θ) = sin(θ) sec(θ)

216 08 Analytical Trigonometry Let s work from the right hand side: tan(θ) = sin(θ)sec(θ) = sin(θ) 1 cos(θ) Reciprocal identity for sec(θ) = sin(θ) cos(θ) = tan(θ) Quotient Identity for tan(θ) Hence we have shown that the left hand side is equal to the right hand side, so we have verified the identity holds for all values of θ. Example 80 Verify the identity. Assume that all quantities are defined (tan(t) sec(t))(tan(t) + sec(t)) = 1 Here, we ll work from the left hand side: (tan(t) sec(t))(tan(t) + sec(t)) = 1 tan (t) sec (t) Expand (using a technique like FOIL) tan (t) (1 + tan (t)) Pythagorean Identity (1 + tan (t) = sec (t)) tan (t) 1 tan (t) Distribute the negative 1 tan (t) tan (t) = 0 Hence we have shown that the left hand side is equal to the right hand side, so we have verified the identity holds for all values of t.

217 6.1 The Pythagorean Identities 09 Example 81 Verify the identity. Assume that all quantities are defined sec(t) 1 tan(t) = 1 cos(t) sin(t) Here, we ll work from the left hand side. Since the right hand side is in terms of sines and cosines, we ll first rewrite the left hand side in terms of those functions as well: 1 cos(t) 1 sin(t) cos(t) 1 cos(t) 1 sin(t) cos(t) cos(t) cos(t) 1 cos(t) cos(t) ( ) sin(t) (1)(cos(t)) (cos(t)) cos(t) Multiply by cos(t) to clear away a layer of the fractions. cos(t) Distribute 1 cos(t) sin(t) Simplify Hence we have shown that the left hand side is equal to the right hand side, so we have verified the identity holds for all values of t.

218 10 Analytical Trigonometry Example 8 Verify the identity. Assume that all quantities are defined sin(θ) 1 cos(θ) = 1 + cos(θ) sin(θ) It is debatable which side of the identity is more complicated. One thing which stands out is that the denominator on the left hand side is 1 cos(θ), while the numerator of the right hand side is 1+cos(θ). This suggests the strategy of starting with the left hand side and multiplying the numerator and denominator by the quantity 1 + cos(θ). The Pythagorean Identity comes to our aid once more when we rearrange it to reveal 1 cos (θ) = sin (θ): sin(θ) 1 cos(θ) = sin(θ) (1 cos(θ)) (1 + cos(θ)) (1 + cos(θ)) = sin(θ)(1 + cos(θ)) (1 cos(θ))(1 + cos(θ)) = sin(θ)(1 + cos(θ)) 1 cos (θ) = sin(θ)(1 + cos(θ)) sin (θ) = sin(θ)(1 + cos(θ)) = 1 + cos(θ) sin(θ) sin(θ) sin(θ) In the example above, we see that multiplying 1 cos(θ) by 1 + cos(θ) produces a difference of squares that can be simplified to one term using the Pythagorean Identity. This is exactly the same kind of phenomenon that occurs when we multiply expressions such as 1 by 1 + or 4i by + 4i. In algebra, these sorts of expressions were called conjugates. Verifying trigonometric identities requires a healthy mix of tenacity and inspiration. You will need to spend many hours struggling with them just to become proficient in the basics. Like many things in life, there is no short-cut here there is no complete algorithm for verifying identities. Nevertheless, a summary of some strategies which may be helpful (depending on the situation) is provided below and ample practice is provided for you in the Exercises.

219 6.1 The Pythagorean Identities 11 Definition 56: Strategies for Verifying Identities Try working on the more complicated side of the identity. Use the Reciprocal and Quotient Identities in Theorem to write functions on one side of the identity in terms of the functions on the other side of the identity. Simplify the resulting complex fractions. Add rational expressions with unlike denominators by obtaining common denominators. Use the Pythagorean Identities in Theorem to exchange sines and cosines, secants and tangents, cosecants and cotangents, and simplify sums or differences of squares to one term. Multiply numerator and denominator by Pythagorean Conjugates in order to take advantage of the Pythagorean Identities. If you find yourself stuck working with one side of the identity, try starting with the other side of the identity and see if you can find a way to bridge the two parts of your work. Try something. Do algebra and have faith, even if you can t see the path forward. The more you work with identities, the better you ll get with identities.

220 1 Analytical Trigonometry Exercises In the following exercises, verify each identity. Assume that all quantities are defined. 1. cos(θ) sec(θ) = cot(t) csc (t) 1 = tan(t). tan(t) cos(t) = sin(t) cos (θ) + 4 sin (θ) = 4. sin(θ) csc(θ) = cos (t) sin (t) = 8 4. tan(t) cot(t) = tan (t) = tan(t) sec (t) tan(t) 5. csc(x) cos(x) = cot(x) 19. sin 5 (x) = ( 1 cos (x) ) sin(x) 6. sin(t) cos (t) = sec(t) tan(t) 0. sec 10 (t) = ( 1 + tan (t) ) 4 sec (t) 7. cos(θ) sin (θ) = csc(θ) cot(θ) 1. cos (x) tan (x) = tan(x) sin(x) cos(x) sin(x) cos(x) = sec(x) + tan(x). sec 4 (t) sec (t) = tan (t) + tan 4 (t) 9. 1 cos(θ) sin(θ) = csc(θ) cot(θ). cos(θ) + 1 cos(θ) 1 = 1 + sec(θ) 1 sec(θ) 10. cos(t) 1 sin (t) = sec(t) 4. sin(t) + 1 sin(t) 1 = 1 + csc(t) 1 csc(t) 11. sin(x) 1 cos (x) = csc(x) 5. 1 cot(x) 1 + cot(x) = tan(x) 1 tan(x) sec(t) 1 + tan (t) = cos(t) 6. 1 tan(t) 1 + tan(t) = cos(t) sin(t) cos(t) + sin(t) csc(θ) 1 + cot (θ) = sin(θ) tan(x) sec (x) 1 = cot(x) 7. tan(θ) + cot(θ) = sec(θ) csc(θ) 8. csc(t) sin(t) = cot(t) cos(t)

221 6.1 The Pythagorean Identities 1 9. cos(x) sec(x) = tan(x) sin(x) 0. cos(x)(tan(x) + cot(x)) = csc(x) csc(t) cot(t) 1 csc(t) + cot(t) = cot(t) cos(θ) 1 tan(θ) + sin(θ) = sin(θ) + cos(θ) 1 cot(θ) 1. sin(t)(tan(t) + cot(t)) = sec(t) 7. 1 sec(t) + tan(t) = sec(t) tan(t). 1 1 cos(θ) cos(θ) = csc (θ) 8. 1 sec(x) tan(x) = sec(x) + tan(x). 1 sec(t) = csc(t) cot(t) sec(t) cos(θ) = csc (θ) + csc(θ) cot(θ) 4. 1 csc(x) = sec(x) tan(x) csc(x) cos(x) = csc (x) csc(x) cot(x)

222 14 Analytical Trigonometry 6. The Sum and Difference Identities Learning Objectives In this section you will: Derive the sum and difference identities Use the sum/difference identities to evaluate trig functions Use the sum/difference identities to simplify expressions Use the sum/difference identities to verify identities The trigonometric identities we will examine in this section can be traced to a Persian astronomer who lived around 950 AD, but the ancient Greeks discovered these same formulas much earlier and stated them in terms of chords. These are special equations or postulates, true for all values input to the equations, and with innumerable applications. Deriving the Sum and Difference Identities Our first task is to find an identity for expressions like cos(α + β) and cos(α β). Be careful when thinking about this; remember, trig expressions are a form of function notation, not algebra, so we can t just distribute the cosine!! We ll start with cos(α β). To help understand what we re dealing with, we ll arbitrarily decide that α > β, and draw each angle in standard position on the unit circle.

223 6. The Sum and Difference Identities 15 We want to try to create an identity for the difference between these two angles, α β. To help develop the identity, we ll trace that difference in red, and look at the distance between the endpoints where it intersects the unit circle. Before we move on, let s use the difference formula to describe d. Since we re drawing angles α and β on the unit circle, we know that the coordinates where their terminal sides intersect the circle are (cos(α), sin(α)) and (cos(β), sin(β)), respectively, so we can use these as our x- and y-values in the distance formula. = (y y 1 ) + (x x 1 ) d = (cos(α) cos(β)) + (sin(α) sin(β)) = cos (α) cos(α) cos(β) + cos (β) + sin (α) sin(α) sin(β) + sin (β) You may notice some identities lurking in that formula. We can apply the Pythagorean identity to help simplify. = cos (α) + sin (α) + cos (β) + sin (β) cos(α) cos(β) sin(α) sin(β) d = cos(α) cos(β) sin(α) sin(β) That s all we can do now, so it s time for a change in perspective. Right now, the way we re picturing it, our

224 16 Analytical Trigonometry α β angle is floating out in the coordinate plane; we can rotate it around so that it s in standard position. We ll temporarily name α β as γ. Then, the coordinates where γ s terminal side intersects the unit circle are (cos(γ), sin(γ)), and the coordinates where its intial side intersect the unit circle along the positive x-axis are (1, 0). From this perspective, then, we can write another formula for the length of d as follows. d = (cos(γ) 1) + (sin(γ) 0) d = cos (γ) cos(γ) sin (γ) Again, we can use the Pythagorean Theorem to simplify. d = cos (γ) + sin (γ) + 1 cos(γ) d = cos(γ) Now we have two expressions both describing the same distance d, marked with * and ** above. We can set these equal to each other and simplify as follows. cos(α) cos(β) sin(α) sin(β) = cos(γ) cos(α) cos(β) sin(α) sin(β) = cos(γ) Square both sides = Subtract from both sides cos(α) cos(β) sin(α) sin(β) = cos(γ) = Divide all terms by - cos(α) cos(β) + sin(α) sin(β) = cos(γ)

225 6. The Sum and Difference Identities 17 All that is left is for us to recall that γ is actually α β, and now we have our first identity of the section: cos(α β) = cos(α) cos(β) + sin(α) sin(β) We can turn this into an identity for cos(α + β) with a clever rewrite and our knowledge that cosine is even and therefore cos( x) = cos(x), and sine is odd and therefore sin( x) = sin(x). cos(α + β) = cos(α ( β)) Rewrite as a difference = cos(α) cos( β) + sin(α) sin( β) Apply the cosine difference identity = cos(α) cos(β) sin(α) sin(β) Apply the even/odd identities First, we ll take note of these identities, then we ll see some ways that we can apply them. Definition 57: Cosine sum and difference identities Cosine sum identity cos(α + β) = cos(α) cos(β) sin(α) sin(β) for all angles α and β. Cosine difference identity cos(α β) = cos(α) cos(β) + sin(α) sin(β) for all angles α and β.

226 18 Analytical Trigonometry Example 8 Find the exact value of cos ( ) π 1 In order to use sum and difference identities to find cos (15 ), we need to write π as a sum or 1 difference of angles whose cosines and sines we know. One way to do so is to write π 1 = π π 1 1 =. We find: π 4 π 6 cos ( ( π 1) = cos π π ) 4 6 Rewrite the angle in terms of special angles = cos ( π 4 ) cos ( π 6 ) + sin ( π 4 ) sin ( π 6 ) apply the cosine difference identity ( ) ( ) ( ) (1 ) = + evaluate the sines and cosines using special unit circle values = simplify. Check Point 61 Use the sum and difference identities to find the exact value. You may have need of the quotient, reciprocal or even/odd identities as well. ( ) π cos 1

227 6. The Sum and Difference Identities 19 Example 84 Suppose α is a Quadrant I angle with sin(α) = 5 Find the exact value of cos(α + β). and β is a Quadrant IV angle with sec(β) = 4. We know cos(α + β) = cos(α) cos(β) sin(α) sin(β). cosines of α and β to complete the problem. We are given sin(α) = 5 Pythagorean Identity cos (α) = 1 sin (α) = 1 ( 5 positive root since α is a Quadrant I angle. Hence, we need to find the sines and, so our first task is to find cos(α). We can quickly get cos(α) using the ) = We get cos(α) = 4, choosing the 5 Next, we need the sin(β) and cos(β). Since sec(β) = 4, we immediately get cos(β) = 1 4 the Reciprocal and Quotient Identities. courtesy of To get sin(β), we employ the Pythagorean Identity: sin (β) = 1 cos (β) = 1 ( 1 since β is a Quadrant IV angle, we get sin(β) = ) = Here, Finally, we get: cos(α + β) = cos(α) cos(β) sin(α) sin(β) = ( ( 4 1 ( 5) 4) 5) ( ) 15 = Check Point 6 Find the exact value of cos(α β) if cos(α) = 1 5 QII and the terminal side of β lies in QI. and cos(β) = and the terminal side of α lies in 10

228 0 Analytical Trigonometry Example 85 Verify the identity cos ( π θ) = sin(θ). Apply the cosine difference identity: ( π ) cos θ ( π ) ( π ) = cos cos (θ) + sin sin (θ) = (0) (cos(θ)) + (1) (sin(θ)) = sin(θ). The Cofunction Identities The identity verified above, namely, cos ( π θ) = sin(θ), is the first of the celebrated cofunction identities. From sin(θ) = cos ( π θ), we get: sin ( π θ) = cos ( π [ π θ]) = cos(θ), which says, in words, that the co sine of an angle is the sine of its co mplement. Now that these identities have been established for cosine and sine, the remaining circular functions follow suit. The remaining proofs are left as exercises. Definition 58: The Cofunction Identities For all applicable angles θ: ( π ) cos θ = sin(θ) ( π ) csc θ = sec(θ) ( π ) sin θ = cos(θ) ( π ) tan θ = cot(θ) ( π ) sec θ = csc(θ) ( π ) cot θ = tan(θ)

229 6. The Sum and Difference Identities 1 The Sine Sum and Difference Identities The Cofunction Identities enable us to derive the sum and difference formulas for sine. We first convert to sine to cosine and expand: ( π ) sin(α + β) = cos (α + β) ([ π ] ) = cos α β ( π ) ( π ) = cos α cos(β) + sin α sin(β) = sin(α) cos(β) + cos(α) sin(β) We can derive the difference formula for sine by rewriting sin(α β) as sin(α + ( β)) and using the sum formula and the Even / Odd Identities. Again, we leave the details to the reader. Definition 59: Sine sum and difference identities Sine sum identity: sin(α + β) = sin(α) cos(β) + cos(α) sin(β) for all angles α and β. Sine difference identity sin(α β) = sin(α) cos(β) cos(α) sin(β) for all angles α and β. We try out these new identities in the next examples.

230 Analytical Trigonometry Example 86 Find the exact value of sin ( ) 19π 1. As in before, we need to write the angle 19π as a sum or difference of common angles. The denominator of 1 suggests a combination of angles with denominators and 4. One such combination is 1 19π 1 = 4π + π. Applying the sine sum identity, we get 4 ( ) ( 19π 4π sin = sin 1 + π ) 4 ( ) 4π ( π ) ( 4π = sin cos + cos 4 = ( ) ( ) + ( 1 ) ) ( ) ( π ) sin 4 = 6 4 Check Point 6 Use the sum and difference identities to find the exact value. You may have need of the quotient, reciprocal or even/odd identities as well. ( ) π cos = 1 Example 87 Suppose α is a Quadrant II angle with sin(α) = 5, and β is a Quadrant III angle with tan(β) =. 1 Find the exact value of sin(α β). In order to find sin(α β) using the sine difference identity, we need to find cos(α) and both cos(β)

231 6. The Sum and Difference Identities and sin(β). To find cos(α), we use the Pythagorean Identity cos (α) = 1 sin (α) = 1 ( ) 5 1 = 144. We get 169 cos(α) = 1, the negative, here, owing to the fact that α is a Quadrant II angle. 1 We now set about finding sin(β) and cos(β). We have several ways to proceed at this point, but since there isn t a direct way to get from tan(β) = to either sin(β) or cos(β), we opt for a more geometric approach. Since β is a Quadrant III angle with tan(β) = =, we know the point Q(x, y) = ( 1, ) is on 1 the terminal side of β as illustrated below. Note that even though tan(β) = sin(β), we cannot take cos(β) sin(β) = and cos(β) = 1. Recall that sin(β) and cos(β) are the y and x coordinates on a specific circle, the Unit Circle. As we ll see shortly, ( 1, ) lies on a circle of 5, so not the Unit Circle. We find r = x + y = ( 1) + ( ) = 5, so sin(β) = 5 = 5 5 and cos(β) = 1 5 = 5 5.

232 4 Analytical Trigonometry At last, we have sin(α β) = sin(α) cos(β) cos(α) sin(β) = ( 5 1) ( ) 5 5 ( 1 1 ) ( ) 5 5 = Check Point 64 Suppose that α and β are angles given as follows: sin(α) = 7 and α is an angle in quadrant I. 5 cos(β) = 1 and β is an angle in quadrant II. Compute the exact value of sin(α + β). 1

233 6. The Sum and Difference Identities 5 A Special Note About Tangents Most Trigonometry Books at this point will create a special set of sum and difference identities for tangents. However, to do this, at some point they end up multiplying by 1 cos(α) cos(β) 1 cos(α) cos(β). The result of that operation, which is often glossed over, is that the domain of the new expression changes, and we end up with an identity where the domains of the left side and the right side of the equation don t match. Therefore, the new expression isn t really an identity because it doesn t apply to all angles α and β! It is for this reason that in this text, we are choosing to not create new identities for tangent or cotangent sums and differences. We can survive using what we know about sine and cosine, and the quotient identities. In other words, we will settle for the following identities. tan(α + β) = sin(α + β) sin(α) cos(β) + cos(α) sin(β) = cos(α + β) cos(α) cos(β) sin(α) sin(β) tan(α β) = sin(α β) sin(α) cos(β) cos(α) sin(β) = cos(α β) cos(α) cos(β) + sin(α) sin(β) cot(α + β) = cos(α + β) sin(α + β) = cos(α) cos(β) sin(α) sin(β) sin(α) cos(β) + cos(α) sin(β) cot(α β) = cos(α β) sin(α β) = cos(α) cos(β) + sin(α) sin(β) sin(α) cos(β) cos(α) sin(β) Example 88 Suppose α is a Quadrant II angle with sin(α) = 5, and β is a Quadrant III angle with tan(β) =. 1 Find the exact value of tan(α β). From our work above, we know that sin(α) = 5 1 Then, since we have tan(α β) = 1, cos(α) = 1, sin(β) = 5 and cos(β) = sin(α β) sin(α) cos(β) cos(α) sin(β) = cos(α β) cos(α) cos(β) + sin(α) sin(β) tan(α β) = (5/1)( 5 5/5) ( 1/1)( 5/5) ( 1/1)( 5/5) + (5/1)( 5/5)

234 6 Analytical Trigonometry = Check Point 65 Find the exact value of tan(α β) if sin(α) = 1 5 and cos(β) = 8, and the terminal side of α lies in QIII and the terminal side of β lies in QII.

235 6. The Sum and Difference Identities 7 Using the Sum and Difference Identities to Simplify Expressions and Verify Identities One of the main applications of these identities in other math classes (like Calculus) is their usefulness in rearranging mathematical statements. To practice this skill, here we ll practice both simplifying expressions and verifying identities. Example 89 Simplify the following expressions. a. sin(x) cos(7x) cos(x) sin(7x) One of the challenges of this chapter is training ourselves to recognize these identities in the wild. Here, we want to focus on the structure of the sines and cosines, and not necessarily the angles themselves yet. Notice that each term has a mix of sines and cosines. This makes it similar in structure to sin(α β) = sin(α) cos(β) cos(α) sin(β). Let α = x and β = 7x, and then we can condense this expression into a single sine: sin(x) cos(7x) cos(x) sin(7x) = sin(x 7x). We can do even better with a bit of algebra and the application of the fact that sine is odd. = sin( 4x) = sin(4x) b. sin(x) sin(7x) cos(x) cos(7x) At first glance, this looks similar in structure to the cosine sum identity: cos(α + β) = cos(α) cos(β) sin(α) sin(β). Notice, though, that the subtraction is backwards; in our expression, we re subtracting the cosines

236 8 Analytical Trigonometry from the sines, and we want it the other way around for our identity. We can use that favored, dirty algebra trick of old: factor out a negative to reverse the order of subtraction: sin(x) sin(7x) cos(x) cos(7x) = ( sin(x) sin(7x) + cos(x) cos(7x)) = (cos(x) cos(7x) sin(x) sin(7x)). Now we apply the identity cos(α + β) = cos(α) cos(β) sin(α) sin(β) with α = x and β = 7x and simplify. = cos(x + 7x) = cos(10x) Example 90 Simplify the following expression. tan(π x) tan(π x) = sin(π x) cos(π x) Quotient Identity = sin(π) cos(x) cos(π) sin(x) cos(π) cos(x) + sin(π) sin(x) Sum/Difference Identities = 0 cos(x) ( 1) sin(x) 1 cos(x) + 0 sin(x) Evaluate known values = sin(x) cos(x) Simplify = tan(x) Quotient Identity

237 6. The Sum and Difference Identities 9 Example 91 Write the following expression as a single cosine in the form F (t) = A cos(bt + C) + D. cos(t) sin(t) This is a new kind of challenge to us, and it turns out, this kind of skill is used in subjects like differential equations and physics! Let s start by teasing apart our goal cosine function, F (t) = A cos(bt + C) + D. Let s suppose that D = 0 for now, and let B =, since the expression we re given has a t in it. Then we apply the cosine sum identity to what s left: A cos(t + C) = A(cos(t) cos(c) sin(t) sin(c)) = A cos(c) cos(t) A sin(c) sin(t). To match this up with cos(t) sin(t), we want A cos(c) = 1 and A sin(c) =. With some algebra, we have cos(c) = 1 A and sin(c) =. We can use the Pythagorean Identity to help solve A for A: cos (C) + sin (C) = 1 ( ) ( ) 1 + = 1 A A 1 A + A = = A A = 4 A = ± We ll choose A to be, and then see what value of C that produces for us. Since A =, we now know that cos(c) = 1 and sin(c) =.

238 0 Analytical Trigonometry Aren t those familiar? We can let C = π to produce those sine and cosine values. Therefore, our final answer is ( F(t) = cos t + π ). We can check our answer using the sum formula for cosine : F (t) = cos ( t + π ) = [ cos(t) cos ( ) ( π sin(t) sin π )] [ = cos(t) ( ) ( )] 1 sin(t) = cos(t) sin(t). A couple of remarks about the previous example are in order. First, had we chosen A = instead of A = as we worked through it, our final answers would have looked different. The reader is encouraged to rework the example using A = to see what these differences are, and then for a challenging exercise, use identities to show that the formulas are all equivalent. Example 9 Verify the identity. sin(α + β) + sin(α β) = sin(α) cos(β) sin(α + β) + sin(α β) = sin(α) cos(β) + cos(α) sin(β) + sin(α) cos(β) cos(α) sin(β) Sine sum/difference identities = sin(α) cos(β) + cos(α) sin(β) + sin(α) cos(β) cos(α) sin(β) cancel = sin(α) cos(β) + sin(α) cos(β) = sin(α) cos(β) combine like terms

239 6. The Sum and Difference Identities 1 Example 9 Verify the identity. = tan(α) tan(β) sin(α β) cos(α) cos(β) sin(α β) cos(α) cos(β) = sin(α) cos(β) cos(α) sin(β) cos(α) cos(β) Sine sum identity = sin(α) cos(β) cos(α) sin(β) cos(α) cos(β) cos(α) cos(β) = sin(α) cos(β) cos(α) cos(α) sin(β) cos(α) cos(β) cos(β) Split up fraction subtraction Simplify = sin(α) cos(α) sin(β) cos(β) = tan(α) tan(β) Quotient identities

240 Analytical Trigonometry 6..1 Exercises In the following exercises, use the Sum and Difference Identities to find the exact value. You may have need of the Quotient, Reciprocal or Even / Odd Identities as well. 1. cos ( ) 1π 1. sin ( ) 11π 1. tan ( ) 1π 1 4.cos ( ) 7π 1 5. tan ( ) 17π 1 6. sin ( ) π 1 7. cot ( ) 11π 1 8. csc ( ) 5π 1 9. sec ( π 1 ) 10. If α is a Quadrant IV angle with cos(α) = 5 5, and sin(β) = 10 10, where π < β < π, find a. cos(α + β) b. sin(α + β) c. tan(α + β) d. cos(α β) e. sin(α β) f. tan(α β) 11. If csc(α) =, where 0 < α < π, and β is a Quadrant II angle with tan(β) = 7, find a. cos(α + β) b. sin(α + β) c. tan(α + β) d. cos(α β) e. sin(α β) f. tan(α β) 1. If sin(α) = 5, where 0 < α < π 1, and cos(β) = 1 where π < β < π, find a. sin(α + β) b. cos(α β) c. tan(α β) 1. If sec(α) = 5, where π < α < π, and tan(β) = 4 7, where π < β < π, find a. csc(α β) b. sec(α + β) c. cot(α + β)

241 6. The Sum and Difference Identities In the following exercises, show that the function is a sinusoid by rewriting it in the forms F(t) = A cos(bt + C) + D and G(t) + A sin(bt + C) + D for B > 0 and 0 C < π. 14. f (t) = sin(t) + cos(t) f (t) = sin(t) cos(t) 16. f (t) = sin(t) + cos(t) 17. f (t) = 1 sin(t) cos(t) 18. f (t) = cos(t) sin(t) 19. f (t) = cos(t) sin(t) f (t) = 1 cos(5t) sin(5t) 1. f (t) = 6 cos(t) 6 sin(t). f (t) = 5 sin(t) 5 cos(t). f (t) = sin ( ( t 6) cos t ) 6 In the following exercises, verify the identity. 4. sin ( θ + π ) = cos(t) 5. cos ( θ π ) = sin(t) 6. cos(θ π) = cos(θ) 7. sin(π θ) = sin(θ) 8. tan ( θ + π ) = cot(θ) 9. sin(α + β) + sin(α β) = sin(α) cos(β) 0. sin(α + β) sin(α β) = cos(α) sin(β) 1. cos(α + β) + cos(α β) = cos(α) cos(β). cos(α + β) cos(α β) = sin(α) sin(β) sin(α + β) 1 + cot(α) tan(β). = sin(α β) 1 cot(α) tan(β) cos(α + β) 1 tan(α) tan(β) 4. = cos(α β) 1 + tan(α) tan(β) tan(α + β) sin(α) cos(α) + sin(β) cos(β) 5. = tan(α β) sin(α) cos(α) ( sin(β) ) cos(β) ( ) sin(t + h) sin(t) sin(h) cos(h) 1 6. = cos(t) + sin(t) h ( h ) h( ) cos(t + h) cos(t) cos(h) 1 sin(h) 7. = cos(t) sin(t) h ( ) ( h ) h tan(t + h) tan(t) tan(h) sec (t) 8. = h h 1 tan(t) tan(h) 9. Is there only one way to evaluate cos ( ) 5π 4? Explain and give examples.

242 4 Analytical Trigonometry 6. Double and Half Angle Identities Learning Objectives In this section you will: Derive and Apply the Double Angle Identities Derive and Apply the Angle Reduction Identities Derive and Apply the Half Angle Identities The Double Angle Identities We ll dive right in and create our next set of identities, the double angle identities. All of these can be found by applying the sum identities from last section. Let s start with cosine. cos(θ) cos(θ) = cos(θ + θ) cos(θ) = cos(θ) cos(θ) sin(θ) sin(θ) cos(θ) = cos (θ) sin (θ) It s quite common to also use the Pythagorean identity to rewrite this in a couple of other ways. If we take cos (θ) + sin (θ) = 1 and solve for sin (θ), we have sin (θ) = 1 cos (θ). Substituting this into our double angle formula (marked with a *), we get cos(θ) = cos (θ) (1 cos (θ) cos(θ) = cos (θ) 1 + cos (θ) cos(θ) = cos (θ) 1. We can also rearrange the Pythagorean identity by solving for cos (θ) to get cos (θ) = 1 sin (θ). Then * becomes cos(θ) = 1 sin (θ) sin (θ) cos(θ) = 1 sin (θ).

243 6. Double and Half Angle Identities 5 The sine double angle identity has less variety. We again start with the sum identity to build this new double angle identity. sin(θ) sin(θ) = sin(θ + θ) sin(θ) = sin(θ) cos(θ) + cos(θ) sin(θ) sin(θ) = sin(θ) cos(θ) We summarize the identities here. Next we ll work on using them! Definition 60: The Double Angle Identities For all angles θ... cos (θ) sin (θ) cos(θ) = cos (θ) 1 1 sin (θ) sin(θ) = sin(θ) cos(θ) Example 94 Suppose P(, 4) lies on the terminal side of θ when θ is plotted in standard position. Find cos(θ) and sin(θ) and determine the quadrant in which the terminal side of the angle θ lies when it is plotted in standard position. We sketch the terminal side of θ below. We think of a reference triangle formed with x = and y = 4, so the hypotenuse is ( ) + 4 = r Hence, cos(θ) = 5 and sin(θ) = = r r = 5 r = 5

244 6 Analytical Trigonometry Our cosine double angle identities give us three different formulas to choose from to find cos(θ). Using the first formula, we get: cos(θ) = cos (θ) sin (θ) = ( 5) ( 4 5) = 7. For sin(θ), we 5 get sin(θ) = sin(θ) cos(θ) = ( ) ( ) = 4 5. Since both cosine and sine of θ are negative, the terminal side of θ, when plotted in standard position, lies in Quadrant III. Check Point 66 Suppose (1, 6) lies on the terminal side of angle x when x is drawn in standard position. Find sin(x) and cos(x). In which quadrant does the terminal side of angle x lie?

245 6. Double and Half Angle Identities 7 Example 95 Given that tan(θ) = 4 and π a. sin(θ) b. cos(θ) c. tan(θ) < θ < π, find the following: We can draw a reference triangle with this tangent and use it to find the sine and cosine of θ. Not that since π < θ < π, the terminal side of θ falls in QII. The hypotenuse of this triangle is r = ( 4) + r = r = 5 r = 5 so sin(θ) = 5 and cos(θ) = 4. Now we have everything we need to answer the rest of the 5 question. sin(θ) = sin(θ) cos(θ) ( ) ( sin(θ = 4 ) 5 5 sin(θ = 4 5

246 8 Analytical Trigonometry cos(θ) = cos (θ) sin (θ) ( cos(θ) = 4 ( ) 5) 5 cos(θ) = cos(θ) = 7 5 tan(θ) = sin(θ) cos(θ) tan(θ) = 4/5 7/5 tan(θ) = 4 7 Check Point 67 Given that tan(θ) = where π < θ < π, find the exact value of cos(θ). Check Point 68 Given that csc(θ) = 1 5 where π < θ < π, find the exact value of sin(θ).

247 6. Double and Half Angle Identities 9 Example 96 Simplify. tan(θ) 1 + tan (θ) tan(θ) 1+tan (θ) = tan(θ) sec (θ) Pythagorean Identity = sin(θ)/ cos(θ) 1/ cos (θ) Quotient and Reciprocal Identities = sin(θ) cos(θ) cos (θ) 1 Fraction division = sin(θ) cos(θ) cos (θ) Simplify 1 = sin(θ) cos(θ) = sin(θ) Sine double angle identity

248 40 Analytical Trigonometry Example 97 Verify the identity. sin(x) 1 + cos(x) tan (x) = tan (x) sin(x) 1+cos(x) tan (x) = sin(x) cos(x) 1+ cos (x) 1 tan (x) Double angle identities = sin(x) cos(x) cos (x) tan (x) Simplify = sin(x) cos(x) tan (x) Simplify = tan(x) tan (x) Quotient Identity = tan (x) Simplify

249 6. Double and Half Angle Identities 41 Example 98 Verify the identity. cos(x) = 4 cos (x) cos(x) We start with a clever rewrite, and then proceed with actual identities. cos(x) = cos(x + x) = cos(x) cos(x) sin(x) sin(x) cosine of a sum identity = ( cos (x) 1) cos(x) sin(x) cos(x) sin(x) double angle identities = cos (x) cos(x) sin (x) cos(x) simplify = cos (x) cos(x) (1 cos (x)) cos(x) Pythagorean identity = cos (x) cos(x) cos(x) + cos (x) distribute = 4 cos (x) cos(x) Example 99 Rewrite cos( arcsin(x)) as a function of x and state the domain. Start by writing t = arcsin(x) so that t lies in the interval [ π, π ] with sin(t) = x. We aim to express cos ( arcsin(x)) = cos(t) in terms of x. Thanks to our double angle identities, we have three choices for rewriting cos(t): cos(t) = cos (t) sin (t), cos(t) = cos (t) 1 and cos(t) = 1 sin (t). Since we know x = sin(t), we choose: cos ( arcsin(x)) = cos(t) = 1 sin (t) = 1 x. Hence, g(x) = cos ( arcsin(x)) = 1 x. The domain of arcsin(x) is [ 1, 1], and since there are no domain restrictions on cosine, the

250 4 Analytical Trigonometry domain of g is [ 1, 1]. The Angle Reduction Identities It turns out, an important skill in calculus is going to be taking trigonometric expressions with powers and writing them without powers. We know this is a vague goal right now, but trust us, it will be very valuable later in your math career. To accomplish this goal, we ll start by rewriting some of our double angle identities. /a cos(θ) = cos (θ) 1 Since the idea here is to take a power and rewrite it as not-a-power, we ll isolate the cos (θ) in this identity. cos(θ) + 1 = cos (θ) cos (θ) = cos(θ) + 1 Similarly, we can start with the double angle identity cos(θ) = 1 sin (θ) and isolate the sin (θ). cos(θ) = 1 sin (θ) cos(θ) + sin (θ) = 1 sin (θ) = 1 cos(θ) sin (θ) = 1 cos(θ) Because we start with a squared trig function and end without any powers, we call these the Power Reduction Formulas.

251 6. Double and Half Angle Identities 4 Definition 61: Power Reduction Formulas For all angles θ cos (θ) = 1 + cos(θ) sin (θ) = 1 cos(θ) Let s do a couple of examples so that we can see these formulas in action. Example 100 Rewrite sin (θ) cos (θ) as a sum and difference of cosines to the first power. We begin with a straightforward application of the power reduction formulas: sin (θ) cos (θ) = ( ) ( ) 1 cos(θ) 1 + cos(θ) = 1 ( 1 cos (θ) ) 4 = cos (θ) Next, we apply the power reduction formula to cos (θ) to finish the reduction sin (θ) cos (θ) = cos (θ) ( ) 1 + cos((θ)) = = cos(4θ) = cos(4θ)

252 44 Analytical Trigonometry Example 101 Rewrite cos 4 (x) as a sum and difference of cosines to the first power. Again, we ll start with a clever rewrite, then apply the reduction formula for cosine twice. cos 4 (x) = (cos (x)) = ( ) 1+cos(x) Reduction Formula = cos(x) cos (x) Expand We still have a power of a cosine, so we need to apply a reduction formula again. Here, we want to think of the θ as being x, so cos (θ) = 1+cos(θ) will be cos (x) = 1+cos( x) when we apply the reduction formula. ( ) = cos(x) cos(4x) reduction formula 4 = cos(x) cos(4x) distribute 8 = cos(x) + 1 cos(4x) Combine like terms 8 Check Point 69 Rewrite the expression cos (6t) with an exponent no higher than 1 using the reduction formulas. The Half Angle Identities Another application of the Power Reduction Formulas is the Half Angle Formulas. To start, we apply the Power Reduction Formula tocos ( ) θ cos ( θ ) = 1 + cos ( ( )) θ = 1 + cos(θ). We can obtain a formula for cos ( θ ) by extracting square roots. In a similar fashion, we may obtain a half angle formula for sine, and by using a quotient formula, obtain a half angle formula for tangent.

253 6. Double and Half Angle Identities 45 Definition 6: Half Angle Identities For all angles θ, ( ) θ 1 + cos(θ) cos = ± ( ) θ 1 cos(θ) sin = ± where the choice of ± depends on the quadrant in which the terminal side of θ lies. Example 10 Use a half angle formula to find the exact value of cos ( π 1). To use the half angle formula, we note that π 1 = π 6 and since π is a Quadrant I angle, its cosine is 1 positive. Thus we have cos ( ( ) π 1 + cos π 1 = + 6 = 1 + ) = 1 + = + + = 4 Back in Section 6., we found cos (15 ) = 6+ by using the difference formula for cosine. The 4 reader is encouraged to prove that these two expressions are equal algebraically. Check Point 70 Use half angle formulas to find the exact value of cos ( 5π 1 ).

254 46 Analytical Trigonometry Example 10 Suppose π t 0 with cos(t) = 5. Find sin ( t ). If π t 0, then π t 0, which means t corresponds to a Quadrant IV angle. Hence, sin ( t ) < 0, so we choose the negative root formula from our half angle identities: ( ) t 1 cos (t) 1 ( ) 5 sin = = = = 10 = 5 Example 104 Given that tan(α) = 8 15 and that the terminal side of α lies in QIII, find the exact value of sin ( α ). Using the given information, we can draw the triangle shown below. Theorem, we find the hypotenuse to be 17. Using the Pythagorean Therefore, sin(α) = 8 15 and cos(α) =

255 6. Double and Half Angle Identities 47 Before we continue, we must remember that although α is in QIII, we need to locate α. Since then π < α < π π < α < π 4 so α is in QII. That means sin ( ) α will be positive. We finish with the sine half angle identity, taking the positive root. ( α ) 1 cos(α) sin = 1 ( 15/17) = /17 = 16 = 17 = 4 17 = Check Point 71 If tan(θ) = 7 4 where π < θ < π, find the exact value of cos ( θ ).

256 48 Analytical Trigonometry 6..1 Exercises In the following exercises, use the given information about θ to find the exact values of sin(θ) sin ( ) θ cos(θ) cos ( ) θ tan(θ) tan ( ) θ 1. sin(θ) = 7 5 where π < θ < π. cos(θ) = 8 5 where 0 < θ < π. tan(θ) = 1 5 where π < θ < π 4. csc(θ) = 4 where π < θ < π 5. cos(θ) = 5 where 0 < θ < π 6. sin(θ) = 4 5 where π < θ < π 7. cos(θ) = 1 1 where π < θ < π 8. sin(θ) = 5 1 where π < θ < π 9. sec(θ) = 5 where π < θ < π 10. tan(θ) = where π < θ < π In the following exercises, use the Half Angle Identities to find the exact value. You may have need of the Quotient, Reciprocal or Even / Odd Identities as well. 11. cos ( ) 7π 1 1. sin ( ) π 1 1. cos ( ) π sin ( ) 5π tan ( ) 7π 8 In the following ( exercises, ( find the exact value or state that it is undefined. 16. sin arcsin 4 )) ( ( 5)) sin arccsc sin ( ( arctan ()) ( )) 19. cos arcsin ( ( 5 )) 5 0. cos arcsec 7 1. cos ( arccot ( 5 )) ( ) arctan(). sin

257 6. Double and Half Angle Identities 49 In the following exercises, rewrite each of the following composite functions as algebraic functions of x and state the domain.. f (x) = sin ( arctan (x)) 4. f (x) = sin ( arccos (x)) 5. f (x) = cos ( arctan (x)) 6. f (x) = sin(arccos(x)) ( ( x )) 7. f (x) = sin arccos ( ( 5 x )) 8. f (x) = cos arcsin 9. f (x) = cos (arctan (x)) In the following exercises, verify the identity. Assume all quantities are defined. 0. (cos(θ) + sin(θ)) = 1 + sin(θ) 1. (cos(θ) sin(θ)) = 1 sin(θ). tan(t) = 1 1 tan(t) 1 1+tan(t). csc(θ) = cot(θ)+tan(θ) 4. 8 sin 4 (x) = cos(4x) 4 cos(x) cos 4 (x) = cos(4x) + 4 cos(x) + 6. sin(θ) = sin(θ) 4 sin (θ) 7. sin(4θ) = 4 sin(θ) cos (θ) 4 sin (θ) cos(θ) 8. sin (t) cos 4 (t) = + cos(t) cos(4t) cos(6t) 9. sin 4 (t) cos (t) = cos(t) cos(4t) + cos(6t) 40. cos(4θ) = 8 cos 4 (θ) 8 cos (θ) cos(8θ) = 18 cos 8 (θ) 56 cos 6 (θ) cos 4 (θ) cos (θ) + 1 cos(x) 4. sec(x) = cos(x) + sin(x) + sin(x) cos(x) sin(x) 1 4. cos(θ) sin(θ) + 1 cos(θ) + sin(θ) = cos(θ) cos(θ) cos(θ) sin(θ) 1 cos(θ) + sin(θ) = sin(θ) cos(θ) 45. sin(x) = tan(x) 1+tan (x) 46. cos(α) = 1 tan (α) 47. tan(x) = 1+tan (α) sin(x) cos(x) cos (x) ( sin (x) 1 ) = cos(x) + sin 4 (x) 49. sin(x) = sin(x) cos (x) sin (x) 50. cos(x) = cos (x) sin (x) cos(x) cos(t) sin(t) cos(t) = cos(t) sin(t) 1 Suppose θ is a Quadrant I angle with sin(θ) = x. Verify the following formulas 5. cos(θ) = 1 x

258 50 Analytical Trigonometry 5. sin(θ) = x 1 x 54. cos(θ) = 1 x Suppose θ is a Quadrant I angle with tan(θ) = x. Verify the following formulas cos(θ) = x + 1 x 56. sin(θ) = x + 1 x 57. sin(θ) = x cos(θ) = 1 x x + 1

259 6.4 Solving Equations Involving Identities Solving Equations Involving Identities Learning Objectives In this section you will: Solve trigonometric equations involving identities When we first encountered trigonometric equations in Section 5.6, each of the problems featured one circular function. If an equation involves two different circular functions or if the equation contains the same circular function but with different arguments, we will need to employ identities and Algebra to reduce the equation to a simpler form. We demonstrate these techniques in the following examples. Example 105 Solve the following equation and list the solutions which lie in the interval [0, π). sec (θ) = tan(θ) + We see immediately in the equation sec (θ) = tan(θ) + that there are two different circular functions present, so we look for an identity to express both sides in terms of the same function. We use the Pythagorean Identity sec (θ) = 1 + tan (θ) to exchange sec (θ) for tangents. What results is a quadratic in disguise, which we can solve using a quick substitution. sec (θ) = tan(θ) tan (θ) = tan(θ) + (Since sec (θ) = 1 + tan (θ).) tan (θ) tan(θ) = 0 u u = 0 Let u = tan(θ). (u + 1)(u ) = 0 This gives u = 1 or u =. Since u = tan(θ), we have tan(θ) = 1 or tan(θ) =.

260 5 Analytical Trigonometry From tan(θ) = 1, we get θ = π +πk for integers k. To solve tan(θ) =, we employ the arctangent 4 function and get θ = arctan() + πk for integers k. From the first set of solutions, we get θ = π 4 and θ = 7π 4 as our answers which lie in [0, π). We get θ = arctan() and θ = π + arctan() 4.49 as answers from our second set of solutions which lie in [0, π). Example 106 Solve the following equation and list the solutions which lie in the interval [0, π). cos(t) = cos(t) The good news is that in the equation cos(t) = cos(t), we have the same circular function, cosine, throughout. The bad news is that we have different arguments, t and t. Using the double angle identity cos(t) = cos (t) 1 results in another quadratic in disguise: cos(t) = cos(t) cos (t) 1 = cos(t) (Since cos(t) = cos (t) 1.) cos (t) cos(t) + 1 = 0 u u + 1 = 0 Let u = cos(t). (u 1)(u 1) = 0 We get u = 1 or u = 1, so cos(t) = 1 or cos(t) = 1. Solving cos(t) = 1, we get t = π + πk or t = 5π + πk for integers k. From cos(t) = 1, we get t = πk for integers k. The answers which lie in [0, π) are t = 0, π, and 5π.

261 6.4 Solving Equations Involving Identities 5 Example 107 Solve the following equation and list the solutions which lie in the interval [0, π). cos(t) = cos(t) To solve cos(t) = cos(t), we take a cue from the previous problem and look for an identity to rewrite cos(t) in terms of cos(t). We can start with a sum identity and work for there. only cos(t) in our expression: Remember, the goal is to end up with cos(t) = cos(t + t) = cos(t) cos(t) sin(t) sin(t) Cosine sum identity = ( cos (t) 1) cos(t) sin(t) cos(t) sin(t) Double angle identities = cos (t) cos(t) sin (t) cos(t) = cos (t) cos(t) (1 cos (t)) cos(t) Pythagorean Identity = cos (t) cos(t) cos(t) + cos (t) = 4 cos (t) cos(t) This transforms the equation into a polynomial in terms of cos(t). cos(t) = cos(t) 4 cos (t) cos(t) = cos(t) cos (t) cos(t) = 0 4u u = 0 Let u = cos(t). Using what we know about polynomials, we factor 4u u as (u 1) ( 4u + 4u + ) and set each factor equal to 0. We get either u 1 = 0 or 4u + u + = 0, and since the discriminant of the latter is negative, the

262 54 Analytical Trigonometry only real solution to 4u u = 0 is u = 1. Since u = cos(t), we get cos(t) = 1, so t = πk for integers k. The only solution which lies in [0, π) is t = 0. Example 108 Solve the following equation and list the solutions which lie in the interval [0, π). cos(x) = cos(5x) While we could approach solving the equation cos(x) = cos(5x) in the same manner as we did the previous two problems, we choose instead to showcase the utility of the Sum to Product Identities. From cos(x) = cos(5x), we get cos(5x) cos(x) = 0, and it is the presence of 0 on the right hand side that indicates a switch to a product would be a good move. Using the Sum-to-Product identities, we rewrite cos(5x) cos(x) as sin ( 5x+x ) ( sin 5x x ) = sin(4x) sin(x). Hence, our original equation cos(x) = cos(5x) is equivalent to sin(4x) sin(x) = 0. From sin(4x) sin(x) = 0, thanks to the Zero Product Property, we get either sin(4x) = 0 or sin(x) = 0. Solving sin(4x) = 0 gives x = π k for integers k, and the solution to sin(x) = 0 is x = πk for 4 integers k. The second set of solutions is contained in the first set of solutions, a cos(5x) = cos(x) is x = π k for integers k. 4 so our final solution to

263 6.4 Solving Equations Involving Identities 55 There are eight of these answers which lie in [0, π): x = 0, π 4, π, π 4, π, 5π 4, π and 7π 4. a As always, when in doubt, write it out! Example 109 Solve the following equation and list the solutions which lie in the interval [0, π). sin(x) = cos(x) div id= mydiv5 style= display:none In the equation sin(x) = cos(x), we not only have different circular functions involved, but we also have different arguments to contend with. Using the double angle identity sin(x) = sin(x) cos(x) makes all of the arguments the same and we proceed to gather all of the nonzero terms on one side of the equation and factor. sin(x) = cos(x) sin(x) cos(x) = cos(x) (Since sin(x) = sin(x) cos(x).) sin(x) cos(x) cos(x) = 0 cos(x)( sin(x) ) = 0 We get cos(x) = 0 or sin(x) =. From cos(x) = 0, we obtain x = π sin(x) =, we get x = π + πk or x = π + πk for integers k. + πk for integers k. From The answers which lie in [0, π) are x = π, π, π and π.

264 56 Analytical Trigonometry Example 110 Solve the following equation and list the solutions which lie in the interval [0, π). sin(x) cos ( x ) + cos(x) sin ( x ) = 1 Unlike the previous problem, there seems to be no quick way to get the circular functions or their arguments to match in the equation sin(x) cos ( x ) + cos(x) sin ( x ) = 1. If we stare at it long enough, however, we realize that the left hand side is the expanded form of the sum formula for sin ( x + x ). Hence, our original equation is equivalent to sin ( x) = 1. Solving, we find x = π + 4π k for integers k. Two of these solutions lie in [0, π): x = π and x = 5π. Example 111 Solve the following equation and list the solutions which lie in the interval [0, π). cos(x) sin(x) = With the absence of double angles or squares, there doesn t seem to be much we can do with the equation cos(x) sin(x) =. However, since the frequencies of the sine and cosine terms are the same, we can rewrite the left hand side of this equation as a sinusoid. To fit f (x) = cos(x) sin(x) to the form A sin(ωt + φ) + B, we find A =, B = 0, ω = 1 and φ = 5π 6. Hence, we can rewrite the equation cos(x) sin(x) = as sin ( x + 5π 6 ) ( ) =, or sin x + 5π 6 = 1.

265 6.4 Solving Equations Involving Identities 57 Solving, we get x = π + πk for integers k. Only one of our solutions, x = 5π, which corresponds to k = 1, lies in [0, π) Exercises In the following exercises, solve the equation, giving the exact solutions which lie in [0, π) 1. sin (θ) = cos (θ). sin (t) = sin (t). sin (x) = cos (x) 4. cos (θ) = sin (θ) 5. cos (t) = cos (t) 6. cos(x) = 5 cos(x) 7. cos(θ) + cos(θ) + = 0 8. cos(t) = 5 sin(t) 9. cos(x) = sin(x) sec (θ) = tan(θ) 11. tan (t) = 1 sec(t) 1. cot (x) = csc(x) 1. sec(θ) = csc(θ) 14. cos(t) csc(t) cot(t) = 6 cot (t) 15. sin(x) = tan(x) 16. cot 4 (θ) = 4 csc (θ) cos(t) + csc (t) = tan (x) = tan (x) 19. tan (θ) = sec (θ) 0. cos (t) = cos (t) 1. tan(x) cos(x) = 0. csc (θ) + csc (θ) = 4 csc(θ) + 4. tan(t) = 1 tan (t) 4. tan (x) = sec (x) 5. sin(6θ) cos(θ) = cos(6θ) sin(θ) 6. sin(t) cos(t) = cos(t) sin(t) 7. cos(x) cos(x) + sin(x) sin(x) = 1 8. cos(5θ) cos(θ) sin(5θ) sin(θ) = 9. sin(t) + cos(t) = 1 0. sin(x) + cos(x) = 1 1. cos(θ) sin(θ) = 1. sin(t) + cos(t) = 1. cos(x) sin(x) = 4. sin(θ) cos(θ) = 5. cos(t) = cos(5t) 6. cos(4x) = cos(x) 7. sin(5θ) = sin(θ) 8. cos(5t) = cos(t) 9. sin(6x) + sin(x) = tan(x) = cos(x)

266 58 Analytical Trigonometry

267 Appendix A Algebra Review One purpose of this Algebra Review Appendix is to support a co-requisite approach to teaching College Algebra or Precalculus. 1 Our goal is to provide instructors with supplemental material linked to the main textbook that can be used to support students who have minor gaps in their pre-college mathematical backgrounds. To that end, we have written a collection of somewhat independent sections designed to review the concepts, skills and vocabulary that we believe are prerequisite to a rigorous, college-level Precalculus course. This review is not designed to teach the material to students who have never seen it before so the presentation is more succinct and the exercise sets are shorter than those usually found in an Intermediate Algebra or high school Algebra II text. Some of this material (like adding fractions and plotting points) is used throughout the text but, where appropriate, we have referenced specific sections of the main body of the Precalculus text in an effort to assist faculty who would like to assign the Appendix as just in time review reading to their students. An outline of the chapter with short descriptions of each section is given below: Section A.1 (Basic Set Theory and Interval Notation) contains a brief summary of the set theory terminology used throughout the text including sets of real numbers and interval notation. Section A. (Real Number Arithmetic) lists the properties of real number arithmetic. Section A. (The Cartesian Plane) discusses the basic notions of plotting points in the plane, reflections and symmetry. We then develop the Distance Formula and the Midpoint Formula. Section A.4 (Linear Equations and Inequalities) focuses on solving linear equations and linear inequalities from a strictly algebraic perspective. The geometry of graphing lines in the plane is deferred until Section A.5 (Graphing Lines). Section A.5 (Graphing Lines) starts by defining the slope of a line between two points and then develops the point-slope form and the slope-intercept form of equations for lines. Horizontal and vertical lines are 1 Please read The New Preface for details as to why we decided to organize our book in order to support this pedagogy. Also, to us Precalculus = College Algebra + College Trigonometry without the formalization of limits. This distinction is not universally agreed upon so we felt the need to point it out. You know, the stuff students mess up all of the time like fractions and negative signs. The collection is close to exhaustive and is definitely exhausting!

268 60 Algebra Review discussed as are parallel and perpendicular lines. This material is required for Section?? (Constant and Linear Functions). Section A.6 (Systems of Two Linear Equations in Two Unknowns) is a review of the basic terminology and techniques related to solving systems of two linear equations that each have the same two variables in them. We start Chapter?? (Systems of Equations and Matrices) assuming students know these techniques. Section A.7 (Absolute Value Equations and Inequalities) begins with a definition of absolute value as a distance. Fundamental properties of absolute value are listed and then basic equations and inequalities involving absolute value are solved using the distance definition and those properties. Absolute value is revisited in much greater depth in Section?? (Absolute Value Functions). Section A.8 (Polynomial Arithmetic) covers the addition, subtraction, multiplication and division of polynomials as well as the vocabulary which is used extensively when the graphs of polynomials are studied in Chapter?? (Polynomials). Section A.9 (Basic Factoring Techniques) contains pretty much what it says: basic factoring techniques and how to solve equations using those techniques along with the Zero Product Property of Real Numbers. Section A.10 (Quadratic Equations) discusses solving quadratic equations using the technique of completing the square and by using the Quadratic Formula. Equations that are quadratic in form are also discussed. This material is revisited in Section?? (Quadratic Functions). Section A.11 (Complex Numbers) covers the basic arithmetic of complex numbers and the solving of quadratic equations with complex solutions. It s required for Section?? (Complex Zeros and the Fundamental Theorem of Algebra). Section A.1 (Rational Expressions and Equations) starts with the basic arithmetic of rational expressions and the simplifying of compound fractions. Solving equations by clearing denominators and the handling of negative integer exponents are presented but the graphing of rational functions is deferred until Chapter?? (Rational Functions). Section A.1 (Radicals and Equations) covers simplifying radicals as well as the solving of basic equations involving radicals. Students should be familiar with this material before starting Chapter?? (Root and Radical Functions). Section A.14 (Variation) looks at a variety of equations from Science and Engineering. It s a self-contained section that can be covered at any time.

269 A.1 Basic Set Theory and Interval Notation 61 A.1 Basic Set Theory and Interval Notation A.1.1 Some Basic Set Theory Notions We begin this section with the definition of a concept that is central to all of Mathematics. A set is a well-defined collection of objects which are called the elements of the set. Here, well-defined means that it is possible to determine if something belongs to the collection or not, without prejudice. For example, the collection of letters that make up the word smolko is well-defined and is a set, but the collection of the worst Math teachers in the world is not well-defined and therefore is not a set. 1 In general, there are three ways to describe sets and those methods are listed below. Ways to Describe Sets 1. The Verbal Method: Use a sentence to describe the elements the set.. The Roster Method: Begin with a left brace {, list each element of the set only once and then end with a right brace }.. The Set-Builder Method: A combination of the verbal and roster methods using a dummy variable such as x and conditions on that variable. Let S be the set described verbally as the set of letters that make up the word smolko. A roster description of S is {s, m, o, l, k}. Note that we listed o only once, even though it appears twice in the word smolko. Also, the order of the elements doesn t matter, so {k, l, m, o, s} is also a roster description of S. A setbuilder description of S is: {x x is a letter in the word smolko }. The way to read this is The set of elements x such that x is a letter in the word smolko. In each of the above cases, we may use the familiar equals sign = and write S = {s, m, o, l, k} or S = {x x is a letter in the word smolko }. Notice that m is in S but many other letters, such as q, are not in S. We express these ideas of set inclusion and exclusion mathematically using the symbols m S (read m is in S ) and q / S (read q is not in S ). More precisely, we have the following. Let A be a set. If x is an element of A then we write x A which is read x is in A. If x is not an element of A then we write x / A which is read x is not in A. Now let s consider the set C = {x x is a consonant in the word smolko }. A roster description of C is C = {s, m, l, k}. Note that by construction, every element of C is also in S. We express this relationship by stating that the set C is a subset of the set S, which is written in symbols as C S. The more formal definition is given at the top of the next page. 1 For a more thought-provoking example, consider the collection of all things that do not contain themselves - this leads to the famous paradox known as Russell s Paradox.

270 6 Algebra Review Given sets A and B, we say that the set A is a subset of the set B and write A B if every element in A is also an element of B. In our previous example, C S yet not vice-versa since o S but o / C. Additionally, the set of vowels V = {a, e, i, o, u}, while it does have an element in common with S, is not a subset of S. (As an added note, S is not a subset of V, either.) We could, however, build a set which contains both S and V as subsets by gathering all of the elements in both S and V together into a single set, say U = {s, m, o, l, k, a, e, i, u}. Then S U and V U. The set U we have built is called the union of the sets S and V and is denoted S V. Furthermore, S and V aren t completely different sets since they both contain the letter o. The intersection of two sets is the set of elements (if any) the two sets have in common. In this case, the intersection of S and V is {o}, written S V = {o}. We formalize these ideas below. Suppose A and B are sets. The intersection of A and B is A B = {x x A and x B} The union of A and B is A B = {x x A or x B (or both)} The key words in Definition A.1.1 to focus on are the conjunctions: intersection corresponds to and meaning the elements have to be in both sets to be in the intersection, whereas union corresponds to or meaning the elements have to be in one set, or the other set (or both). Please note that this mathematical use of the word or differs than how we use or in spoken English. In Math, we use the inclusive or which allows for the element to be in both sets. At a restaurant if you re asked Do you want fries or a salad? you must pick one and only one. This is known as the exclusive or and it plays a role in other Math classes. For our purposes it is good enough to say that for an element to belong to the union of two sets it must belong to at least one of them. Returning to the sets C and V above, C V = {s, m, l, k, a, e, i, o, u}. Their intersection, however, creates a bit of notational awkwardness since C and V have no elements in common. While we could write C V = {}, this sort of thing happens often enough that we give the set with no elements a name. The Empty Set is the set which contains no elements and is denoted. That is, = {} = {x x x}. As promised, the empty set is the set containing no elements since no matter what x is, x = x. Like the number 0, the empty set plays a vital role in mathematics. We introduce it here more as a symbol of convenience as opposed to a contrivance 4 because saying that C V = is unambiguous whereas {} looks like a typographical error. A nice way to visualize the relationships between sets and set operations is to draw a Venn Diagram. A Venn Diagram for the sets S, C and V is drawn at the top of the next page. Which just so happens to be the same set as S V. Sadly, the full extent of the empty set s role will not be explored in this text. 4 Actually, the empty set can be used to generate numbers - mathematicians can create something from nothing!

271 A.1 Basic Set Theory and Interval Notation 6 U C s m l k o a e i u S V A Venn Diagram for C, S and V. In the Venn Diagram above we have three circles - one for each of the sets C, S and V. We visualize the area enclosed by each of these circles as the elements of each set. Here, we ve spelled out the elements for definitiveness. Notice that the circle representing the set C is completely inside the circle representing S. This is a geometric way of showing that C S. Also, notice that the circles representing S and V overlap on the letter o. This common region is how we visualize S V. Notice that since C V =, the circles which represent C and V have no overlap whatsoever. All of these circles lie in a rectangle labeled U for the universal set. A universal set contains all of the elements under discussion, so it could always be taken as the union of all of the sets in question, or an even larger set. In this case, we could take U = S V or U as the set of letters in the entire alphabet. The reader may well wonder if there is an ultimate universal set which contains everything. The short answer is no and we refer you once again to Russell s Paradox. The usual triptych of Venn Diagrams indicating generic sets A and B along with A B and A B is given below. U U U A B A B A B A B A B Sets A and B. A B is shaded. A B is shaded. The one major limitation of Venn Diagrams is that they become unwieldy if more than four sets need to be drawn simultaneously within the same universal set. This idea is explored in the Exercises.

272 64 Algebra Review A.1. Sets of Real Numbers The playground for most of this text is the set of Real Numbers. Much of the real world can be quantified using real numbers: the temperature at a given time, the revenue generated by selling a certain number of products and the maximum population of Sasquatch which can inhabit a particular region are just three basic examples. A succinct, but nonetheless incomplete 5 definition of a real number is given below. A real number is any number which possesses a decimal representation. The set of real numbers is denoted by the character R. Certain subsets of the real numbers are worthy of note and are listed below. In fact, in more advanced texts, 6 the real numbers are constructed from some of these subsets. Special Subsets of Real Numbers 1. The Natural Numbers: N = {1,,,...} The periods of ellipsis... here indicate that the natural numbers contain 1,, and so forth.. The Whole Numbers: W = {0, 1,,...}.. The Integers: Z = {...,,, 1, 0, 1,,,...} = {0, ±1, ±, ±,...}. a 4. The Rational Numbers: Q = { a b a Z and b Z where b 0}. Rational numbers are the ratios of integers where the denominator is not zero. It turns out that another way to describe the rational numbers b is: Q = {x x possesses a repeating or terminating decimal representation} 5. The Irrational Numbers: P = {x x R but x / Q}. c That is, an irrational number is a real number which isn t rational. Said differently, P = {x x possesses a decimal representation which neither repeats nor terminates} a The symbol ± is read plus or minus and it is a shorthand notation which appears throughout the text. Just remember that x = ± means x = or x =. b See Section??. c Examples here include number π (See Section B.1), and Note that every natural number is a whole number which, in turn, is an integer. Each integer is a rational number (take b = 1 in the above definition for Q) and since every rational number is a real number 7 the sets N, W, Z, Q, and R are nested like Matryoshka dolls. More formally, these sets form a subset chain: N W Z Q R. The reader is encouraged to sketch a Venn Diagram depicting R and all of the subsets mentioned above. It is time to put all of this together in an example. 5 Math pun intended! 6 See, for instance, Landau s Foundations of Analysis. 7 Thanks to long division!

273 A.1 Basic Set Theory and Interval Notation Write a roster description for P = { n n N} and E = {n n Z}.. Write a verbal description for S = {x x R}.. Let A = { 117, 4, 0.000, }. 5 (a) Which elements of A are natural numbers? Rational numbers? Real numbers? (b) Find A W, A Z and A P. 4. What is another name for N Q? What about Q P? Solution. 1. To find roster descriptions for each of these sets, we need to list their elements. Starting with the set P = { n n N}, we substitute natural number values n into the formula n. For n = 1 we get 1 =, for n = we get = 4, for n = we get = 8 and for n = 4 we get 4 = 16. Hence P describes the powers of, so a roster description for P is P = {, 4, 8, 16,...} where the... indicates the that pattern continues. 8 Proceeding in the same way, we generate elements in E = {n n Z} by plugging in integer values of n into the formula n. Starting with n = 0 we obtain (0) = 0. For n = 1 we get (1) =, for n = 1 we get ( 1) = for n =, we get () = 4 and for n = we get ( ) = 4. As n moves through the integers, n produces all of the even integers. 9 A roster description for E is E = {0, ±, ±4,...}.. One way to verbally describe S is to say that S is the set of all squares of real numbers. While this isn t incorrect, we d like to take this opportunity to delve a little deeper. 10 What makes the set S = {x x R} a little trickier to wrangle than the sets P or E above is that the dummy variable here, x, runs through all real numbers. Unlike the natural numbers or the integers, the real numbers cannot be listed in any methodical way. 11 Nevertheless, we can select some real numbers, square them and get a sense of what kind of numbers lie in S. For x =, x = ( ) = 4 so 4 is in S, as are ( ) = 9 4 and ( 117) = 117. Even things like ( π) and ( ) are in S. So suppose s S. What can be said about s? We know there is some real number x so that s = x. Since x 0 for any real number x, we know s 0. This tells us that everything in S is a non-negative real number. 1 This begs the question: are all of the non-negative real numbers in S? Suppose n is a non-negative real number, that is, n 0. If n were in S, there would be a real number 8 This isn t the most precise way to describe this set - it s always dangerous to use... since we assume that the pattern is clearly demonstrated and thus made evident to the reader. Formulas are more precise because the pattern is clear. 9 This shouldn t be too surprising, since an even integer is defined to be an integer multiple of. 10 Think of this as an opportunity to stop and smell the mathematical roses. 11 This is a nontrivial statement. Interested readers are directed to a discussion of Cantor s Diagonal Argument. 1 This means S is a subset of the non-negative real numbers.

274 66 Algebra Review x so that x = n. As you may recall, we can solve x = n by extracting square roots : x = ± n. Since n 0, n is a real number. 1 Moreover, ( n) = n so n is the square of a real number which means n S. Hence, S is the set of non-negative real numbers.. (a) The set A contains no natural numbers. 14 Clearly 4 is a rational number as is 117 (which can 5 be written as 117 ). It s the last two numbers listed in A, and , 1 that warrant some discussion. First, recall that the line over the digits 00 in (called the vinculum) indicates that these digits repeat, so it is a rational number. 15 As for the number , the... indicates the pattern of adding an extra 0 followed by a is what defines this real number. Despite the fact there is a pattern to this decimal, this decimal is not repeating, so it is not a rational number - it is, in fact, an irrational number. All of the elements of A are real numbers, since all of them can be expressed as decimals (remember that 4 5 = 0.8). (b) The set A W = {x x A and x W} is another way of saying we are looking for the set of numbers in A which are whole numbers. Since A contains no whole numbers, A W =. Similarly, A Z is looking for the set of numbers in A which are integers. Since 117 is the only integer in A, A Z = { 117}. For the set A P, as discussed in part (a), the number is irrational, so A P = { }. 4. The set N Q = {x x N or x Q} is the union of the set of natural numbers with the set of rational numbers. Since every natural number is a rational number, N doesn t contribute any new elements to Q, so N Q = Q. 16 For the set Q P, we note that every real number is either rational or not, hence Q P = R, pretty much by the definition of the set P. As you may recall, we often visualize the set of real numbers R as a line where each point on the line corresponds to one and only one real number. Given two different real numbers a and b, we write a < b if a is located to the left of b on the number line, as shown below. a b The real number line with two numbers a and b where a < b. While this notion seems innocuous, it is worth pointing out that this convention is rooted in two deep properties of real numbers. The first property is that R is complete. This means that there are no holes or gaps in the real number line. 17 Another way to think about this is that if you choose any two distinct (different) real numbers, and look between them, you ll find a solid line segment (or interval) consisting of infinitely many real numbers. The next result tells us what types of numbers we can expect to find. 1 This is called the square root closed property of the non-negative real numbers. 14 Carl was tempted to include 0.9 in the set A, but thought better of it. See Section?? for details. 15 So = In fact, anytime A B, A B = B and vice-versa. See the exercises. 17 Alas, this intuitive feel for what it means to be complete is as good as it gets at this level. Completeness is given a much more precise meaning later in courses like Analysis and Topology.

275 A.1 Basic Set Theory and Interval Notation 67 Density Property of Q and P in R Between any two distinct real numbers, there is at least one rational number and one irrational number. It then follows that between any two distinct real numbers there will be infinitely many rational and infinitely many irrational numbers. The root word dense here communicates the idea that rationals and irrationals are thoroughly mixed into R. The reader is encouraged to think about how one would find both a rational and an irrational number between, say, and 1. Once you ve done that, try doing the same thing for the numbers 0.9 and 1. ( Try is the operative word, here. 18 ) The second property R possesses that lets us view it as a line is that the set is totally ordered. This means that given any two real numbers a and b, either a < b, a > b or a = b which allows us to arrange the numbers from least (left) to greatest (right). This property is given below. Law of Trichotomy If a and b are real numbers then exactly one of the following statements is true: a < b a > b a = b Segments of the real number line are called intervals. They play a huge role not only in this text but also in the Calculus curriculum so we need a concise way to describe them. We start by examining a few examples of the interval notation associated with some specific sets of numbers. Set of Real Numbers Interval Notation Region on the Real Number Line {x 1 x < } [1, ) 1 {x 1 x 4} [ 1, 4] 1 4 {x x 5} (, 5] 5 {x x > } (, ) As you can glean from the table, for intervals with finite endpoints we start by writing left endpoint, right endpoint. We use square brackets, [ or ], if the endpoint is included in the interval. This corresponds to a filled-in or closed dot on the number line to indicate that the number is included in the set. Otherwise, we use parentheses, ( or ) that correspond to an open circle which indicates that the endpoint is not part of the set. If the interval does not have finite endpoints, we use the symbol to indicate that the interval extends indefinitely to the left and the symbol to indicate that the interval extends indefinitely to the right. Since infinity is a concept, and not a number, we always use parentheses when using these 18 Again, see Section?? for details.

276 68 Algebra Review symbols in interval notation, and use the appropriate arrow to indicate that the interval extends indefinitely in one or both directions. We summarize all of the possible cases in one convenient table below. 19 Let a and b be real numbers with a < b. Interval Notation Set of Real Numbers Interval Notation Region on the Real Number Line {x a < x < b} (a, b) a b {x a x < b} [a, b) a b {x a < x b} (a, b] a b {x a x b} [a, b] a b {x x < b} (, b) b {x x b} (, b] b {x x > a} (a, ) a {x x a} [a, ) a R (, ) Intervals of the forms (a, b), (, b) and (a, ) are said to be open intervals. [a, b], (, b] and [a, ) are said to be closed intervals. Those of the forms Unfortunately, the words open and closed are not antonyms here because the empty set and the set (, ) are simultaneously open and closed 0 while the intervals (a, b] and [a, b) are neither open nor closed. The inclusion or exclusion of an endpoint might seem like a terribly small thing to fuss about but these sorts of technicalities in the language become important in Calculus so we feel the need to put this material in the Precalculus book. 19 The importance of understanding interval notation in this book and also in Calculus cannot be overstated so please do yourself a favor and memorize this chart. 0 You don t need to worry about that fact until you take an advanced course in Topology.

277 A.1 Basic Set Theory and Interval Notation 69 We close this section with an example that ties together some of the concepts presented earlier. Specifically, we demonstrate how to use interval notation along with the concepts of union and intersection to describe a variety of sets on the real number line. In many sections of the text to come you will need to be fluent with this notation so take the time to study it deeply now.

278 70 Algebra Review 1. Express the following sets of numbers using interval notation. (a) {x x or x } (b) {x x < and x 8 5 } (c) {x x ±} (d) {x 1 < x or x = 5}. Let A = [ 5, ) and B = (1, ). Find A B and A B. Solution. 1. (a) The best way to proceed here is to graph the set of numbers on the number line and glean the answer from it. The inequality x corresponds to the interval (, ] and the inequality x corresponds to the interval [, ). The or in {x x or x } tells us that we are looking for the union of these two intervals, so our answer is (, ] [, ). (, ] [, ) (b) For the set {x x < and x 8 5 }, we need the real numbers less than (to the left of) that are simultaneously greater than (to the right of) 8 5, including 8 5 but excluding. This yields {x x < and x 8 5 } = [ 8 5, ). 8 5 [ 8 5, ) (c) For the set {x x ±}, we proceed as before and exclude both x = and x = from our set. (Refer back to page 66 for a discussion about x = ±) This breaks the number line into three intervals, (, ), (, ) and (, ). Since the set describes real numbers which come from the first, second or third interval, we have {x x ±} = (, ) (, ) (, ). (, ) (, ) (, ) (d) Graphing the set {x 1 < x or x = 5} yields the interval ( 1, ] along with the single number 5. While we could express this single point as [5, 5], it is customary to write a single point as a singleton set, so in our case we have the set {5}. This means that our final answer is written {x 1 < x or x = 5} = ( 1, ] {5}. 1 5 ( 1, ] {5}

279 A.1 Basic Set Theory and Interval Notation 71. We start by graphing A = [ 5, ) and B = (1, ) on the number line. To find A B, we need to find the numbers common to both A and B; in other words, we need to find the overlap of the two intervals. Clearly, everything between 1 and is in both A and B. However, since 1 is in A but not in B, 1 is not in the intersection. Similarly, since is in B but not in A, it isn t in the intersection either. Hence, A B = (1, ). To find A B, we need to find the numbers in at least one of A or B. Graphically, we shade A and B along with it. Notice here that even though 1 isn t in B, it is in A, so it s in the union along with all of the other elements of A between 5 and 1. A similar argument goes for the inclusion of in the union. The result of shading both A and B together gives us A B = [ 5, ). 5 1 A = [ 5, ), B = (1, ) 5 1 A B = (1, ) 5 1 A B = [ 5, )

280 7 Algebra Review A.1. Exercises 1. Find a verbal description for O = {n 1 n N}. Find a roster description for X = {z z Z} {. Let A =, 1.0, 5, 0.57, 1.,, , 0 } 10, 117 (a) List the elements of A which are natural numbers. (b) List the elements of A which are irrational numbers. (c) Find A Z (d) Find A Q 4. Fill in the chart below. Set of Real Numbers Interval Notation Region on the Real Number Line {x 1 x < 5} [0, ) {x 5 < x 0} 7 (, ) {x x } 5 7 (, 9) {x x } 4

281 A.1 Basic Set Theory and Interval Notation 7 In Exercises 5-10, find the indicated intersection or union and simplify if possible. Express your answers in interval notation. 5. ( 1, 5] [0, 8) 6. ( 1, 1) [0, 6] 7. (, 4] (0, ) 8. (, 0) [1, 5] 9. (, 0) [1, 5] 10. (, 5] [5, 8) In Exercises 11 -, write the set using interval notation. 11. {x x 5} 1. {x x 1} 1. {x x, 4} 14. {x x 0, } 15. {x x, } 16. {x x 0, ±4} 17. {x x 1 or x 1} 18. {x x < and x } 19. {x x or x > 0} 0. {x x and x > } 1. {x x > or x = ±1}. {x < x < 1 and x 4} For Exercises - 8, use the blank Venn Diagram below with A, B, and C in it as a guide to help you shade the following sets. U A B C. A C 4. B C 5. (A B) C 6. (A B) C 7. A (B C) 8. (A B) (A C) 9. Explain how your answers to problems 7 and 8 show A (B C) = (A B) (A C). Phrased differently, this shows intersection distributes over union. Discuss with your classmates if union distributes over intersection. Use a Venn Diagram to support your answer. 0. Show that A B if and only if A B = B. 1. Let A = {1,, 5, 7, 9}, B = {, 4, 6, 8, 10}, C = {1, 6, 9} and D = {, 7, 10}. Draw one Venn Diagram that shows all four of these sets. What sort of difficulties do you encounter?

282 74 Algebra Review A.1.4 Answers 1. O is the odd natural numbers.. X = {0, 1, 4, 9, 16,...}. (a) 0 = and (b) and { (c), 0 } 10, 117 { (d), 1.0, } 0, 0.57, 1., 5 10, 117 Set of Real Numbers Interval Notation Region on the Real Number Line {x 1 x < 5} [ 1, 5) 1 5 {x 0 x < } [0, ) {x < x 7} (, 7] 0 7 {x 5 < x 0} ( 5, 0] 5 0 {x < x < } (, ) {x 5 x 7} [5, 7] {x x } (, ] {x x < 9} (, 9) {x x > 4} (4, ) {x x } [, )

283 A.1 Basic Set Theory and Interval Notation ( 1, 5] [0, 8) = [0, 5] 6. ( 1, 1) [0, 6] = ( 1, 6] 7. (, 4] (0, ) = (0, 4] 8. (, 0) [1, 5] = 9. (, 0) [1, 5] = (, 0) [1, 5] 10. (, 5] [5, 8) = {5} 11. (, 5) (5, ) 1. (, 1) ( 1, ) 1. (, ) (, 4) (4, ) 14. (, 0) (0, ) (, ) 15. (, ) (, ) (, ) 16. (, 4) ( 4, 0) (0, 4) (4, ) 17. (, 1] [1, ) 18. [, ) 19. (, ] (0, ) { 1} {1} (, ). (, 4) (4, 1). A C 4. B C U U A B A B C C

284 76 Algebra Review 5. (A B) C 6. (A B) C U U A B A B C C 7. A (B C) 8. (A B) (A C) U U A B A B C C 9. Yes, A (B C) = (A B) (A C). U A B C

285 A. Real Number Arithmetic 77 A. Real Number Arithmetic In this section we list the properties of real number arithmetic. This is meant to be a succinct, targeted review so we ll resist the temptation to wax poetic about these axioms and their subtleties and refer the interested reader to a more formal course in Abstract Algebra. There are two primary operations one can perform with real numbers: addition and multiplication. We ll start with the properties of addition. Properties of Real Number Addition Closure: For all real numbers a and b, a + b is also a real number. Commutativity: For all real numbers a and b, a + b = b + a. Associativity: For all real numbers a, b and c, a + (b + c) = (a + b) + c. Identity: There is a real number 0 so that for all real numbers a, a + 0 = a. Inverse: For all real numbers a, there is a real number a such that a + ( a) = 0. Definition of Subtraction: For all real numbers a and b, a b = a + ( b). Next, we give real number multiplication a similar treatment. Recall that we may denote the product of two real numbers a and b a variety of ways: ab, a b, a(b), (a)(b) and so on. We ll refrain from using a b for real number multiplication in this text with one notable exception in Definition A.. Properties of Real Number Multiplication Closure: For all real numbers a and b, ab is also a real number. Commutativity: For all real numbers a and b, ab = ba. Associativity: For all real numbers a, b and c, a(bc) = (ab)c. Identity: There is a real number 1 so that for all real numbers a, a 1 = a. Inverse: For all real numbers a 0, there is a real number 1 ( 1 a such that a a ) Definition of Division: For all real numbers a and b 0, a b = a b = a ( 1 b ). = 1. While most students and some faculty tend to skip over these properties or give them a cursory glance at best, 1 it is important to realize that the properties stated above are what drive the symbolic manipulation in all of Algebra. When listing a tally of more than two numbers, 1++ for example, we don t need to specify the order in which those numbers are added. Notice though, try as we might, we can add only two numbers at a time and it is the associative property of addition which assures us that we could organize this sum as (1+)+ or 1+(+). This brings up a note about grouping symbols. Recall that parentheses and brackets 1 Not unlike how Carl approached all the Elven poetry in The Lord of the Rings.

286 78 Algebra Review are used in order to specify which operations are to be performed first. In the absence of such grouping symbols, multiplication (and hence division) is given priority over addition (and hence subtraction). For example, 1 + = = 7, but (1 + ) = = 9. As you may recall, we can distribute the across the addition if we really wanted to do the multiplication first: (1 + ) = 1 + = + 6 = 9. More generally, we have the following. The Distributive Property and Factoring For all real numbers a, b and c: Distributive Property: a(b + c) = ab + ac and (a + b)c = ac + bc. Factoring: a ab + ac = a(b + c) and ac + bc = (a + b)c. a Or, as Carl calls it, reading the Distributive Property from right to left. It is worth pointing out that we didn t really need to list the Distributive Property both for a(b +c) (distributing from the left) and (a+b)c (distributing from the right), since the commutative property of multiplication gives us one from the other. Also, factoring is really the same equation as the distributive property, just read from right to left. These are the first of many redundancies in this section, and they exist in this review section for one reason only - in our experience, many students see these things differently so we will list them as such. It is hard to overstate the importance of the Distributive Property. For example, in the expression 5( + x), without knowing the value of x, we cannot perform the addition inside the parentheses first; we must rely on the distributive property here to get 5( + x) = x = x. The Distributive Property is also responsible for combining like terms. Why is x + x = 5x? Because x + x = ( + )x = 5x. We continue our review with summaries of other properties of arithmetic, each of which can be derived from the properties listed above. First up are properties of the additive identity 0. Suppose a and b are real numbers. Properties of Zero Zero Product Property: ab = 0 if and only if a = 0 or b = 0 (or both) Note: This not only says that 0 a = 0 for any real number a, it also says that the only way to get an answer of 0 when multiplying two real numbers is to have one (or both) of the numbers be 0 in the first place. ) Zeros in Fractions: If a 0, 0 a = 0 ( 1 a = 0. Note: The quantity a 0 is undefined.a a The expression 0 is technically an indeterminant form as opposed to being strictly undefined meaning that with Calculus 0 we can make some sense of it in certain situations. We ll talk more about this in Chapter??.

287 A. Real Number Arithmetic 79 The Zero Product Property drives most of the equation solving algorithms in Algebra because it allows us to take complicated equations and reduce them to simpler ones. For example, you may recall that one way to solve x +x 6 = 0 is by factoring the left hand side of this equation to get (x )(x +) = 0. From here, we apply the Zero Product Property and set each factor equal to zero. This yields x = 0 or x + = 0 so x = or x =. This application to solving equations leads, in turn, to some deep and profound structure theorems in Chapter??. Next up is a review of the arithmetic of negatives. On page 79 we first introduced the dash which we all recognize as the negative symbol in terms of the additive inverse. For example, the number (read negative ) is defined so that + ( ) = 0. We then defined subtraction using the concept of the additive inverse again so that, for example, 5 = 5+( ). In this text we do not distinguish typographically between the dashes in the expressions 5 and even though they are mathematically quite different. In the expression 5, the dash is a binary operation (that is, an operation requiring two numbers) whereas in, the dash is a unary operation (that is, an operation requiring only one number). You might ask, Who cares? Your calculator does - that s who! In the text we can write = 6 but that will not work in your calculator. Instead you d need to type to get 6 where the first dash comes from the +/ key and the second dash comes from the subtraction key. Given real numbers a and b we have the following. Properties of Negatives Additive Inverse Properties: a = ( 1)a and ( a) = a Products of Negatives: ( a)( b) = ab. Negatives and Products: ab = (ab) = ( a)b = a( b). Negatives and Fractions: If b is nonzero, a b = a b = a a and b b = a b. Distributing Negatives: (a + b) = a b and (a b) = a + b = b a. Factoring Negatives: a a b = (a + b) and b a = (a b). a Or, as Carl calls it, reading Distributing Negatives from right to left. An important point here is that when we distribute negatives, we do so across addition or subtraction only. This is because we are really distributing a factor of 1 across each of these terms: (a + b) = ( 1)(a + b) = ( 1)(a) + ( 1)(b) = ( a) + ( b) = a b. Negatives do not distribute across multiplication: ( ) ( ) ( ). Instead, ( ) = ( ) () = () ( ) = 6. The same sort of thing goes for fractions: 5 can be written as 5 or, but not 5 5. Don t worry. We ll review this in due course. And, yes, this is our old friend the Distributive Property! We re not just being lazy here. We looked at many of the big publishers Precalculus books and none of them use different dashes, either.

288 80 Algebra Review Speaking of fractions, we now review their arithmetic. Properties of Fractions Suppose a, b, c and d are real numbers. Assume them to be nonzero whenever necessary; for example, when they appear in a denominator. Identity Properties: a = a 1 and a a = 1. Fraction Equality: a b = c d if and only if ad = bc. Multiplication of Fractions: a b c d = ac bd. In particular: a b c = a b c 1 = ac b Note: A common denominator is not required to multiply fractions! Division a of Fractions: a b c d = a b d c = ad bc. In particular: 1 a b = b a and a b c = a b c 1 = a b 1 c = a bc Note: A common denominator is not required to divide fractions! Addition and Subtraction of Fractions: a b ± c b = a ± c b. Note: A common denominator is required to add or subtract fractions! Equivalent Fractions: a b = ad bd, since a b = a b 1 = a b d d = ad bd Note: The only way to change the denominator is to multiply both it and the numerator by the same nonzero value because we are, in essence, multiplying the fraction by 1. Reducing b Fractions: a d b d = a ad, since b bd = a b d d = a b 1 = a b. In particular, ab ab = a since b b = ab 1 b = a b 1 b = a 1 = a and b a a b = ( 1) (a b) (a = 1. b) Note: We may only cancel common factors from both numerator and denominator. a The old invert and multiply or fraction gymnastics play. b Or Canceling Common Factors - this is really just reading the previous property from right to left. Students make so many mistakes with fractions that we feel it is necessary to pause the narrative for a moment and offer you the following examples. Please take the time to read these carefully. In the main body of the text we will skip many of the steps shown here and it is your responsibility to understand the arithmetic behind the computations we use throughout the text. We deliberately limited these examples to nice numbers (meaning that the numerators and denominators of the fractions are small integers) and will discuss more complicated matters later. In the upcoming example, we will make use of the Fundamental Theorem of Arithmetic which essentially says that every natural number has a unique prime factorization. Thus lowest terms is clearly defined when reducing the fractions you re about to see.

289 A. Real Number Arithmetic 81 Perform the indicated operations and simplify. By simplify here, we mean to have the final answer written in the form a b where a and b are integers which have no common factors. Said another way, we want a b in lowest terms (() + 1)( ( )) 5(4 7) 4 () ( ) ( ) ( ) ( ) ( 1 ) ( ) 4 ( ) Solution. 1. It may seem silly to start with an example this basic but experience has taught us not to take much for granted. We start by finding the lowest common denominator and then we rewrite the fractions using that new denominator. Since 4 and 7 are relatively prime, meaning they have no factors in common, the lowest common denominator is 4 7 = = = = 1 8 Equivalent Fractions Multiplication of Fractions Addition of Fractions The result is in lowest terms because 1 and 8 are relatively prime so we re done.. We could begin with the subtraction in parentheses, namely 47 7, and then subtract that result 0 from 5. It s easier, however, to first distribute the negative across the quantity in parentheses and 1 then use the Associative Property to perform all of the addition and subtraction in one step. 4 The lowest common denominator 5 for all three fractions is ( ) 5 = = = = Distribute the Negative Equivalent Fractions Multiplication of Fractions Addition and Subtraction of Fractions The numerator and denominator are relatively prime so the fraction is in lowest terms and we have our final answer. 4 See the remark on page 79 about how we add We could have used 1 0 = 1080 as our common denominator but then the numerators would become unnecessarily large. It s best to use the lowest common denominator.

290 8 Algebra Review. What we are asked to simplify in this problem is known as a complex or compound fraction. Simply put, we have fractions within a fraction. 6 The longest division line 7 acts as a grouping symbol, quite literally dividing the compound fraction into a numerator (containing fractions) and a denominator (which in this case does not contain fractions). The first step to simplifying a compound fraction like this one is to see if you can simplify the little fractions inside it. To that end, we clean up the fractions in the numerator as follows = = = ( ) Properties of Negatives Distribute the Negative We are left with a compound fraction with decimals. We could replace.1 with 1 but that would 100 make a mess. 8 It s better in this case to eliminate the decimal by multiplying the numerator and denominator of the fraction with the decimal in it by 100 (since = 1 is an integer) as shown below = = We now perform the subtraction in the numerator and replace 0.1 with 1 in the denominator. This 100 will leave us with one fraction divided by another fraction. We finish by performing the division by a fraction is multiplication by the reciprocal trick and then cancel any factors that we can = = = = = = 50 1 The last step comes from the factorizations = 4 50 and 98 = We are given another compound fraction to simplify and this time both the numerator and denominator contain fractions. As before, the longest division line acts as a grouping symbol to separate the 6 Fractionception, perhaps? 7 Also called a vinculum. 8 Try it if you don t believe us.

291 A. Real Number Arithmetic 8 numerator from the denominator ( ) 4 ( ) = ( ) 4 ( ( ) ( )) Hence, one way to proceed is as before: simplify the numerator and the denominator then perform the division by a fraction is the multiplication by the reciprocal trick. While there is nothing wrong with this approach, we ll use our Equivalent Fractions property to rid ourselves of the compound nature of this fraction straight away. The idea is to multiply both the numerator and denominator by the lowest common denominator of each of the smaller fractions - in this case, 4 5 = 10. ( ) ( ) 10 4 ( ( ) ( )) = ( ( ) ( )) Equivalent Fractions ( ) ( ) 1 7 (10) (10) 5 4 = ( ) ( ) Distributive Property 1 7 (1)(10) + (10) 5 4 = = = (1 4) (7 5) 10 + (1 7) 88 5 = = 5 04 Multiply fractions Factor and cancel Since 5 = 11 and 04 = 17 have no common factors our result is in lowest terms which means we are done. 5. This fraction may look simpler than the one before it, but the negative signs and parentheses mean that we shouldn t get complacent. Again we note that the division line here acts as a grouping symbol. That is,

292 84 Algebra Review (() + 1)( ( )) 5(4 7) 4 () = ((() + 1)( ( )) 5(4 7)) (4 ()) This means that we should simplify the numerator and denominator first, then perform the division last. We tend to what s in parentheses first, giving multiplication priority over addition and subtraction. (() + 1)( ( )) 5(4 7) 4 () = = (4 + 1)( + ) 5( ) 4 6 (5)(0) = = 15 Properties of Negatives Since 15 = 5 and have no common factors, we are done. 6. In this problem, we have multiplication and subtraction. Multiplication takes precedence so we perform it first. Recall that to multiply fractions, we do not need to obtain common denominators; rather, we multiply the corresponding numerators together along with the corresponding denominators. Like the previous example, we have parentheses and negative signs for added fun! ( ) ( ) ( 4 5 ) ( 1 ) 1 5 = ( 1) 5 1 = = = 65 = 6 65 Multiply fractions Properties of Negatives Add numerators Since 64 = 7 and 65 = 5 1 have no common factors, our answer 6 is in lowest terms and 65 we are done. Of the issues discussed in the previous set of examples none causes students more trouble than simplifying compound fractions. We presented two different methods for simplifying them: one in which we simplified the overall numerator and denominator and then performed the division and one in which we removed the compound nature of the fraction at the very beginning. We encourage the reader to go back and use both methods on each of the compound fractions presented. Keep in mind that when a compound fraction is encountered in the rest of the text it will usually be simplified using only one method and we may not choose your favorite method. Feel free to use the other one in your notes.

293 A. Real Number Arithmetic 85 Next, we review exponents and their properties. Recall that can be written as because exponential notation expresses repeated multiplication. In the expression, is called the base and is called the exponent. In order to generalize exponents from natural numbers to the integers, and eventually to rational and real numbers, it is helpful to think of the exponent as a count of the number of factors of the base we are multiplying by 1. For instance, = 1 (three factors of two) = 1 ( ) = 8. From this, it makes sense that 0 = 1 (zero factors of two) = 1. What about? The in the exponent indicates that we are taking away three factors of two, essentially dividing by three factors of two. So, = 1 (three factors of two) = 1 ( ) = We summarize the properties of integer exponents below. Properties of Integer Exponents Suppose a and b are nonzero real numbers and n and m are integers. Product Rules: (ab) n = a n b n and a n a m = a n+m. ( a ) n a n an Quotient Rules: = and b bn a m = an m. Power Rule: (a n ) m = a nm. Negatives in Exponents: a n = 1 a n. ( a ) ( n n b In particular, = = b a) bn a n and 1 a n = an. Zero Powers: a 0 = 1. Note: The expression 0 0 is an indeterminate form. a Powers of Zero: For any natural number n, 0 n = 0. Note: The expression 0 n for integers n 0 is not defined. a See the comment regarding 0 on page = 1 8. While it is important the state the Properties of Exponents, it is also equally important to take a moment to discuss one of the most common errors in Algebra. It is true that (ab) = a b (which some students refer to as distributing the exponent to each factor) but you cannot do this sort of thing with addition. That is, in general, (a + b) a + b. (For example, take a = and b = 4.) The same goes for any other powers.

294 86 Algebra Review With exponents now in the mix, we can now state the Order of Operations Agreement. Order of Operations Agreement When evaluating an expression involving real numbers: 1. Evaluate any expressions in parentheses (or other grouping symbols).. Evaluate exponents.. Evaluate multiplication and division as you read from left to right. 4. Evaluate addition and subtraction as you read from left to right. We note that there are many useful mnemonic devices for remembering the order of operations. a a Our favorite is Please entertain my dear auld Sasquatch. For example, + 4 = + 16 = + 48 = 50. Where students get into trouble is with things like. If we think of this as 0, then it is clear that we evaluate the exponent first: = 0 = 0 9 = 9. In general, we interpret a n = (a n ). If we want the negative to also be raised to a power, we must write ( a) n instead. To summarize, = 9 but ( ) = 9. Of course, many of the properties we ve stated in this section can be viewed as ways to circumvent the order of operations. We ve already seen how the distributive property allows us to simplify 5( + x) by performing the indicated multiplication before the addition that s in parentheses. Similarly, consider trying to evaluate The Order of Operations Agreement demands that the exponents be dealt with first, however, trying to compute 017 is a challenge, even for a calculator. One of the Product Rules of Exponents, however, allow us to rewrite this product, essentially performing the multiplication first, to get: = = 8. Let s take a break and enjoy another example. Perform the indicated operations and simplify. 1.. (4 )( 4) (4) (4 ). 1( 5)( 5 + ) 4 + 6( 5) ( 4)( 5 + ) 5 ( ) 5 51 ( ) ( 6 1 ) ( ) Solution. 1. We begin working inside the parentheses then deal with the exponents before working through the other operations. As we saw in Example A., the division here acts as a grouping symbol, so we

295 A. Real Number Arithmetic 87 save the division to the end. (4 )( 4) (4) (4 ) = = ()(8) (4) ()(8) 16 () = = = 0. As before, we simplify what s in the parentheses first, then work our way through the exponents, multiplication, and finally, the addition. 1( 5)( 5 + ) 4 + 6( 5) ( 4)( 5 + ) 5 = 1( 5)( ) 4 + 6( 5) ( 4)( ) 5 ( ) ( ) 1 1 = 1( 5) ( ) 4 + 6( 5) ( 4) ( ) ( ) ( ) = 1( 5) + 6(5)( 4) 16 ( ) ( ) 1 1 = ( 60) + ( 600) 16 = 60 ( ) = = = = 4 = 60 4 = 15. The Order of Operations Agreement mandates that we work within each set of parentheses first, giving precedence to the exponents, then the multiplication, and, finally the division. The trouble with this approach is that the exponents are so large that computation becomes a trifle unwieldy. What we observe, however, is that the bases of the exponential expressions, and 4, occur in both the numerator and denominator of the compound fraction. This gives us hope that we can use some of the Properties of Exponents (the Quotient Rule, in particular) to help us out. Our first step here is to invert and multiply. We see immediately that the 5 s cancel after which we group the powers of

296 88 Algebra Review together and the powers of 4 together and apply the properties of exponents. ( ) ( ) = = = ( ) 1 = = 4 = ( ) = 9 = We have yet another instance of a compound fraction so our first order of business is to rid ourselves of the compound nature of the fraction like we did in Example A.. To do this, however, we need to tend to the exponents first so that we can determine what common denominator is needed to simplify the fraction. ( ) 1 ( ) ( ) ( ) ( ) = ( ) = ( ) = ( ) ( ) ( ) = = ( ) = Since 10 and 119 have no common factors, we are done. = ( ) = One of the places where the properties of exponents play an important role is in the use of Scientific Notation. The basis for scientific notation is that since we use decimals (base ten numerals) to represent real numbers, we can adjust where the decimal point lies by multiplying by an appropriate power of 10. This allows scientists and engineers to focus in on the significant digits 9 of a number - the nonzero values - and adjust for the decimal places later. For instance, 61 = and 0.0 =. 10. Notice here that we revert to using the familiar to indicate multiplication. 10 In general, we arrange the real number so exactly one non-zero digit appears to the left of the decimal point. We make this idea precise in the following: A real number is written in Scientific Notation if it has the form ±n.d 1 d k where n is a natural number, d 1, d, etc., are whole numbers, and k is an integer. 9 Awesome pun! 10 This is the notable exception we alluded to earlier.

297 A. Real Number Arithmetic 89 On calculators, scientific notation may appear using an E or EE as opposed to the symbol. For instance, while we will write in the text, the calculator may display 6.0 E or 6.0 EE. Perform the indicated operations and simplify. Write your final answer in scientific notation, rounded to two decimal places. ( ) ( ) ( ) 100 Solution. 1. As mentioned earlier, the point of scientific notation is to separate out the significant parts of a calculation and deal with the powers of 10 later. In that spirit, we separate out the powers of 10 in both the numerator and the denominator and proceed as follows ( ) ( ) = (6.66)(.14) = = = = We are asked to write our final answer in scientific notation, rounded to two decimal places. To do this, we note that = , so = = = Our final answer, rounded to two decimal places, is We could have done that whole computation on a calculator so why did we bother doing any of this by hand in the first place? The answer lies in the next example.. If you try to compute ( ) 100 using most hand-held calculators, you ll most likely get an overflow error. It is possible, however, to use the calculator in combination with the properties of exponents to compute this number. Using properties of exponents, we get: ( ) 100 = (.1) ( ) 100 = ( ) ( ) = ( ) ( ) = = To two decimal places our answer is

298 90 Algebra Review We close our review of real number arithmetic with a discussion of roots and radical notation. Just as subtraction and division were defined in terms of the inverse of addition and multiplication, respectively, we define roots by undoing natural number exponents. Let a be a real number and let n be a natural number. If n is odd, then the principal n th root of a (denoted n a) is the unique real number satisfying ( n a ) n = a. If n is even, n a is defined similarly provided a 0 and n a 0. The number n is called the index of the root and the number a is called the radicand. For n =, we write a instead of a. The reasons for the added stipulations for even-indexed roots in Definition A. can be found in the Properties of Negatives. First, for all real numbers, x even power 0, which means it is never negative. Thus if a is a negative real number, there are no real numbers x with x even power = a. This is why if n is even, n a only exists if a 0. The second restriction for even-indexed roots is that n a 0. This comes from the fact that x even power = ( x) even power, and we require n a to have just one value. So even though 4 = 16 and ( ) 4 = 16, we require 4 16 = and ignore. Dealing with odd powers is much easier. For example, x = 8 has one and only one real solution, namely x =, which means not only does 8 exist, there is only one choice, namely 8 =. Of course, when it comes to solving x 51 = 117, it s not so clear that there is one and only one real solution, let alone that the solution is Such pills are easier to swallow once we ve thought a bit about such equations graphically, 11 and ultimately, these things come from the completeness property of the real numbers mentioned earlier. We list properties of radicals below as a theorem as opposed to a definition since they can be justified using the properties of exponents. Theorem A.1. Properties of Radicals: Let a and b be real numbers and let m and n be natural numbers. If n a and n b are real numbers, then Product Rule: n ab = n a n b Quotient Rule: n a b = n a n b, provided b 0. Power Rule: n a m = ( n a ) m The proof of Theorem A.1 is based on the definition of the principal n th root and the Properties of Exponents. To establish the product rule, consider the following. If n is odd, then by definition n ab is the unique real number such that ( n ab) n = ab. Given that ( n a n b) n = ( n a) n ( n b) n = ab as well, it must be the case n that ab = n a n b. If n is even, then n ab is the unique non-negative real number such that ( n ab) n = ab. n Note that since n is even, a and n b are also non-negative thus n a n b 0 as well. Proceeding as n above, we find that ab = n a n b. The quotient rule is proved similarly and is left as an exercise. The power rule results from repeated application of the product rule, so long as n a is a real number to start with. 1 We leave that as an exercise as well. 11 See Chapter??. 1 Otherwise we d run into an interesting paradox. See Section A.11.

299 A. Real Number Arithmetic 91 We pause here to point out one of the most common errors students make when working with radicals. Obviously 9 =, 16 = 4 and = 5 = 5. Thus we can clearly see that 5 = 5 = = + 4 = 7 because we all know that 5 7. The authors urge you to never consider distributing n roots or exponents. It s wrong and no good will come of it because in general a + b n a + n b. Since radicals have properties inherited from exponents, they are often written as such. We define rational exponents in terms of radicals in the box below. Let a be a real number, let m be an integer and let n be a natural number. a 1 n a m n = n a whenever n a is a real number. a = ( n a ) m = n a m whenever n a is a real number. a If n is even we need a 0. It would make life really nice if the rational exponents defined in Definition A. had all of the same properties that integer exponents have as listed on page 87 - but they don t. Why not? Let s look at an example to see what goes wrong. Consider the Product Rule which says that (ab) n = a n b n and let a = 16, b = 81 and n = 1 4. Plugging the values into the Product Rule yields the equation (( 16)( 81))1/4 = ( 16) 1/4 ( 81) 1/4. The left side of this equation is 196 1/4 which equals 6 but the right side is undefined because neither root is a real number. Would it help if, when it comes to even roots (as signified by even denominators in the fractional exponents), we ensure that everything they apply to is non-negative? That works for some of the rules - we leave it as an exercise to see which ones - but does not work for the Power Rule. Consider the expression ( a /) /. Applying the usual laws of exponents, we d be tempted to simplify this as ( a /) / = a = a 1 = a. However, if we substitute a = 1 and apply Definition A., we find ( 1) / = ( 1 ) = ( 1) = 1 so that ( ( 1) /) / = 1 / = ( 1 ) = 1 = 1. Thus in this case we have ( a / ) / a even though all of the roots were defined. It is true, however, that ( a / ) / = a and we leave this for the reader to show. The moral of the story is that when simplifying powers of rational exponents where the base is negative or worse, unknown, it s usually best to rewrite them as radicals. 1 Perform the indicated operations and simplify. 1. ( 4) ( 4) 4()( ) (). ( 54) 4.. ( ) ( 1 ) ( ) 1/ ( ) ( 1 ) ( ) / Much to Jeff s chagrin. He s fairly traditional and therefore doesn t care much for radicals.

300 9 Algebra Review Solution. 1. We begin in the numerator and note that the radical here acts a grouping symbol, 14 so our first order of business is to simplify the radicand. ( 4) ( 4) 4()( ) () = ( 4) 16 4()( ) () = ( 4) 16 4( 6) () = ( 4) 16 ( 4) () = ( 4) () = ( 4) 40 () As you may recall, 40 can be factored using a perfect square as 40 = 4 10 so we use the product rule of radicals to write 40 = 4 10 = 4 10 = 10. This lets us factor a out of both terms in the numerator, eventually allowing us to cancel it with a factor of in the denominator. ( 4) 40 () = ( 4) 10 () = 10 () = ( 10) () = 4 10 () = ( 10) () = 10 Since the numerator and denominator have no more common factors, 15 we are done.. Once again we have a compound fraction, so we first simplify the exponent in the denominator to see which factor we ll need to multiply by in order to clean up the fraction. 14 The line extending horizontally from the square root symbol is, you guessed it, another vinculum. 15 Do you see why we aren t canceling the remaining s?

301 A. Real Number Arithmetic 9 ( ) ( ) = 1 = = = ( ) ( ) = ( ) 1 ( ) ( ) = 1 1 ( ) ( ( )) = ( ) ( ) 1 9 ( ) ( ) = =. Working inside the parentheses, we first encounter. While the isn t a perfect cube, 16 we may think of = ( 1)(). Since ( 1) = 1, which is a perfect cube, we may write = ( 1)() = 1 =. When it comes to 54, we may write it as ( 7)() = 7 =. So, 54 = ( ) = +. At this stage, we can simplify + =. You may remember this as being called combining like radicals, but it is in fact just another application of the distributive property: + = ( 1) + = ( 1 + ) =. Putting all this together, we get: ( 54) = ( + ) = ( ) = ( ) = 4 = 4 4 There are no perfect integer cubes which are factors of 4 (apart from 1, of course), so we are done. 16 Of an integer, that is!

302 94 Algebra Review 4. We start working in the parentheses and get a common denominator to subtract the fractions: 9 4 = = = 4 The denominators in the fractional exponents are odd, so we can proceed by using the properties of exponents: ( ) 1/ ( ) ( ) ( ) / ( ) 1/ ( ) ( ) ( ) / = + ( 4 ) 4 4 ( ) ( ) ( ) ( ) 1/ / = + (4) 1/ 4 ( ) ( ) ( ) ( ) ( ) 1/ 9 1 (4) / = + (4) 1/ 4 ( ) / = ( )1/ 4 1/ / 4 ( ) / = ( )1/ 4 1/ + 4 / ( ) / = ( )1/ 4 1/ + 4/ ( ) / At this point, we could start looking for common denominators but it turns out that these fractions reduce even further. Since 4 =, 4 1/ = ( ) 1/ = /. Similarly, 4 / = ( ) / = 4/. The expressions ( ) 1/ and ( ) / contain negative bases so we proceed with caution and convert them back to radical notation to get: ( ) 1/ = = = 1/ and ( ) / = ( ) = ( ) = ( ) = /. Hence: ( ) 1/ + 4/ = ( 1/ ) + 4/ 4 1/ ( ) / / / = 1 ( 1/ ) + 1 4/ / 1 / = 1 / ( 1/ ) + 1 / 4/ 1 = 1/ ( 1/ ) + 1/ 1/ = 1/ 1/ + 1/ 1/ = 0 We close this section with a note about simplifying. In the preceding examples we used nice numbers because we wanted to show as many properties as we could per example. This then begs the question What happens when the numbers are not nice? Unfortunately, the answer is Not much simplifying can be done. Take, for example, 7 π π + 4 = π π 11 π 11 Sadly, that s as good as it gets.

303 A. Real Number Arithmetic 95 A..1 Exercises In Exercises 1 -, perform the indicated operations and simplify ( + ) ( ) ( ) 7. () (4 1) ( ) 5( ) ( ) () 7 () () (( 1) 1) (( 1) + 1) 1. ( ) 4 1 ( ) ( ) ( ) 6 ( ) 4 1 ( ( 5 ) ) ( ( 5 ) ) ( ) ( 8) / 9 / 4. ( 9 5. ( 4) + (5 ) 6. ( ( 1)) + ( 1 ) 7. ( 5 5) + ( 18 8) ) / () 4()( 1) () 0. ( 4) + ( 4) 4(1)( 1) (1) 1. ( 5)( 5 + 1) 1 + ( 5) ( 1)( 5 + 1). (4) (4) ( 1 ) ((4) + 1) 1/ (). ( 7) 1 ( 7) + ( 7) ( 1 ) (1 ( 7)) / ( 1) 4. With the help of your calculator, find ( ) 117. Write your final answer, using scientific notation, rounded to two decimal places. (See Example A..) 5. Prove the Quotient Rule and Power Rule stated in Theorem A Discuss with your classmates how you might attempt to simplify the following. (a) (b) (c) π + 7 π

304 96 Algebra Review A.. Answers Undefined Undefined = 6/

305 A. The Cartesian Plane 97 A. The Cartesian Plane A..1 The Cartesian Coordinate Plane In order to visualize the pure excitement that is Precalculus, we need to unite Algebra and Geometry. Simply put, we must find a way to draw algebraic things. Let s start with possibly the greatest mathematical achievement of all time: the Cartesian Coordinate Plane. 1 Imagine two real number lines crossing at a right angle at 0 as drawn below. y x 4 The horizontal number line is usually called the x-axis while the vertical number line is usually called the y-axis. As with things in the real world, however, it s best not to get too caught up with labels. Think of x and y as generic label placeholders, in much the same way as the variables x and y are placeholders for real numbers. The letters we choose to identify with the axes depend on the context. For example, if we were plotting the relationship between time and the number of Sasquatch sightings, we might label the horizontal axis as the t-axis (for time ) and the vertical axis the N-axis (for number of sightings.) As with the usual number line, we imagine these axes extending off indefinitely in both directions. Having two number lines allows us to locate the positions of points off of the number lines as well as points on the lines themselves. For example, consider the point P on the next page. To use the numbers on the axes to label this point, we imagine dropping a vertical line from the x-axis to P and extending a horizontal line from the y-axis to P. This process is sometimes called projecting the point P to the x- (respectively y-) axis. We then describe the point P using the ordered pair (, 4). The first number in the ordered pair is called the abscissa or x-coordinate and the second is called the ordinate or y-coordinate. Again, the names of the coordinates 1 So named in honor of René Descartes. Usually extending off towards infinity is indicated by arrows, but here, the arrows are used to indicate the direction of increasing values of x and y.

306 98 Algebra Review can vary depending on the context of the application. If, as in the previous paragraph, the horizontal axis represented time and the vertical axis represented the number of Sasquatch sightings, the first coordinate would be called the t-coordinate and the second coordinate would be the N-coordinate. What s important is that we maintain the convention that the abscissa (first coordinate) always corresponds to the horizontal position, while the ordinate (second coordinate) always corresponds to the vertical position. Taken together, the ordered pair (, 4) comprise the Cartesian coordinates of the point P. In practice, the distinction between a point and its coordinates is blurred; for example, we often speak of the point (, 4). We can think of (, 4) as instructions on how to reach P from the origin (0, 0) by moving units to the right and 4 units downwards. Notice that the order in the ordered pair is important, as are the signs of the numbers in the pair. If we wish to plot the point ( 4, ), we would move to the left 4 units from the origin and then move upwards units, as below on the right. y y 4 1 ( 4, ) x x 4 P 4 P(, 4) When we speak of the Cartesian Coordinate Plane, we mean the set of all possible ordered pairs (x, y) as x and y take values from the real numbers. Below is a summary of some basic, but nonetheless important, facts about Cartesian coordinates. Important Facts about the Cartesian Coordinate Plane (a, b) and (c, d) represent the same point in the plane if and only if a = c and b = d. (x, y) lies on the x-axis if and only if y = 0. (x, y) lies on the y-axis if and only if x = 0. The origin is the point (0, 0). It is the only point common to both axes. Also called the rectangular coordinates of P see Section?? for more details.

307 A. The Cartesian Plane 99 Plot the following points: A(5, 8), B ( 5, ), C( 5.8, ), D(4.5, 1), E(5, 0), F (0, 5), G( 7, 0), H(0, 9), O(0, 0).(The letter O is almost always reserved for the origin.) Solution. To plot these points, we start at the origin and move to the right if the x-coordinate is positive; to the left if it is negative. Next, we move up if the y-coordinate is positive or down if it is negative. If the x-coordinate is 0, we start at the origin and move along the y-axis only. If the y-coordinate is 0 we move along the x-axis only. 9 y B ( 5, ) G( 7, 0) C( 5.8, ) A(5, 8) F (0, 5) O(0, 0) E(5, 0) D(4.5, 1) H(0, 9) x The axes divide the plane into four regions called quadrants. They are labeled with Roman numerals and proceed counterclockwise around the plane: y Quadrant II x < 0, y > Quadrant I x > 0, y > x Quadrant III x < 0, y < 0 4 Quadrant IV x > 0, y < 0

308 00 Algebra Review For example, (1, ) lies in Quadrant I, ( 1, ) in Quadrant II, ( 1, ) in Quadrant III and (1, ) in Quadrant IV. If a point other than the origin happens to lie on the axes, we typically refer to that point as lying on the positive or negative x-axis (if y = 0) or on the positive or negative y-axis (if x = 0). For example, (0, 4) lies on the positive y-axis whereas ( 117, 0) lies on the negative x-axis. Such points do not belong to any of the four quadrants. One of the most important concepts in all of Mathematics is symmetry. 4 There are many types of symmetry in Mathematics, but three of them can be discussed easily using Cartesian Coordinates. Two points (a, b) and (c, d) in the plane are said to be symmetric about the x-axis if a = c and b = d symmetric about the y-axis if a = c and b = d symmetric about the origin if a = c and b = d Schematically, y Q( x, y) P(x, y) 0 x R( x, y) S(x, y) In the above figure, P and S are symmetric about the x-axis, as are Q and R; P and Q are symmetric about the y-axis, as are R and S; and P and R are symmetric about the origin, as are Q and S. Let P be the point (, ). Find the points which are symmetric to P about the: 1. x-axis. y-axis. origin Check your answer by plotting the points. Solution. The figure after Definition A..1 gives us a good way to think about finding symmetric points in terms of taking the opposites of the x- and/or y-coordinates of P(, ). 1. To find the point symmetric about the x-axis, we replace the y-coordinate of with its opposite to get (, ). 4 According to Carl. Jeff thinks symmetry is overrated.

309 A. The Cartesian Plane 01. To find the point symmetric about the y-axis, we replace the x-coordinate of with its opposite ( ) = to get (, ).. To find the point symmetric about the origin, we replace both the x- and y-coordinates with their opposites to get (, ). y P(, ) (, ) x (, ) (, ) One way to visualize the processes in the previous example is with the concept of a reflection. If we start with our point (, ) and pretend that the x-axis is a mirror, then the reflection of (, ) across the x-axis would lie at (, ). If we pretend that the y-axis is a mirror, the reflection of (, ) across that axis would be (, ). If we reflect across the x-axis and then the y-axis, we would go from (, ) to (, ) then to (, ), and so we would end up at the point symmetric to (, ) about the origin. We summarize and generalize this process below. To reflect a point (x, y) about the: Reflections x-axis, replace y with y. y-axis, replace x with x. origin, replace x with x and y with y. A.. Distance in the Plane Another fundamental concept in Geometry is the notion of length. If we are going to unite Algebra and Geometry using the Cartesian Plane, then we need to develop an algebraic understanding of what distance in the plane means. Before we can do that, we need to state what we believe is the most important theorem in all of Geometry: The Pythagorean Theorem.

310 0 Algebra Review Theorem A.. The Pythagorean Theorem: The triangle ABC shown below is a right triangle if and only if a + b = c B c a A b C A proof of this theorem will be given in Section C.1. The theorem actually says two different things. If we know that a + b = c then the angle C must be a right angle. If we know geometrically that C is already a right angle then we have that a + b = c. We need the latter statement in the discussion which follows. Suppose we have two points, P (x 0, y 0 ) and Q (x 1, y 1 ), in the plane. By the distance d between P and Q, we mean the length of the line segment joining P with Q. (Remember, given any two distinct points in the plane, there is a unique line containing both points.) Our goal now is to create an algebraic formula to compute the distance between these two points. Consider the generic situation below on the left. d Q (x 1, y 1 ) d Q (x 1, y 1 ) P (x 0, y 0 ) P (x 0, y 0 ) (x 1, y 0 ) With a little more imagination, we can envision a right triangle whose hypotenuse has length d as drawn above on the right. From the latter figure, we see that the lengths of the legs of the triangle are x 1 x 0 and y 1 y 0 so the Pythagorean Theorem gives us x 1 x 0 + y 1 y 0 = d (x 1 x 0 ) + (y 1 y 0 ) = d (Do you remember why we can replace the absolute value notation with parentheses?) By extracting the square root of both sides of the second equation and using the fact that distance is never negative, we get Equation A.1. The Distance Formula: The distance d between the points P (x 0, y 0 ) and Q (x 1, y 1 ) is: d = (x 1 x 0 ) + (y 1 y 0 ) A couple of remarks about Equation A.1 are in order. First, it is not always the case that the points P and Q lend themselves to constructing such a triangle. If the points P and Q are arranged vertically or horizontally, or describe the exact same point, we cannot use the above geometric argument to derive the distance formula. It is left to the reader in Exercise 1 to verify Equation A.1 for these cases. Second, distance is a length. So, technically, the number we obtain from the distance formula has some attached units of length. In this text, we ll adopt the convention that the phrase units refers to some generic units of

311 A. The Cartesian Plane 0 length. 5 Our next example gives us an opportunity to test drive the distance formula as well as brush up on some arithmetic and prerequisite algebra. Find and simplify the distance between the following sets of points: 1. P(, ) and Q(1, ). R ( 1, ( ) and S 4, ) 1 5. T (, 0) and V ( 1, 5) 4. O(0, 0) and P(x, y). Solution. In each case, we apply the distance formula, Equation A.1 with the first point listed taken as (x 0, y 0 ) and the second point taken as (x 1, y 1 ) With (, ) = (x 0, y 0 ) and (1, ) = (x 1, y 1 ), we get d = (x 1 x 0 ) + (y 1 y 0 ) = (1 ( )) + ( ) = = 45 = 9 5 = 9 5 For nonnegative numbers, ab = a b. = 5 So the distance is 5 units.. With ( 1, ) = (x0, y 0 ) and ( 4, 5) 1 = (x1, y 1 ), we get d = = = = = = = (x 1 x 0 ) + (y 1 y 0 ) ( 1 ( 4 ) ) Get common denominators to add and subtract fractions. ( 1 ) ( ) Since ( ) a 5 b = a, b 0. b For nonnegative numbers, a b = a b, b 0. So the distance is units. 5 As a result, we ll measure area with square units, or units and volume with cubic units, or units. 6 This choice is completely arbitrary. The reader is encouraged to work these examples taking the first point listed as (x 1, y 1 ) and the second point listed as (x 0, y 0 ) and verifying the distance works out to be the same. Can you see why the order of the subtraction in Equation A.1 ultimately doesn t matter?

312 04 Algebra Review. With (, 0) = (x 0, y 0 ) and ( 1, 5) = (x 1, y 1 ), we get d = (x 1 x 0 ) + (y 1 y 0 ) ( 1 ) ( ) = + 5 ( 0) ( ) ( ) = Simplify the radicals to get like terms. ( ) ( ) = + 5 = Since ( a) = a and (b a) = b ( a). = 48 = 4 So the distance is 4 units. 4. With (0, 0) = (x 0, y 0 ) and (x, y) = (x 1, y 1 ), we get d = = (x 1 x 0 ) + (y 1 y 0 ) (x 0) + (y 0) = x + y As tempting as it may look, x + y does not, in general, reduce to x + y or even x + y. So, in this case, the best we can do is state that the distance is x + y units. Related to finding the distance between two points is the problem of finding the midpoint of the line segment connecting two points. Given two points, P (x 0, y 0 ) and Q (x 1, y 1 ), the midpoint M of P and Q is defined to be the point on the line segment connecting P and Q whose distance from P is equal to its distance from Q. Q (x 1, y 1 ) P (x 0, y 0 ) M If we think of reaching M by going halfway over and halfway up we get the following formula. Equation A.. The Midpoint Formula: The midpoint M of the line segment connecting P (x 0, y 0 ) and Q (x 1, y 1 ) is: ( x0 + x 1 M =, y 0 + y ) 1

313 A. The Cartesian Plane 05 If we let d denote the distance between P and Q, we leave it as Exercise 1 to show that the distance between P and M is d/ which is the same as the distance between M and Q. This suffices to show that Equation A. gives the coordinates of the midpoint. Find the midpoint of the line segment connecting the following pairs of points: 1. P(, ) and Q(1, ). R ( 1, ( ) and S 4, ) 1 5. T (, 0) and V ( 1, 5) 4. O(0, 0) and P(x, y). Solution. As with Example A.., in each case, we apply the midpoint formula, Equation A. with the first point listed taken as (x 0, y 0 ) and the second point taken as (x 1, y 1 ). 7 We also note that midpoints are points, which means all of our answers should be ordered pairs. 1. With (, ) = (x 0, y 0 ) and (1, ) = (x 1, y 1 ), we get M = = = ( x0 + x 1, y 0 + y ) 1 ( ( ) + 1, + ( ) ) = ( 1 ), 0 ( 1, 0 ) The midpoint is ( 1, 0).. With ( 1, ) = (x0, y 0 ) and ( 4, 5) 1 = (x1, y 1 ), we get M = = = = ( x0 + x 1, y 0 + y ) 1 ( ) 1 + 4, (( 1 + ) ( 4 4, + ) ) ( 5 8, 1 ) 0 Simplify compound fractions. The midpoint is ( 5 8, 1 0). 7 As in Example A.., this choice is also completely arbitrary. The reader is encouraged to work these examples taking the first point listed as (x 1, y 1 ) and the second point listed as (x 0, y 0 ) and verifying the midpoint works out to be the same. Can you see why the order of the points in Equation A. doesn t matter?

314 06 Algebra Review. With (, 0) = (x 0, y 0 ) and ( 1, 5) = (x 1, y 1 ), we get ( x0 + x 1 M =, y 0 + y ) 1 ( + 1 =, 0 + ) 5 ( + =, 5 + ) 5 ( ) 5 =, ( ) The midpoint is, 5. Simplify radicals to get like terms. 4. With (0, 0) = (x 0, y 0 ) and (x, y) = (x 1, y 1 ), we get M = = = ( x0 + x 1, y 0 + y ) 1 ( x + 0, y + 0 ) ( x, y ) The midpoint is ( x, y ). We close with a more abstract application of the Midpoint Formula. We will expand upon this example in Example A.5 in Section A.5. If a b, show that the line y = x equally divides the line segment with endpoints (a, b) and (b, a). Solution. To prove the claim, we use Equation A. to find the midpoint M = = ( a + b, b + a ) ( a + b, a + b ) Since the x and y coordinates of this point are the same, we find that the midpoint lies on the line y = x, as required.

315 A. The Cartesian Plane 07 A.. Exercises 1. Plot and label the points A(, 7), B(1., ), C(π, 10), D(0, 8), E( 5.5, 0), F ( 8, 4), G(9., 7.8) and H(7, 5) in the Cartesian Coordinate Plane given below. y x For each point given in Exercise 1 above Identify the quadrant or axis in/on which the point lies. Find the point symmetric to the given point about the x-axis. Find the point symmetric to the given point about the y-axis. Find the point symmetric to the given point about the origin. In Exercises - 10, find the distance d between the points and the midpoint M of the line segment which connects them.. (1, ), (, 5) 4. (, 10), ( 1, )

316 08 Algebra Review ( ) ( ) ( 1 5., 4,, 1 6., ) ( ) 7,, ( , 6 ) (, 11 ) 5 5, (, ), ( 8, 1 ) 5 9. ( 45, 1 ), ( 0, 7 ) (. 10., 1 ) ( ),, Let s assume that we are standing at the origin and the positive y-axis points due North while the positive x-axis points due East. Our Sasquatch-o-meter tells us that Sasquatch is miles West and 4 miles South of our current position. What are the coordinates of his position? How far away is he from us? If he runs 7 miles due East what would his new position be? 1. Verify the Distance Formula A.1 for the cases when: (a) The points are arranged vertically. (Hint: Use P(a, y 0 ) and Q(a, y 1 ).) (b) The points are arranged horizontally. (Hint: Use P(x 0, b) and Q(x 1, b).) (c) The points are actually the same point. (You shouldn t need a hint for this one.) 1. Verify the Midpoint Formula by showing the distance between P(x 1, y 1 ) and M and the distance between M and Q(x, y ) are both half of the distance between P and Q. 14. Show that the points A, B and C below are the vertices of a right triangle. (a) A(, ), B( 6, 4), and C(1, 8) (b) A(, 1), B(4, 0) and C(0, ) 15. Find a point D(x, y) such that the points A(, 1). B(4, 0), C(0, ) and D are the corners of a square. Justify your answer. 16. Suppose the distance between C(h, k) and P(x, y) is r. Use the distance formula to show (x h) + (y k) = r We will see this formula (and its cousins) in Chapter??. 17. Let P(x, y) be a point in the plane and let Q be the result of reflecting P about the x-axis, y-axis, or origin. Show the distance from the origin to P is the same as the distance from the origin to Q. 18. Let O(0, 0) (that is, O is the origin), P(, 1), Q( 4, ), and R(6, ). (a) Find the distance from O to P and from O to Q. What do you notice? (b) Find the distance from O to P and from O to R. What do you notice?

317 A. The Cartesian Plane 09 (c) For a generic point P(x, y), let Q(kx, ky) be the point obtained from P by multiplying both the x and y coordinates of P by the same number, k. Show the distance from O to Q is exactly k times the distance from O to P. Explain what these results mean geometrically. (We ll revisit this in Theorem?? in Section??.) 19. In this exercise, we explore some of the properties of distance. For brevity, we ll adopt the notation d(p, Q) to denote the distance between points P and Q. (a) (Non-negative Property) Explain why d(p, Q) 0 for any two points in the plane. (b) (Symmetric Property) Explain why d(p, Q) = d(q, P) for any two points in the plane. (c) (Identity Property) Show that d(p, Q) = 0 if and only if P and Q are the same point. NOTE: The phrase if and only if means you need to show two things: If P and Q are the same point, then d(p, Q) = 0. If d(p, Q) = 0, then P and Q are the same point. (d) (Triangle Inequality) The Triangle Inequality says that for any triangle, the sum of the lengths of two sides of a triangle always exceeds the length of the third. Use the Triangle Inequality to show that for any three points P, Q, and R, d(p, R) d(p, Q) + d(q, R) Under what conditions does d(p, R) = d(p, Q) + d(q, R)? 0. (Another way to measure distance.) In this text, we defined the distance between two points as the length of the line segment connecting the two points. Depending on the situation, however, there may be better ways to describe how far one location is from another. Consider the situation below on the left. Suppose P and Q are locations on a city grid, and a taxi is hailed at point P to travel to point Q. In this situation, diagonal movement is impossible, 8 so the taxi is limited to traveling horizontally and vertically. Q (x 1, y 1 ) Q y 1 y 0 P (x 0, y 0 ) P x 1 x 0 From the diagram, we see the horizontal distance is x 1 x 0 and the vertical distance is y 1 y 0, so the total distance the taxi needs to travel to get from P to Q is given by: We call d T the taxi distance from P to Q. 8 Maybe discouraged or difficult would be better word choices. d T = x 1 x 0 + y 1 y 0

318 10 Algebra Review (a) Let P(, ) and Q(4, ). Find the distance, d from P to Q and the taxi distance, d T from P to Q. Repeat this exercise with several points of your own choosing. Which is larger, d or d T? (b) Using the notation of Exercise 19, show that d(p, Q) d T (P, Q) for any two points P and Q in the plane. (The Triangle Inequality is useful once again here.) Under what conditions is d(p, Q) = d T (P, Q)? (c) Repeat Exercise 19 with the taxi distance, d T. (You may need to skip ahead to Exercise?? in Section?? to verify the Triangle Inequality piece.) (d) Think about ways to define a midpoint using the taxi distance. What would your formula be? To help you get started, play around with the origin (0, 0) as one point and the point (4, ) as the other. 1. The world is not flat. 9 Thus the Cartesian Plane cannot possibly be the end of the story. Discuss with your classmates how you would extend Cartesian Coordinates to represent the three dimensional world. What would the Distance and Midpoint formulas look like, assuming those concepts make sense at all? 9 There are those who disagree with this statement. Look them up on the Internet some time when you re bored.

319 A. The Cartesian Plane 11 A..4 Answers 1. The required points A(, 7), B(1., ), C(π, 10), D(0, 8), E( 5.5, 0), F( 8, 4), G(9., 7.8), and H(7, 5) are plotted in the Cartesian Coordinate Plane below. y 9 8 D(0, 8) H(7, 5) F ( 8, 4) 4 C(π, 10) E( 5.5, 0) x B(1., ) A(, 7) G(9., 7.8)

320 1 Algebra Review. (a) The point A(, 7) is in Quadrant III symmetric about x-axis with (, 7) symmetric about y-axis with (, 7) symmetric about origin with (, 7) (c) The point C(π, 10) is in Quadrant I symmetric about x-axis with (π, 10) symmetric about y-axis with ( π, 10) symmetric about origin with ( π, 10) (e) The point E( 5.5, 0) is on the negative x-axis symmetric about x-axis with ( 5.5, 0) symmetric about y-axis with (5.5, 0) symmetric about origin with (5.5, 0) (g) The point G(9., 7.8) is in Quadrant IV symmetric about x-axis with (9., 7.8) symmetric about y-axis with ( 9., 7.8) symmetric about origin with ( 9., 7.8). d = 5 units, M = ( 1, 7 5. d = 6 units, M = ( 1, 7. d = 74 units, M = ( 1 10, ) d = ( 8 units, M = 4 ) 5, 5 ) ) (b) The point B(1., ) is in Quadrant IV symmetric about x-axis with (1., ) symmetric about y-axis with ( 1., ) symmetric about origin with ( 1., ) (d) The point D(0, 8) is on the positive y-axis symmetric about x-axis with (0, 8) symmetric about y-axis with (0, 8) symmetric about origin with (0, 8) (f) The point F( 8, 4) is in Quadrant II symmetric about x-axis with ( 8, 4) symmetric about y-axis with (8, 4) symmetric about origin with (8, 4) (h) The point H(7, 5) is in Quadrant I symmetric about x-axis with (7, 5) symmetric about y-axis with ( 7, 5) symmetric about origin with ( 7, 5) 4. d = 4 10 units, M = (1, 4) 6. d = 7 units, M = ( 5 6, ) d = ( ) 5 units, M =, 10. d = units, M = (0, 0) 11. (, 4), 5 miles, (4, 4) 14. (a) The distance from A to B is AB = 1, the distance from A to C is AC = 5, and the distance from B to C is BC = 65. Since ( 1 ) + ( 5 ) = ( 65 ), we are guaranteed by the converse of the Pythagorean Theorem that the triangle is a right triangle. (b) Show that AC + BC = AB

321 A.4 Linear Equations and Inequalities 1 A.4 Linear Equations and Inequalities In the introduction to this chapter we said that we were going to review the concepts, skills and vocabulary we believe are prerequisite to a rigorous, college-level Precalculus course. So far, we ve presented a lot of vocabulary and concepts but we haven t done much to refresh the skills needed to survive in the Precalculus wilderness. Thus over the course of the next few sections we will focus our review on the Algebra skills needed to solve basic equations and inequalities, with one brief detour in Section A.5 where we discuss graphing lines in the plane. In general, equations and inequalities fall into one of three categories: conditional, identity or contradiction, depending on the nature of their solutions. A conditional equation or inequality is true for only certain real numbers. For example, x + 1 = 7 is true precisely when x =, and w 4 is true precisely when w 7. An identity is an equation or inequality that is true for all real numbers. For example, x = 1 + x 4 + x or t t +. A contradiction is an equation or inequality that is never true. Examples here include x 4 = x + 7 and a 1 > a +. As you may recall, solving an equation or inequality means finding all of the values of the variable, if any exist, which make the given equation or inequality true. This often requires us to manipulate the given equation or inequality from its given form to an easier form. For example, if we re asked to solve (x ) = 7x + (x + 1), we get x = 1, but not without a fair amount of algebraic manipulation. In order to obtain the correct answer(s), however, we need to make sure that whatever maneuvers we apply are reversible in order to guarantee that we maintain a chain of equivalent equations or inequalities. Two equations or inequalities are called equivalent if they have the same solutions. We summarize these legal moves in the box below. Procedures which Generate Equivalent Equations Add (or subtract) the same real number to (from) both sides of the equation. Multiply (or divide) both sides of the equation by the same nonzero real number. a Procedures which Generate Equivalent Inequalities Add (or subtract) the same real number to (from) both sides of the equation. Multiply (or divide) both sides of the equation by the same positive real number. b a Multiplying both sides of an equation by 0 collapses the equation to 0 = 0, which doesn t do anybody any good. b Remember that if you multiply both sides of an inequality by a negative real number, the inequality sign is reversed: 4, but ( )() ( )(4). A.4.1 Linear Equations The first equations we wish to review are linear equations as defined below. An equation is said to be linear in a variable x if it can be written in the form ax = b where a and b are expressions which do not involve x and a 0.

322 14 Algebra Review One key point about Definition A.4.1 is that the exponent on the unknown x in the equation is 1, that is x = x 1. Our main strategy for solving linear equations is summarized below. Strategy for Solving Linear Equations In order to solve an equation which is linear in a given variable, say x: 1. Isolate all of the terms containing x on one side of the equation, putting all of the terms not containing x on the other side of the equation.. Factor out the x and divide both sides of the equation by its coefficient. We illustrate this process with a collection of examples below. Solve the following equations for the indicated variable. Check your answer. 1. Solve for x: x 6 = 7x + 4. Solve for t: 1.7t = t 4. Solve for a: 1 18 (7 4a) + = a 4 a 1 4. Solve for y: 8y + 1 = 7 1(5 y) 5. Solve for x: x 1 = x Solve for y: x(4 y) = 8y Solution. 1. The variable we are asked to solve for is x so our first move is to gather all of the terms involving x on one side and put the remaining terms on the other. 1 x 6 = 7x + 4 (x 6) 7x + 6 = (7x + 4) 7x + 6 Subtract 7x, add 6 x 7x = 7x 7x Rearrange terms 4x = 10 x 7x = ( 7)x = 4x 4x 10 = 4 4 Divide by the coefficient of x x = 5 Reduce to lowest terms To check our answer, we substitute x = 5 into each side of the orginial equation to see the equation is satisfied. Sure enough, ( ) 5 6 = 7 and 7 ( ) = 7.. In our next example, the unknown is t and we not only have a fraction but also a decimal to wrangle. 1 In the margin notes, when we speak of operations, e.g., Subtract 7x, we mean to subtract 7x from both sides of the equation. The from both sides of the equation is omitted in the interest of spacing.

323 A.4 Linear Equations and Inequalities 15 Fortunately, with equations we can multiply both sides to rid us of these computational obstacles: t 1.7t = 4 ( ) t 40( 1.7t) = () 40(1.7t) = 40t t = 10t Multiply by 40 Distribute (10 68t) + 68t = 10t + 68t Add 68t to both sides 10 = 78t 68t + 10t = ( )t = 78t 10 = 78t Divide by the coefficient of t = t 0 1 = t Reduce to lowest terms To check, we again substitute t = 0 1 into each side of the original equation. We find that 1.7 ( 0 ( ) ( 17 0 ) 10 1 = 5 (0/1) and = = 5 as well. 1 1) =. To solve this next equation, we begin once again by clearing fractions. The least common denominator here is 6: 1 18 (7 4a) + = a 4 a ( ) ( 1 1 a 6 (7 4a) + = a ) 1 Multiply by 6 6 6a 6(4 a) (7 4a) + (6)() = Distribute 18 1 (7 4a) + 7 = 1a (4 a) Distribute 14 8a + 7 = 1a 1 + a 86 8a = 15a 1 1a + a = (1 + )a = 15a (86 8a) + 8a + 1 = (15a 1) + 8a + 1 Add 8a and a + 8a = 15a + 8a Rearrange terms 98 = a 15a + 8a = (15 + 8)a = a 98 = a 98 = a Divide by the coefficient of a The check, as usual, involves substituting a = 98 into both sides of the original equation. The reader is encouraged to work through the (admittedly messy) arithmetic. Both sides work out to

324 16 Algebra Review 4. The square roots may dishearten you but we treat them just like the real numbers they are. Our strategy is the same: get everything with the variable (in this case y) on one side, put everything else on the other and divide by the coefficient of the variable. We ve added a few steps to the narrative that we would ordinarily omit just to help you see that this equation is indeed linear. 8y + 1 = 7 1(5 y) 8y + 1 = 7 1(5) + 1y Distribute 8y + 1 = 7 ( )5 + ( )y 1 = 4 = 8y + 1 = y (8y + 1) 1 y = ( y ) 1 y Subtract 1 and y 8y y = y y Rearrange terms (8 )y = y = 6 10 See note below 6y 6 = y = 5 y = ( 5) 5 y = Divide 6 Factor and cancel In the list of computations above we marked the row 6y = 6 10 with a note. That s because we wanted to draw your attention to this line without breaking the flow of the manipulations. The equation 6y = 6 10 is in fact linear according to Definition A.4.1: the variable is y, the value of A is 6 and B = Checking the solution, while not trivial, is good mental exercise. Each side works out to be Proceeding as before, we simplify radicals and clear denominators. Once we gather all of the terms containing x on one side and move the other terms to the other, we factor out x to identify its

325 A.4 Linear Equations and Inequalities 17 coefficient then divide to get our answer. x 1 x 1 ( ) x 1 (x 1) = x = 5x = 5 = ( 5x + 4 ) Multiply by = (5x ) + 4 Distribute x 1 = 10x + 8 (x 1) 10x + 1 = (10x + 8) 10x + 1 Subtract 10x, add 1 x 10x = 10x 10x Rearrange terms x 10x = 9 ( 10 )x = 9 Factor ( 10 )x 10 = x = Divide by the coefficient of x The reader is encouraged to check this solution - it isn t as bad as it looks if you re careful! Each side works out to be If we were instructed to solve our last equation for x, we d be done in one step: divide both sides by (4 y) - assuming 4 y 0, that is. Alas, we are instructed to solve for y, which means we have some more work to do. x(4 y) = 8y 4x xy = 8y Distribute (4x xy) + xy = 8y + xy Add xy 4x = (8 + x)y Factor In order to finish the problem, we need to divide both sides of the equation by the coefficient of y which in this case is 8 + x. This expression contains a variable so we need to stipulate that we may perform this division only if 8 + x 0, or, in other words, x 8. Hence, we write our solution as: y = 4x, provided x x What happens if x = 8? Substituting x = 8 into the original equation gives ( 8)(4 y) = 8y or + 8y = 8y. This reduces to = 0, which is a contradiction. This means there is no solution when x = 8, so we ve covered all the bases. Checking our answer requires some Algebra we haven t reviewed yet in this text, but the necessary skills should be lurking somewhere in the mathematical mists of your mind. The adventurous reader is invited to plug y = 4x 8+x equation and show that both sides work out to x x+8. into the original

326 18 Algebra Review A.4. Linear Inequalities We now turn our attention to linear inequalities. Unlike linear equations which admit at most one solution, the solutions to linear inequalities are generally intervals of real numbers. While the solution strategy for solving linear inequalities is the same as with solving linear equations, we need to remind ourselves that, should we decide to multiply or divide both sides of an inequality by a negative number, we need to reverse the direction of the inequality. (See the footnote in the box on page 15.) In the example below, we work not only some simple linear inequalities in the sense there is only one inequality present, but also some compound linear inequalities which require us to revisit the notions of intersection and union. Solve the following inequalities for the indicated variable. 1. Solve for x: 7 8x 4x + 1. Solve for y: 4 7 y. Solve for t: t 1 4 t < 6t Solve for x: 5 + 7x 4x Solve for w: w or w Solution. 1. We begin by clearing denominators and gathering all of the terms containing x to one side of the inequality and putting the remaining terms on the other. 7 8x ( ) 7 8x 4x + 1 < 6 (4x + 1) Multiply by (7 8x) (4x) + (1) Distribute 7 8x 8x + (7 8x) + 8x 8x + + 8x Add 8x, subtract 7 8x + 8x 8x + 8x + Rearrange terms 5 16x 8x + 8x = (8 + 8)x = 16x x Divide by the coefficient of x x 5 We get 16 x or, said differently, x We express this set of real numbers as (, 16] 5. Though not required to do so, we could partially check our answer by substituting x = 5 and a few 16 other values in our solution set (x = 0, for instance) to make sure the inequality holds. (It also isn t a bad idea to choose an x > 5, say x = 1, to see that the inequality doesn t hold there.) The 16 only real way to actually show that our answer works for all values in our solution set is to start with Using set-builder notation, our set of solutions here is {x x 5 16 }.

327 A.4 Linear Equations and Inequalities 19 x 5 and reverse all of the steps in our solution procedure to prove it is equivalent to our original 16 inequality.. We have our first example of a compound inequality. The solutions to must satisfy 4 7 y 4 7 y and < 6 7 y One approach is to solve each of these inequalities separately, then intersect their solution sets. While this method works (and will be used later for more complicated problems), our variable y appears only in the middle expression so we can proceed by working both inequalities at once: 4 7 y ( ) ( ) 7 y < (7 y) < 4 (7 y) < 4 < 6 < 4(6) Multiply by 4 (7) y < 4 Distrbute 14 y < 4 14 (14 y) 14 < 4 14 Subtract y < y > y > 5 Divide by the coefficient of y Reverse inequalities Our final answer is ( ] y > 5, or, said differently, 5 < y. In interval notation, this is 5, 11. We could check the reasonableness of our answer as before, and the reader is encouraged to do so.. We have another compound inequality and what distinguishes this one from our previous example is that t appears on both sides of both inequalities. In this case, we need to create two separate inequalities and find all of the real numbers t which satisfy both t 1 4 t and 4 t < 6t + 1. The first inequality, t 1 4 t, reduces to t 5 or t 5. The second inequality, 4 t < 6t + 1, becomes < 7t which reduces to t > 7. Thus our solution is all real numbers t with t 5 and t > 7, or, writing this as a compound inequality, < 7 t 5. Using interval notation, we express our solution as ( 7, ] 5. If we intersect the solution sets of the two individual inequalities, we get the answer, too: (, 5 ] ( 7, ) = ( 7, 5 ].

328 0 Algebra Review 4. As before, with this inequality we have no choice but to solve each inequality individually and intersect the solution sets. Starting with the leftmost inequality, we first note that the in the term 7x, the vinculum of the square root extends over the 7 only, meaning the x is not part of the radicand. In order to avoid confusion, we will write 7x as x x 7 4x + 1 (5 + x 7) 4x 5 (4x + 1) 4x 5 Subtract 4x and 5 x 7 4x x 4x Rearrange terms x( 7 4) 4 Factor At this point, we need to exercise a bit of caution because the number 7 4 is negative. 4 When we divide by it the inequality reverses: x( 7 4) 4 x( 7 4) 7 4 x Divide by the coefficient of x Reverse inequalities x x 4 (4 7) We re only half done because we still have the rightmost inequality to solve. Fortunately, that one seems rather mundane: 4x reduces to x 7 without too much incident. Our solution is 4 x 4 4 and x We may be tempted to write 4 4 x 7 and call it a day but that would be 7 4 nonsense! To see why, notice that 4 7 is between and so 4 4 is between 7 4 = and 4 4 = 4. 4 In particular, we get 4 >. On the other hand, 7 <. This means that our solutions have to 7 4 be simultaneously greater than AND less than which is impossible. Therefore, this compound inequality has no solution, which means we did all that work for nothing Our last example is yet another compound inequality but here, instead of the two inequalities being connected with the conjunction and, they are connected with or, which indicates that we need to find the union of the results of each. Starting with w, we get 0.01w 5.1, which gives 6 w 510. The second inequality, w, becomes 0.01w 0.9, which reduces to w 90. Our solution set consists of all real numbers w with w 510 or w 90. In interval notation, this is (, 90] [510, ). 4 Since 4 < 7 < 9, it stands to reason that 4 < 7 < 9 so < 7 <. 5 Much like how people walking on treadmills get nowhere. Math is the endurance cardio of the brain, folks! 6 Don t forget to flip the inequality!

329 A.4 Linear Equations and Inequalities 1 A.4. Exercises In Exercises 1-9, solve the given linear equation and check your answer. 1. x 4 = 4(x ) y = 0 5. t 4 49w 14 7 = 7t + 1. (w ) 5 = 4 15 w = w ( 4w) 6. 7 (4 x) = x 7. t = y = 6 8y 9. 4 (x + 1) = x 7 9 In equations 10-7, solve each equation for the indicated variable. 10. Solve for y: x + y = Solve for x: x + y = 4 1. Solve for C: F = 9 5 C + 1. Solve for x: p =.5x Solve for x: C = 00x Solve for y: x = 4(y + 1) Solve for w: vw 1 = v 17. Solve for v: vw 1 = v 18. Solve for y: x(y ) = y Solve for π: C = πr 0. Solve for V : PV = nrt 1. Solve for R: PV = nrt. Solve for g: E = mgh. Solve for m: E = 1 mv In Exercises 4-7, the subscripts on the variables have no intrinsic mathematical meaning; they re just used to distinguish one variable from another. In other words, treat P 1 and P as two different variables as you would x and y. (The same goes for x and x 0, etc.) 4. Solve for V : P 1 V 1 = P V 5. Solve for t: x = x 0 + at 6. Solve for x: y y 0 = m(x x 0 ) 7. Solve for T 1 : q = mc(t T 1 ) 8. With the help of your classmates, find values for c so that the equation: x 5c = 1 c(x + ) (a) has x = 4 as a solution. (b) has no solution (that is, the equation is a contradiction.) Is it possible to find a value of c so the equation is an identity? Explain.

330 Algebra Review In Exercises 9-46, solve the given inequality. Write your answer using interval notation. 9. 4x 0 0. t 1 < (4t ) R + 1. > R. 7 ( x) x y y m + 1 m x 1 > x t 7 18t y y 7y x t 10 < x < y y < 7 4. x 4 x t > 4t t 44. x or x x 0 or x + 7 < x x > x or x + 5 1

331 A.4 Linear Equations and Inequalities A.4.4 Answers 1. x = t = 1 0. w = y = All real numbers. 6. No solution. 7. t = = y = 1 17 = 7 9. x = y = 4 x or y = x + 1. C = 5 9 (F ) or C = 5 9 F x = C y 11. x = or x = y x = p 15.5 = 15 p or x =.5 5 p + 6. or x = 1 00 C y = x 7 or y = x w = v + 1, provided v v = v 1, provided w. w 18. y = x + 1 x, provided x. 19. π = C, provided r 0. r 0. V = nrt P PV, provided P R =, provided n 0, T 0. nt. g = E E, provided m 0, h 0.. m = mh v, provided v 0 (so v 0). 4. V = P 1V 1, provided P t = x x 0, provided a 0. P a 6. x = y y 0 + mx 0 m 7. T 1 = mct q mc ( ] 9.., 4 ( 4 ), or x = x 0 + y y 0, provided m 0. m or T 1 = T q, provided m 0, c 0. mc ( ) 0., 7 6 [ ) 7 5. (4, ) 6. 18, [ ] ( 1 8., , 19 ] 1. ( ], 1. No solution. 4. (, ) 7. [0, ) 40. ( 1 10, 7 ] 10

332 4 Algebra Review 41. ( 4, 1] 4. {1} = [1, 1] 4. [ 6, 18 ) (, 1] [0, ) 45. (, 7) [4, ) 46. (, )

333 A.5 Graphing Lines 5 A.5 Graphing Lines In Section A.., we concerned ourselves with the finite line segment between two points P and Q. Specifically, we found its length (the distance between P and Q) and its midpoint. In this section, our focus will be on the entire line, and ways to describe it algebraically. Consider the generic situation below. P (x 0, y 0 ) Q (x 1, y 1 ) To give a sense of the steepness of the line, we recall that we can compute the slope of the line as follows. (Read the character as change in.) Equation A.. The slope m of the line containing the points P (x 0, y 0 ) and Q (x 1, y 1 ) is: provided x 1 x 0, that is, x 0. m = y 1 y 0 x 1 x 0 = y x, A couple of notes about Equation A. are in order. First, don t ask why we use the letter m to represent slope. There are many explanations out there, but apparently no one really knows for sure. 1 Secondly, the stipulation x 1 x 0 (or x 0) ensures that we aren t trying to divide by zero. The reader is invited to pause to think about what is happening geometrically when the change in x is 0; the anxious reader can skip along to the next example. Find the slope of the line containing the following pairs of points, if it exists. Plot each pair of points and the line containing them. 1. P(0, 0), Q(, 4). P( 1, ), Q(, 4). P(, ), Q(, ) 4. P(, ), Q(4, ) 5. P(, ), Q(, 1) 6. P(, ), Q(.1, 1) Solution. In each of these examples, we apply the slope formula, Equation A.. y 1. m = = 4 = 4 1 Q P 1 4 x 1 See or for discussions on this topic.

334 6 Algebra Review y 4 Q. m = 4 ( 1) = 4 = 1 P x y 4 P. m = ( ) = 6 4 = x Q 4 y 4. m = 4 ( ) = 0 7 = 0 P Q x y P 5. m = 1 = 4, which is undefined Q x

335 A.5 Graphing Lines 7 y P 6. m = 1.1 = = Q x A few comments about Example A.5 are in order. First, if the slope is positive then the resulting line is said to be increasing, meaning as we move from left to right, the y-values are getting larger. Similarly, if the slope is negative, we say the line is decreasing, since as we move from left to right, the y-values are getting smaller. A slope of 0 results in a horizontal line which we say is constant, since the y-values here remain unchanged as we move from left to right, and an undefined slope results in a vertical line. 4 Second, the larger the slope is in absolute value, the steeper the line. You may recall from Intermediate Algebra that slope can be described as the ratio rise run. For example, if the slope works out to be 1, we can interpret this as a rise of 1 unit upward for every run of units to the right: y 4 over up x In this way, we may view the slope as the rate of change of y with respect to x. From the expression m = y x we get y = m x so that the y-values change m times as fast as the x-values. We ll have more to say about this concept in Section?? when we explore applications of linear functions; presently, we will keep our attention focused on the analytic geometry of lines. To that end, our next task is to find algebraic equations that describe lines and we start with a discussion of vertical and horizontal lines. That is, as we increase the x-values... We ll have more to say about this idea in Section??. 4 Some authors use the unfortunate moniker no slope when a slope is undefined. It s easy to confuse the notions of no slope with slope of 0. For this reason, we will describe slopes of vertical lines as undefined.

336 8 Algebra Review Consider the two lines shown below: V (for V ertical Line) and H (for H orizontal Line). 4 y y x x 4 4 The line H The line V All of the points on the line V have an x-coordinate of. Conversely, any point with an x-coordinate of lies on the line V. Said differently, the point (x, y) lies on V if and only if x =. Because of this, we say the equation x = describes the line V, or, said differently, the graph of the equation x = is the line V. In Section??, we ll spend a great deal of time talking about graphing equations. For now, it suffices to know that a graph of an equation is a plot of all of the points which make the equation true. So to graph x =, we plot all of the points (x, y) which satisfy x = and this gives us our vertical line V. Turning our attention to H, we note that every point on H has a y-coordinate of, and vice-versa. Hence the equation y = describes the line H, or the graph of the equation y = is H. In general: Equation A.4. Equations of Vertical and Horizontal Lines The graph of the equation x = a in the xy-plane is a vertical line through (a, 0). The graph of the equation y = b in the xy-plane is a horizontal line through (0, b). Of course, we may be working on axes which aren t labeled with the usual x s and y s. In this case, we understand Equation A.4 to say horizontal axis label = a describes a vertical line through (a, 0) and vertical axis label = b describes a horizontal line through (0, b). 1. Graph the following equations in the xy-plane: (a) y = (b) x = 117

337 A.5 Graphing Lines 9. Find the equation of each of the given lines. 4 y s x Line L t Line L 1 Solution. 1. Since we re in the familiar xy-plane, the graph of y = is a horizontal line through (0, ), shown below on the left and the graph of x = 117 is a vertical line through ( 117, 0). We scale the x-axis differently than the y-axis to produce the graph below on the right. 4 y y 4 1 (0, ) 1 ( 117, 0) x The line y = x 4 The line x = 117. Since L 1 is a vertical line through (, 0), and the horizontal axis is labeled with x, the equation of L 1 is x =. Since L is a horizontal line through (0, ) and the vertical axis is labeled as s, the equation of this line is s =.

338 0 Algebra Review Using the concept of slope, we can develop equations for the other varieties of lines. Suppose a line has a slope of m and contains the point (x 0, y 0 ). Suppose (x, y) is another point on the line, as indicated below. (x, y) Equation A. yields (x 0, y 0 ) which is known as the point-slope form of a line. m = y y 0 x x 0 m (x x 0 ) = y y 0 y y 0 = m (x x 0 ) Equation A.5. The point-slope form of the line with slope m containing the point (x 0, y 0 ) is the equation y y 0 = m (x x 0 ) A few remarks about Equation A.5 are in order. First, note that if the slope m = 0, then the line is horizontal and Equation A.5 reduces to y y 0 = 0 or y = y 0, as prescribed by Equation A.4. 5 Second, we may need to change the letters in Equation A.5 from x and y depending on the context, so while Equation A.5 should be committed to memory, it should be understood that x refers to whichever variable is used to label the horizontal axis, and y refers to whichever variable is used to label the vertical axis. Lastly, while Equation A.5 is, by far, the easiest way to construct the equation of a line given a point and a slope, more often than not, the equation is solved for y and simplified into the form below. Equation A.6. The slope-intercept form of the line with slope m and y-intercept (0, b) is the equation y = mx + b Equation A.6 is probably 6 a familiar sight from Intermediate Algebra. You may recall from that class that the intercept in slope-intercept comes from the fact that this line intercepts or crosses the y-axis at the point (0, b). 7 If we set the slope, m = 0, we obtain y = b, the formula for Horizontal Lines first introduced in Equation A.4. Hence, any line which has a defined slope m can be represented in both point-slope and slope-intercept forms. The only exceptions are vertical lines. 8 There is one equation - the aptly named general form - which describes every type of line and it is presented on the next page. 5 Here we have y 0 as the constant whereas in the Equation we used the letter b. The form y = constant is what matters. 6 Hopefully? 7 We can verify this algebraically by setting x = 0 in the equation y = mx + b and obtaining y = b. 8 We ll have more to say about this in Section??.

339 A.5 Graphing Lines 1 Equation A.7. Every line may be represented by an equation of the form Ax + By = C, where A, B and C are real numbers for which A and B aren t both zero. This is called a general form of the line. Note the indefinite article a in Equation A.7. The line y = 5 is a general form for the horizontal line through (0, 5), but so are y = 15 and 0.5y =.5. The reader is left to ponder the use of the definite article the in Equations A.5 and A.6. Regardless of which form the equation of a line takes, note that the variables involved are all raised to the first power. 9 For instance, there are no x terms, no y terms or any variables appearing in denominators. Let s look at a few examples. 1. Graph the following equations in the xy-plane: (a) y = x 1 (b) x + 4y =. Find the slope-intercept form of the line containing the points ( 1, ) and (, 1).. Find the slope-intercept form of the equation of the line below: 5 s t Solution. 1. To graph a line, we need just two points on that line. There are several ways to do this, and we showcase two of them here. For the first equation, we recognize that y = x 1 is in slope-intercept form, y = mx + b, with m = and b = 1. This immediately gives us one point on the graph the y-intercept (0, 1). From here, we use the slope m = = and move one unit to the right and three 1 units up, to obtain a second point on the line, (1, ). Connecting these points gives us the graph on the left at the top of the next page. The second equation, x + 4y =, is a general form of a line. To get two points here, we choose convenient values for one of the variables, and solve for the other variable. Choosing x = 0, for example, reduces x + 4y = to 4y =, or y = 4. This means the point ( 0, 4) is on the graph. Choosing y = 0 gives x =, or x =. This gives is a second point on the line, (, 0). 10 Our graph of x + 4y = is on the right at the top of the next page. 9 Recall, x = x 1, y = y 1, etc. 10 You may recall, that this is the x-intercept of the line.

340 Algebra Review 4 1 y y = x 1 x y x + 4y = x. We ll assume we re using the familiar (x, y) axis labels and begin by finding the slope of the line using Equation A.: m = y x = 1 =. Next, we substitute this result, along with one of the ( 1) given points, into the point-slope equation of the line, Equation A.5. We have two options for the point (x 0, y 0 ). We ll use ( 1, ) and leave it to the reader to check that using (, 1) results in the same equation. Substituting into the point-slope form of the line, we get y y 0 = m (x x 0 ) y = (x ( 1)) y = (x + 1) y = x y = x + y = x + 7. We can check our answer by showing that both ( 1, ) and (, 1) are on the graph of y = x + 7 algebraically by showing that the equation holds true when we substitute x = 1 and y = and when x = and y = 1.. From the graph, we see that the points (0, 5) and (5, 0) are on the line, so we may proceed as we did in the previous problem. Here, however, we use t in place of x and s in place of y in accordance to the axis labels given. We find the slope m = s = 0 5 = 1. As before, we have two points to t 5 0 choose from to substitute into the point-slope formula, and, as before, we ll select one of them, (0, 5) and leave the computations with (5, 0) to the reader. s s 0 = m (t t 0 ) s 5 = ( 1) (t 0) s 5 = t s = t + 5. As before we can check this line contains both points (t, s) = (0, 5) and (t, s) = (5, 0) algebraically.

341 A.5 Graphing Lines While every point on a line holds value and meaning, 11 we ve reminded you of certain points, called intercepts, which hold special enough significance to be singled out. Formally, we define these as follows. Given a graph of an equation in the xy-plane: A point on a graph which is also on the x-axis is called an x-intercept of the graph. To determine the x-intercept(s) of a graph, set y = 0 in the equation and solve for x. NOTE: x-intercepts always have the form: (x 0, 0). A point on a graph which is also on the y-axis is called an y-intercept of the graph. To determine the y-intercept(s) of a graph, set x = 0 in the equation and solve for y. NOTE: y-intercepts always have the form: (0, y 0 ). As usual, the labels of the axes in the problem will dictate the labels on the intercepts. If we re working in the vw-plane, for instance, there would be v- and w-intercepts. The last little bit of analytic geometry we need to review about lines are the concepts of parallel and perpendicular lines. Parallel lines do not intersect, 1 and hence, parallel lines necessarily have the same slope. Perpendicular lines intersect at a right (90 ) angle. The relationship between these slopes is somewhat more complicated, and is summarized below. Theorem A.. Suppose line L 1 has slope m 1 and line L has slope m : L 1 and L are parallel (written L 1 L ) if and only if m 1 = m. If m 1 0 and m 0 then L 1 and L are perpendicular (written L 1 L ) if and only if m 1 m = 1. NOTE: m 1 m = 1 is equivalent to m = 1 m 1, so that perpendicular lines have slopes which are opposite reciprocals of one another. y L y L 1 L 1 L 1 L, m 1 = m x L L 1 L, m 1 m = 1 x A few remarks about Theorem A. are in order. First off, the theorem assumes that the slopes of the lines exist. The reader is encouraged to think about the case when one (or both) of the slopes don t exist. Along those same lines, the reader is encouraged to think about why the stipulations m 1 0 and m 0 appear 11 Lines missing points - even one - usually belie some algebraic pathology which we ll discuss in more detail in Chapter??. 1 Well, at least in Euclidean Geometry...

342 4 Algebra Review in the statement regarding slopes of perpendicular lines, and what happens in this case as well. (Think geometrically!) In Exercise 41, you ll prove the assertion about the slopes of perpendicular lines. For now, we accept it as true and use it in the following example. For line y = x 1 and the point (, 4), find: 1. the equation of the line parallel to the given line which contains the given point.. the equation of the line perpendicular to the given line which contains the given point. Check your answers by graphing them, along with the original line, using a graphing utility. Solution. 1. Since y = x 1 is already in slope-intercept form, we have the slope m =. To find the line parallel to this line containing (, 4), we use the point-slope form with m = to get: y y 0 = m (x x 0 ) y 4 = (x ) y 4 = x 6 y = x Algebraically, we can verify that the slope is indeed and that when x = we get y = 4. Using a graphing utility with a window centered at the point (, 4), we graph both y = x 1 and y = x below on the left and observe that they appear to be parallel.. To find the line perpendicular to y = x 1 containing (, 4), we use the slope m = 1 point-slope formula: y y 0 = m (x x 0 ) in the y 4 = 1 (x ) y 4 = 1 x + y = 1 x + 11 Algebraically, we check that the slope is m = 1 and when x = we get y = 4 as required. When checking using our graphing utility, we centered the viewing window at (, 4) and had to square it, removing its default aspect ratio, to truly observe the perpendicular nature of the lines. y = x 1 and y = x 1 and y = x y = 1 x + 11

343 A.5 Graphing Lines 5 Our last example with lines sets up a fourth kind of symmetry which will be revisited in Section??. Show that the points (a, b) and (b, a) in the xy-plane are symmetric about the line y = x. Solution. If a = b then (a, b) = (a, a) = (b, a) and this point lies on the line y = x. 1 To prove the claim for the case when a b, we will show that the line y = x is a perpendicular bisector of the line segment with endpoints (a, b) and (b, a), as illustrated below. y (a, b) (b, a) x To show the perpendicular part, we first note the slope of the line containing (a, b) and (b, a) is m = a b b a = (a b) (b a) = 1 Since the slope of y = x = 1x + 0 is m = 1, we see that the slopes of these two lines are negative reciprocals. Hence, y = x and the line segment with endpoints (a, b) and (b, a) are perpendicular. For the bisector part, we use Equation A. to find the midpoint of the line segment with endpoints (a, b) and (b, a): M = = ( a + b, b + a ) ( a + b, a + b ) Since the x and y coordinates of this point are the same, we find that the midpoint lies on the line y = x. 1 Please ask your instructor if lying on the line counts as being symmetric about the line or not.

344 6 Algebra Review A.5.1 Exercises In Exercises 1-10, find both the point-slope form and the slope-intercept form of the line with the given slope which passes through the given point. 1. m =, P(, 1). m =, P( 5, 8). m = 1, P( 7, 1) 4. m =, P(, 1) 5. m = 1 5, P(10, 4) 6. m = 1, P( 1, 4) 7 7. m = 0, P(, 117) 8. m =, P(0, ) 9. m = 5, P(, ) 10. m = 678, P( 1, 1) In Exercises 11-0, find the slope-intercept form of the line which passes through the given points. 11. P(0, 0), Q(, 5) 1. P( 1, ), Q(, ) 1. P(5, 0), Q(0, 8) 14. P(, 5), Q(7, 4) 15. P( 1, 5), Q(7, 5) 16. P(4, 8), Q(5, 8) 17. P ( 1, ( 4), Q 5, ) P ( 4, ) ( 7, Q 1, ) 19. P (, ), Q (, ) 0. P (, 1 ), Q (, 1 ) In Exercises 1-6, graph the line. Find the slope, y-intercept and x-intercept, if any exist. 1. y = x 1. y = x. y = 4. y = 0 5. y = x y = 1 x 7. Graph v + w = 6 on both the vw- and wv-axes. What characteristics to both graphs share? What s different? 8. Find all of the points on the line y = x + 1 which are 4 units from the point ( 1, ). In Exercises 9-4, you are given a line and a point which is not on that line. Find the line parallel to the given line which passes through the given point.

345 A.5 Graphing Lines 7 9. y = x +, P(0, 0) 0. y = 6x + 5, P(, ) 1. y = x 7, P(6, 0). y = 4 x, P(1, 1). y = 6, P(, ) 4. x = 1, P( 5, 0) In Exercises 5-40, you are given a line and a point which is not on that line. Find the line perpendicular to the given line which passes through the given point. 5. y = 1 x +, P(0, 0) 6. y = 6x + 5, P(, ) 7. y = x 7, P(6, 0) 8. y = 4 x, P(1, 1) 9. y = 6, P(, ) 40. x = 1, P( 5, 0) 41. We shall now prove that y = m 1 x + b 1 is perpendicular to y = m x + b if and only if m 1 m = 1. To make our lives easier we shall assume that m 1 > 0 and m < 0. We can also move the lines so that their point of intersection is the origin without messing things up, so we ll assume b 1 = b = 0. (Take a moment with your classmates to discuss why this is okay.) Graphing the lines and plotting the points O(0, 0), P(1, m 1 ) and Q(1, m ) gives us the following set up. y P O x Q The line y = m 1 x will be perpendicular to the line y = m x if and only if OPQ is a right triangle. Let d 1 be the distance from O to P, let d be the distance from O to Q and let d be the distance from P to Q. Use the Pythagorean Theorem to show that OPQ is a right triangle if and only if m 1 m = 1 by showing d 1 + d = d if and only if m 1 m = 1.

346 8 Algebra Review A.5. Answers 1. y + 1 = (x ) y = x 10. y + 1 = (x + 7) y = x 8. y 8 = (x + 5) y = x 4. y 1 = (x + ) y = x y 4 = 1 (x 10) 5 6. y 4 = 1 (x + 1) 7 y = 1 5 x + 6 y = 1 7 x y 117 = 0 y = y = 5(x ) y = 5x y + = (x 0) y = x 10. y + 1 = 678(x + 1) y = 678x y = 5 x 1. y = 1. y = 8 5 x y = 9 4 x y = y = y = 5 4 x y = x y = x 0. y = x 1. y = x 1 slope: m = y-intercept: (0, 1) y 1 x-intercept: ( 1, 0) 1 1 x 1. y = x slope: m = 1 y-intercept: (0, ) x-intercept: (, 0) 4 1 y x

347 A.5 Graphing Lines 9. y = slope: m = 0 y-intercept: (0, ) x-intercept: none 4 1 y 1 1 x 4. y = 0 slope: m = 0 y-intercept: (0, 0) x-intercept: {(x, 0) x is a real number} 5. y = x + 1 slope: m = y-intercept: ( ) 0, 1 x-intercept: ( 1, 0) y y x x 6. y = 1 x slope: m = 1 y-intercept: ( 0, 1 x-intercept: (1, 0) ) y x 7. w = v + slope: m = w-intercept: (0, ) v-intercept: (, 0) 1 w 1 v v = w + slope: m = v-intercept: (0, ) w-intercept: (, 0) 8. ( 1, 1) and ( 11 5, ) v 1 w

348 40 Algebra Review 9. y = x 0. y = 6x y = x 4. y = 1 x. y = 4. x = 5 5. y = x 6. y = 1 6 x + 7. y = x y = x 4 9. x = 40. y = 0

349 A.6 Systems of Two Linear Equations in Two Unknowns 41 A.6 Systems of Two Linear Equations in Two Unknowns This section of the Appendix combines ideas from Section A.4 and A.5 so that we can start to solve systems of linear equations. Before we get ahead of ourselves, let s review a few definitions. A linear equation in two variables is an equation of the form a 1 x + a y = c where a 1, a and c are real numbers and at least one of a 1 and a is nonzero. For reasons which will become clear when you study Chapter??, we are using subscripts in Definition A.6 to indicate different, but fixed, real numbers and those subscripts have no mathematical meaning beyond that. For example, x y = 0.1 is a linear equation in two variables with a 1 =, a = 1 and c = 0.1. We can also consider x = 5 to be a linear equation in two variables 1 by identifying a 1 = 1, a = 0, and c = 5. If a 1 and a are both 0, then depending on c, we get either an equation which is always true, called an identity, or an equation which is never true, called a contradiction. (If c = 0, then we get 0 = 0, which is always true. If c 0, then we d have 0 0, which is never true.) Even though identities and contradictions have a large role to play throughout Chapter??, we do not consider them linear equations. The key to identifying linear equations is to note that the variables involved are to the first power and that the coefficients of the variables are numbers. Some examples of equations which are non-linear are x + y = 1, xy = 5 and e x + ln(y) = 1. The reader should consider why these do not satisfy Definition A.6. We know from our work is Sections A.5 that the graphs of linear equations are lines. If we couple two or more linear equations together, in effect to find the points of intersection of two or more lines, we obtain a system of linear equations in two variables. Our first example explores the basic techniques for solving these systems. Remember - if we are looking for points in the plane, then both the x and y values are important. This is a key distinction between solving one equation and solving a system of equations. Solve the following systems of equations. Check your answer algebraically and graphically. (Said another way, make sure both x and y are correct!) { x y = 1 y = { x 4y = 6 x 6y = { x + 4y = x y = 5 { 6x + y = 9 4x + y = { x 4y = x 9 + y = 1 x y = 0 x + y = x + y = Solution. 1. Our first system is nearly solved for us. The second equation tells us that y =. To find the corresponding value of x, we substitute this value for y into the the first equation to obtain x = 1, so that x =. Our solution to the system is (, ). To check this algebraically, we substitute x = and y = into each equation and see that they are satisfied. We see () = 1, and =, as required. To check our answer graphically, we graph the lines x y = 1 and y = and verify that they intersect at (, ). 1 Critics may argue that x = 5 is clearly an equation in one variable. It can also be considered an equation in 117 variables with the coefficients of 116 variables set to 0. As with many conventions in Mathematics, the context will clarify the situation.

350 4 Algebra Review. To solve the second system, we use the addition method to eliminate the variable x. We take the two equations as given and add equals to equals to obtain x + 4y = + ( x y = 5) y = This gives us y = 1. We now substitute y = 1 into either of the two equations, say x y = 5, to get x 1 = 5 so that x =. Our solution is (, 1). Substituting x = and y = 1 into the first equation gives ( ) + 4(1) =, which is true, and, likewise, when we check (, 1) in the second equation, we get ( ) 1 = 5, which is also true. Geometrically, the lines x + 4y = and x y = 5 intersect at (, 1). 4 y y (, ) 1 (, 1) x x x y = 1 y = x + 4y = x y = 5. The equations in the third system are more approachable if we clear denominators. We multiply both sides of the first equation by 15 and both sides of the second equation by 18 to obtain the kinder, gentler system { 5x 1y = 1 4x + 6y = 9 Adding these two equations directly fails to eliminate either of the variables, but we note that if we multiply the first equation by 4 and the second by 5, we will be in a position to eliminate the x term 0x 48y = 84 + ( 0x 0y = 45) 78y = 9 From this we get y = 1. We can temporarily avoid too much unpleasantness by choosing to substitute y = 1 into one of the equivalent equations we found by clearing denominators, say into 5x 1y = 1. We get 5x + 6 = 1 which gives x =. Our answer is (, ) 1. At this point, we have no choice in order to check an answer algebraically, we must see if the answer satisfies both of the original equations, so we substitute x = and y = 1 into both x 4y 5 = 7 5 and x 9 + y = 1.

351 A.6 Systems of Two Linear Equations in Two Unknowns 4 We leave it to the reader to verify that the solution is correct. Graphing both of the lines involved with considerable care yields an intersection point of (, 1 ). (The picture is on the next page.) 4. An eerie calm settles over us as we cautiously approach our fourth system. Do its friendly integer coefficients belie something more sinister? We note that if we multiply both sides of the first equation by and both sides of the second equation by, we are ready to eliminate the x 6x 1y = 18 + ( 6x + 1y = 18) 0 = 0 We eliminated not only the x, but the y as well and we are left with the identity 0 = 0. This means that these two different linear equations are, in fact, equivalent. In other words, if an ordered pair (x, y) satisfies the equation x 4y = 6, it automatically satisfies the equation x 6y = 9. This system has infinitely many solutions and one way to describe the solution set to this system is to use the roster method and write {(x, y) x 4y = 6}. While this is correct (and corresponds exactly to what s happening graphically, as we shall see shortly), we take this opportunity to introduce the notion of a parametric solution to a system. Our first step is to solve x 4y = 6 for one of the variables, say y = 1 x. For each value of x, the formula y = 1 x determines the corresponding y-value of a solution. Since we have no restriction on x, it is called a free variable. We let x = t, a so-called parameter, and get y = 1 t. Our set of solutions can then be described as {( t, 1 t } ) < t <. For specific values of t, we can generate solutions. For example, t = 0 gives us the solution ( 0, ) ; t = 117 gives us (117, 57), and while we can check that each of these particular solutions satisfy both equations, the question is how do we check our general answer algebraically? Same as always. We claim that for any real number t, the pair ( t, 1 t ) satisfies both equations. Substituting x = t and y = 1 t into x 4y = 6 gives t 4 ( 1 t ) = 6. Simplifying, we get t t + 6 = 6, which is always true. Similarly, when we make these substitutions in the equation x 6y = 9, we get t 6 ( 1 t ) = 9 which reduces to t t + 9 = 9, so it checks out, too. Geometrically, x 4y = 6 and x 6y = 9 are the same line, which means that they intersect at every point on their graphs. The reader is encouraged to think about how our parametric solution says exactly that. See Section A.1 for a review of this. Note that we could have just as easily chosen to solve x 4y = 6 for x to obtain x = y +. Letting y be the parameter t, we have that for any value of t, x = t +, which gives {(t +, t) < t < }. There is no one correct way to parameterize the solution set, which is why it is always best to check your answer.

352 44 Algebra Review The picture for this system is shown below on the right while the picture for the previous example is shown on the left. y y ( ), 1 x x 4 x 4y 5 = 7 5 x 9 + y = 1 x 4y = 6 x 6y = 9 (Same line.) 5. Multiplying both sides of the first equation by and the both sides of the second equation by, we set the stage to eliminate x 1x + 6y = 18 + ( 1x 6y = 6) 0 = 18 As in the previous example, both x and y dropped out of the equation, but we are left with an irrevocable contradiction, 0 = 18. This tells us that it is impossible to find a pair (x, y) which satisfies both equations; in other words, the system has no solution. Graphically, the lines 6x + y = 9 and 4x + y = 1 are distinct and parallel, so they do not intersect. 6. We can begin to solve our last system by adding the first two equations x y = 0 + (x + y = ) x = which gives x = 1. Substituting this into the first equation gives 1 y = 0 so that y = 1. We seem to have determined a solution to our system, (1, 1). While this checks in the first two equations, when we substitute x = 1 and y = 1 into the third equation, we get (1) + (1) = which simplifies to the contradiction 1 =. Graphing the lines x y = 0, x + y =, and x + y =, we see that the first two lines do, in fact, intersect at (1, 1), however, all three lines never intersect at the same point simultaneously, which is what is required if a solution to the system is to be found.

353 A.6 Systems of Two Linear Equations in Two Unknowns y 1 x 1 1 y x 6x + y = 9 4x + y = 1 y x = 0 y + x = x + y = A few remarks about Example A.6 are in order. Notice that some of the systems of linear equations had solutions while others did not. Those which have solutions are called consistent, those with no solution are called inconsistent. We also distinguish between the two different types of behavior among consistent systems. Those which admit free variables are called dependent and those with no free variables are called independent. 4 Using this new vocabulary, we classify numbers 1, and in Example A.6 as consistent independent systems, number 4 is consistent dependent, and numbers 5 and 6 are inconsistent. 5 The system in 6 above is called overdetermined, since we have more equations than variables. 6 Not surprisingly, a system with more variables than equations is called underdetermined. While the system in number 6 above is overdetermined and inconsistent, there exist overdetermined consistent systems (both dependent and independent) and we leave it to the reader to think about what is happening algebraically and geometrically in these cases. Likewise, there are both consistent and inconsistent underdetermined systems, 7 but a consistent underdetermined system of linear equations is necessarily dependent. 8 We end this section with a story problem. It is an example of a classic mixture problem and should be familiar to most readers. The basic goal here is to create two equations: one which represents and the other which represents stuff + other stuff = total stuff value of stuff + value of other stuff = value of total stuff. 4 In the case of systems of linear equations, regardless of the number of equations or variables, consistent independent systems have exactly one solution. The reader is encouraged to think about why this is the case for linear equations in two variables. Hint: think geometrically. 5 The adjectives dependent and independent apply only to consistent systems they describe the type of solutions. Is there a free variable (dependent) or not (independent)? 6 If we think if each variable being an unknown quantity, then ostensibly, to recover two unknown quantities, we need two pieces of information - i.e., two equations. Having more than two equations suggests we have more information than necessary to determine the values of the unknowns. While this is not necessarily the case, it does explain the choice of terminology overdetermined. 7 We need more than two variables to give an example of the latter. 8 Again, experience with systems with more variables helps to see this here, as does a solid course in Linear Algebra.

354 46 Algebra Review The Dude-Bros want to create a highly caffeinated, yet still drinkable, fruit punch for their annual Disturb the Neighbors BBQ and Dance Competition. They plan to add Sasquatch Sweat TM Energy Drink, which has 100 mg. of caffeine per fluid ounce, to Frooty Giggle Delight TM, which has only mg. of caffeine per fluid ounce. How much of each component is required to make 5 gallons 9 of a fruit punch that has 80 mg. of caffeine per fluid ounce. Solution. Let S stand for the number of fluid ounces of Sasquatch Sweat TM Energy Drink and let F be the number of fluid ounces of Frooty Giggle Delight TM that will be added together. The goal is to make 5 gallons and there are 18 fluid ounces per gallon so the first equation is S + F = 640. That equation describes stuff + other stuff = total stuff measured in fluid ounces. Now we need to consider the value of the stuff - in this case we need to see how much caffeine is being contributed by each component. Each fluid ounce of Sasquatch Sweat TM contains 100 mg. of caffeine so S fluid ounces would contain 100S mg./ of caffeine. Similarly, the F fluid ounces of Frooty Giggle Delight TM add F mg. of caffeine to the total mixture. Thus when we go to express value of stuff + value of other stuff = value of total stuff we need to figure out how much caffeine is supposed to be in the end product. Well, the goal was 5 gallons of punch that had 80 mg. of caffeine per fluid ounce so the Dude-Bros need to end up with = 5100 mg. of caffeine when they re done. Hence the second is equation is 100S + F = By turning the first equation into F = 640 S and substituting that into the second equation we get 100S + (640 S) = 5100 which yields S = fluid ounces. Back-substituting this value of S into the first equation gives 97 us F = fluid ounces. 97 The reader should take the time to verify that S = and F = do indeed satisfy both equations and thus are the solution to the problem. Those are fairly unattractive numbers so we end this example by discussing a way to verify an approximate answer which is reasonable without having to fight with fractions. Round S down to 508 and round F up to 1. Clearly = 640 so the first equation is still satisfied. Notice that = which is really close to Thus the second equation is nearly satisfied which means the values S = 508 and F = 1, while not precise, are reasonable Warning: unit conversion ahead! 10 Just be careful here - sometimes close enough for the Dude-Bros is not good enough for your Professor!

355 A.6 Systems of Two Linear Equations in Two Unknowns 47 A.6.1 Exercises In Exercises 1-8, solve the given system using substitution and/or elimination. Classify each system as consistent independent, consistent dependent, or inconsistent. Check your answers both algebraically and graphically. 1. { x + y = 5 x = 6 { x+y 4 = 5. { y x = 1 y = { x 1 5 y =. x y = 1 { 1 x 1 y = x + 4 y = 1 { x + 4y = y x = 6 { y x = 15 1 x y = { 1 1 x + 1 y = x + 5 y = 7 10 x 0 y = A local buffet charges $7.50 per person for the basic buffet and $9.5 for the deluxe buffet (which includes crab legs.) If 7 diners went out to eat and the total bill was $7.00 before taxes, how many chose the basic buffet and how many chose the deluxe buffet? 10. At The Old Home Fill er Up and Keep on a-truckin Cafe, Mavis mixes two different types of coffee beans to produce a house blend. The first type costs $ per pound and the second costs $8 per pound. How much of each type does Mavis use to make 50 pounds of a blend which costs $6 per pound? 11. Skippy has a total of $10,000 to split between two investments. One account offers % simple interest, and the other account offers 8% simple interest. For tax reasons, he can only earn $500 in interest the entire year. How much money should Skippy invest in each account to earn $500 in interest for the year? 1. A 10% salt solution is to be mixed with pure water to produce 75 gallons of a % salt solution. How much of each are needed? 1. This exercise is a follow-up to Example A.6. Work with your classmates to explain why mixing 4 gallons of Sasquatch Sweat TM Energy Drink and 1 gallon of Frooty Giggle Delight TM would also produce a mixture that was close enough for the Dude-Bros.

356 48 Algebra Review A.6. Answers 1. Consistent independent Solution ( ) 6, 1. Consistent independent Solution ( 16 7, ) Consistent dependent Solution ( t, t + ) for all real numbers t 7. Inconsistent No solution. Consistent independent Solution ( 7, ) 4. Consistent independent Solution ( 49 1, ) Consistent dependent Solution (6 4t, t) for all real numbers t 8. Inconsistent No solution 9. 1 chose the basic buffet and 14 chose the deluxe buffet. 10. Mavis needs 0 pounds of $ per pound coffee and 0 pounds of $8 per pound coffee. 11. Skippy needs to invest $6000 in the % account and $4000 in the 8% account gallons of the 10% solution and 5.5 gallons of pure water.

357 A.7 Absolute Value Equations and Inequalities 49 A.7 Absolute Value Equations and Inequalities In this section, we review some of the basic concepts involving the absolute value of a real number x. There are a few different ways to define absolute value and in this section we choose the following definition. (Absolute value will be revisited in much greater depth in Section?? where we present what one can think of as the precise definition.) Absolute Value as Distance: For every real number x, the absolute value of x, denoted x, is the distance between x and 0 on the number line. More generally, if x and c are real numbers, x c is the distance between the numbers x and c on the number line. For example, 5 = 5 and 5 = 5, since each is 5 units from 0 on the number line: distance is 5 units distance is 5 units Graphically why 5 = 5 and 5 = 5 Computationally, the absolute value makes negative numbers positive, though we need to be a little cautious with this description. While 7 = 7, The absolute value acts as a grouping symbol, so 5 7 = =, which makes sense since 5 and 7 are two units away from each other on the number line: distance is units Graphically why 5 7 = We list some of the operational properties of absolute value below. Theorem A.4. Properties of Absolute Value: Let a and b be real numbers and let n be an integer. a Product Rule: ab = a b Power Rule: a n = a n whenever a n is defined Quotient Rule: a = a b b, provided b 0 a See page 66 if you don t remember what an integer is. The proof of Theorem A.4 is difficult, but not impossible, using the distance definition of absolute value or even the it makes negatives positive notion. It is, however, much easier if one uses the precise definition given in Section?? so we will revisit the proof then. For now, let s focus on how to solve basic equations and inequalities involving the absolute value.

358 50 Algebra Review A.7.1 Absolute Value Equations Thinking of absolute value in terms of distance gives us a geometric way to interpret equations. For example, to solve x =, we are looking for all real numbers x whose distance from 0 is units. If we move three units to the right of 0, we end up at x =. If we move three units to the left, we end up at x =. Thus the solutions to x = are x = ±. units units 0 The solutions to x = are x = ±. Thinking this way gives us the following. Theorem A.5. Absolute Value Equations: Suppose x, y and c are real numbers. x = 0 if and only if x = 0. For c > 0, x = c if and only if x = c or x = c. For c < 0, x = c has no solution. x = y if and only if x = y or x = y. (That is, if two numbers have the same absolute values, they are either the same number or exact opposites of each other.) Theorem A.5 is our main tool in solving equations involving the absolute value, since it allows us a way to rewrite such equations as compound linear equations. Strategy for Solving Equations Involving Absolute Value In order to solve an equation involving the absolute value of a quantity X : 1. Isolate the absolute value on one side of the equation so it has the form X = c.. Apply Theorem A.5. The techniques we use to isolate the absolute value are precisely those we used in Section A.4 to isolate the variable when solving linear equations. Time for some practice. Solve each of the following equations. 1. x 1 = 6. y + 5 = 1. t = w + = 5 5. x 1 = 4x t 1 t + 1 = 0

359 A.7 Absolute Value Equations and Inequalities 51 Solution. 1. The equation x 1 = 6 is of already in the form X = c, so we know that either x 1 = 6 or x 1 = 6. Solving the former gives us at x = 7 and solving the latter yields x = 5. We may check both of these solutions by substituting them into the original equation and showing that the arithmetic works out.. We begin solving y+5 = 1 by isolating the absolute value to put it in the form X = c. y + 5 = 1 y + 5 = Multiply by y + 5 = 1 Subtract y + 5 = 1 Divide by 1 At this point, we have y + 5 = 1 or y + 5 = 1, so our solutions are y = 4 or y = 6. We leave it to the reader to check both answers in the original equation.. As in the previous example, we first isolate the absolute value. Don t let the 5 throw you off - it s just another real number, so we treat it as such: t = 0 t + 1 = 5 Add 5 5 t + 1 = Divide by From here, we have that t + 1 = 5 or t + 1 = 5. The first equation gives t = 5 while 6 the second gives t = 5 thus we list our answers as t = ± 5. The reader should enjoy the 6 6 challenge of substituting both answers into the original equation and following through the arithmetic to see that both answers work. 4. Upon isolating the absolute value in the equation 4 5w + = 5, we get 5w + = 1. At this point, we know there cannot be any real solution. By definition, the absolute value is a distance, and as such is never negative. We write no solution and carry on. 5. Our next equation already has the absolute value expressions (plural) isolated, so we work from the principle that if x = y, then x = y or x = y. Thus from x 1 = 4x + 1 we get two equations to solve: x 1 = 4x + 1, and x 1 = (4x + 1) Notice that the right side of the second equation is (4x + 1) and not simply 4x + 1. Remember, the expression 4x + 1 represents a single real number so in order to negate it we need to negate the entire expression (4x + 1). Moving along, when solving x 1 = 4x + 1, we obtain x = and the solution to x 1 = (4x + 1) is x =. As usual, the reader is invited to check 1 4 these answers by substituting them into the original equation.

360 5 Algebra Review 6. We start by isolating one of the absolute value expressions: t 1 t +1 = 0 gives t 1 = t +1. While this resembles the form x = y, the coefficient in t + 1 prevents it from being an exact match. Not to worry - since is positive, = so t + 1 = t + 1 = (t + 1) = t +. Hence, our equation becomes t 1 = t + which results in the two equations: t 1 = t + and t 1 = (t + ). The first equation gives t = and the second gives t = 1. The reader is encouraged to check both answers in the original equation. A.7. Absolute Value Inequalities We now turn our attention to solving some basic inequalities involving the absolute value. Suppose we wished to solve x <. Geometrically, we are looking for all of the real numbers whose distance from 0 is less than units. We get < x <, or in interval notation, (, ). Suppose we are asked to solve x > instead. Now we want the distance between x and 0 to be greater than units. Moving in the positive direction, this means x >. In the negative direction, this puts x <. Our solutions would then satisfy x < or x >. In interval notation, we express this as (, ) (, ). units units units units 0 0 The solution to x < is (, ) The solution to x > is (, ) (, ) Generalizing this notion, we get the following: Theorem A.6. Inequalities Involving Absolute Value: Let c be a real number. If c > 0, x < c is equivalent to c < x < c. If c 0, x < c has no solution. If c > 0, x > c is equivalent to x < c or x > c. If c 0, x > c is true for all real numbers. If the inequality we re faced with involves or, we can combine the results of Theorem A.6 with Theorem A.5 as needed. Strategy for Solving Inequalities Involving Absolute Value In order to solve an inequality involving the absolute value of a quantity X : 1. Isolate the absolute value on one side of the inequality.. Apply Theorem A.6.

361 A.7 Absolute Value Equations and Inequalities 5 Solve the following inequalities. 1. x 4 5 > 1. 4 x x 1 4 8x x 1 4 8x < x x x 0 Solution. 1. From Theorem A.6, x 4 5 > 1 is equivalent to x 4 5 < 1 or x 4 5 > 1. Solving this compound inequality, we get x < or x > Our answer, in interval notation, is: (, ) ( , ). As with linear inequalities, we can only partially check our answer by selecting values of x both inside and outside of the solution intervals to see which values of x satisfy the original inequality and which do not.. Our first step in solving 4 x+1 4 is to isolate the absolute value. 4 x x Multiply by 4 x Subtract 4 x Divide by, reverse the inequality x Reduce Since we re dealing with instead of just <, we can combine Theorems A.6 and A.5 to rewrite this last inequality as: 1 ( + ) x Subtracting the 1 across both inequalities gives x 1 +, which reduces to x 1+. In interval notation this reads as [, 1+ ].. There are two absolute values in x 1 4 8x 10, so we cannot directly apply Theorem A.6 here. Notice, however, that 4 8x = ( 4)(x 1). Using this, we get: x 1 4 8x 10 x 1 ( 4)(x 1) 10 Factor x 1 4 x 1 10 Product Rule x 1 1 x x 1 10 Subtract 1 x 1 x Divide by 11 and reduce Now we are allowed to invoke Theorems A.5 and A.6 and write the equivalent compound inequality: x or x We get x 1 1 or x, which when written with interval notation becomes ( [, ] 1 1, ). 1 Note the use of parentheses: ( + ) as opposed to +.

362 54 Algebra Review 4. The inequality x 1 4 8x + 10 differs from the previous example in exactly one respect: on the right side of the inequality, we have +10 instead of 10. The steps to isolate the absolute value here are identical to those in the previous example, but instead of obtaining x 1 10 as before, 11 we obtain x This latter inequality is always true. (Absolute value is, by definition, a 11 distance and hence always 0 or greater.) Thus our solution to this inequality is all real numbers. 5. To solve < x 1 5, we rewrite it as the compound inequality: < x 1 and x 1 5. The first inequality, < x 1, can be re-written as x 1 > so it is equivalent to x 1 < or x 1 >. Thus the solution to < x 1 is x < 1 or x >, which in interval notation is (, 1) (, ). For x 1 5, we combine the results of Theorems A.5 and A.6 to get 5 x 1 5 so that 4 x 6, or [ 4, 6]. Our solution to < x 1 5 is comprised of values of x which satisfy both parts of the inequality, so we intersect (, 1) (, ) with [ 4, 6] to get our final answer [ 4, 1) (, 6]. 6. Our first hope when encountering 10x x 0 is that we can somehow combine the two absolute value quantities as we d done in earlier examples. We leave it to the reader to show, however, that no matter what we try to factor out of the absolute value quantities, what remains inside the absolute values will always be different. At this point, we take a step back and look at the equation in a more general way: we are adding two absolute values together and wanting the result to be less than or equal to 0. The absolute value of anything is always 0 or greater, so there are no solutions to: 10x x < 0. Is it possible that 10x x = 0? Only if there is an x where 10x 5 = 0 and 10 5x = 0 at the same time. The first equation holds only when x = 1, while the second holds only when x =. Alas, we have no solution. The astute reader will have noticed by now that the authors have done nothing in the way of explaining why anyone would ever need to know this stuff. Go back and read the New Preface and the introduction to the Appendix. These sections are designed to review skills and concepts that you ve already learned. Thus the deeper applications are in the main body of the text as opposed to here in the Appendix. We close this section with an example of how the properties in Theorem A.4 are used in Calculus. Here, ε is the Greek letter epsilon and it represents a positive real number. Those of you who will be taking Calculus in the future should become very familiar with this type of algebraic manipulation. Do you see why? Not for lack of trying, however!

363 A.7 Absolute Value Equations and Inequalities x < ε 8 4x 4(x ) 4 x 4 4 x 4 x < ε Quotient Rule < ε Factor < ε Product Rule < ε x < 4 ε < 4 ε Multiply by 4

364 56 Algebra Review A.7. Exercises In Exercises 1-18, solve the equation. 1. x = 6. t 1 = w = y = 5. 5m + 1 = x 1 + = x = w 1 = 5 9. t + 4 = v = 1 v x + 1 = x y + 4 = y 1. t = t x + 1 = 4x y = y x x + = z = 5 z w 1 = w + 1 In Exercises 19-0, solve the inequality. Write your answer using interval notation. 19. x t + > w < 0. y 4. z < v + 4 > 1 5. x + 5 < 6. 5t t + 7. w < w 8. 4 y < < w 9 0. > x > 1 1. With help from your classmates, solve: (a) 5 x = 4 (b) 5 x < 4

365 A.7 Absolute Value Equations and Inequalities 57 A.7.4 Answers 1. x = 6 or x = 6. t = or t = y = 1 or y = 1 5. m = 1 or m = x = or x = 8. w = 1 8 or w = 5 8. w = or w = No solution ± 9. t = 10. v = 1 or v = No solution 1. y = 1. t = 1 or t = x = 1 7 or x = y = 0 or y = x = z = [ ] 1, 0. (, 1 7 ) ( ) 8 7, ± 18. w = See footnote 4 1. (, ). (, 1] [, ). No solution 4. (, ) 5. (, 6 5) (6 [ 5, ) 6. 4, ] 4 7. No solution 8. (, ] [6, 11) 9. [, 4) (5, 6] ( 0., ) ( 1 + 1, ) + 1. (a) x =, or x = 1, or x =, or x = 6 (b) (, 1) (, 6) 4 That is, w = + or w = +

366 58 Algebra Review A.8 Polynomial Arithmetic In this section, we review the vocabulary and arithmetic of polynomials. We start by defining what is meant by the word polynomial in general. A more narrow definition of a polynomial function will be given in Chapter??. The general definition suffices for the purposes of this review. A polynomial is a sum of terms each of which is a real number or a real number multiplied by one or more variables to natural number powers. Some examples of polynomials are x +x +4, 7x y + 7x and 6. Things like x, 4x x+1 and 1x / y are not polynomials. In the box below we review some of the terminology associated with polynomials. Polynomial Vocabulary Constant Terms: Terms in polynomials without variables are called constant terms. Coefficient: In non-constant terms, the real number factor in the expression is called the coefficient of the term. Degree: The degree of a non-constant term is the sum of the exponents on the variables in the term; non-zero constant terms are defined to have degree 0. The degree of a polynomial is the highest degree of the nonzero terms. Like Terms: Terms in a polynomial are called like terms if they have the same variables each with the same corresponding exponents. Simplified: A polynomial is said to be simplified if all arithmetic operations have been completed and there are no longer any like terms. Classification by Number of Terms: A simplified polynomial is called a monomial if it has exactly one nonzero term binomial if it has exactly two nonzero terms trinomial if it has exactly three nonzero terms For example, x + x + 4 is a trinomial of degree. The coefficient of x is 1 and the constant term is 4. The polynomial 7x y + 7x is a binomial of degree (x y = x y 1 ) with constant term 0. The concept of like terms really amounts to finding terms which can be combined using the Distributive Property. For example, in the polynomial 17x y xy + 7xy, xy and 7xy are like terms, since they have the same variables with the same corresponding exponents. This allows us to combine these two terms as follows: 17x y xy + 7xy = 17x y + ( )xy + 7xy + 17x y + ( + 7)xy = 17x y + 4xy Note that even though 17x y and 4xy have the same variables, they are not like terms since in the first

367 A.8 Polynomial Arithmetic 59 term we have x and y = y 1 but in the second we have x = x 1 and y = y so the corresponding exponents aren t the same. Hence, 17x y + 4xy is the simplified form of the polynomial. There are four basic operations we can perform with polynomials: addition, subtraction, multiplication and division. The first three of these follow directly from properties of real number arithmetic and will be discussed together. Division, on the other hand, is a bit more complicated and will be discussed separately. A.8.1 Polynomial Addition, Subtraction and Multiplication. Adding and subtracting polynomials comes down to identifying like terms and then adding or subtracting the coefficients of those like terms. Multiplying polynomials comes to us courtesy of the Generalized Distributive Property. Theorem A.7. Generalized Distributive Property: To multiply a quantity of n terms by a quantity of m terms, multiply each of the n terms of the first quantity by each of the m terms in the second quantity and add the resulting n m terms together. In particular, Theorem A.7 says that, before combining like terms, a product of an n-term polynomial and an m-term polynomial will generate (n m)-terms. For example, a binomial times a trinomial will produce six terms some of which may be like terms. Thus the simplified end result may have fewer than six terms but you will start with six terms. A special case of Theorem A.7 is the famous F.O.I.L., listed here: 1 Theorem A.8. F.O.I.L: The terms generated from the product of two binomials: (a + b)(c + d) can be verbalized as follows: Take the sum of the product of the First terms a and c, ac the product of the Outer terms a and d, ad the product of the Inner terms b and c, bc the product of the Last terms b and d, bd. That is, (a + b)(c + d) = ac + ad + bc + bd. Theorem A.7 is best proved using the technique known as Mathematical Induction which is covered in Section??. The result is really nothing more than repeated applications of the Distributive Property so it seems reasonable and we ll use it without proof for now. The other major piece of polynomial multiplication is one of the Power Rules of Exponents from page 87 in Section A., namely a n a m = a n+m. The Commutative and Associative Properties of addition and multiplication are also used extensively. We put all of these properties to good use in the next example. 1 We caved to peer pressure on this one. Apparently all of the cool Precalculus books have FOIL in them even though it s redundant once you know how to distribute multiplication across addition. In general, we don t like mechanical short-cuts that interfere with a student s understanding of the material and FOIL is one of the worst.

368 60 Algebra Review Perform the indicated operations and simplify. 1. ( x x + 1 ) (7x ). 4xz z(xz x + 4). (t + 1)(t 7) 4. ( y ) ( 9y + y + 4 ) 5. ( 4w 1 ) 6. [ (x + h) (x + h) ] ( x x ) Solution. 1. We begin distributing the negative as indicated on page 81 in Section A., then we rearrange and combine like terms: ( x x + 1 ) (7x ) = x x + 1 7x + Distribute = x x 7x Rearrange terms = x 9x + 4 Combine like terms Our answer is x 9x Following in our footsteps from the previous example, we first distribute the z through, then rearrange and combine like terms: 4xz z(xz x + 4) = 4xz z(xz) + z(x) z(4) Distribute = 4xz xz + xz 1z Multiply = xz + xz 1z Combine like terms We get our final answer: xz + xz 1z.. At last, we have a chance to use our F.O.I.L. technique: (t + 1)(t 7) = (t)(t) + (t)( 7) + (1)(t) + (1)( 7) F.O.I.L. = 6t 14t + t 7 Multiply = 6t 11t 7 Combine like terms We get 6t 11t 7 as our final answer. 4. We use the Generalized Distributive Property here, multiplying each term in the second quantity first by y, then by : ( y ) ( 9y + y + 4 ) = y ( 9y ) + y ( y ) + y ( 4 ) ( 9y ) ( y ) ( 4 ) To our surprise and delight, this product reduces to 7y. = 7y + 9y + y 4 9y y 4 8 = 7y + 9y 9y + y 4 y 4 = 7y

369 A.8 Polynomial Arithmetic Exponents do not distribute across powers so we know that ( 4w 1 ) (4w) ( 1 ). Instead, we proceed as follows: ( ( ) ( 4w ) 1 = 4w 1 4w 1 ( = (4w)(4w) + (4w) 1 ) ( + 1 ) ( (4w) + 1 ) ( 1 ) ) F.O.I.L. = 16w w w = 16w 4w Multiply Combine like terms Our (correct) final answer is 16w 4w Our last example has two levels of grouping symbols. We begin simplifying the quantity inside the brackets, expanding (x + h) in the same way we expanded (4w 1 ) in our previous example: (x + h) = (x + h)(x + h) = (x)(x) + (x)(h) + (h)(x) + (h)(h) = x + xh + h When we substitute this into our expression, we envelope it in parentheses, as usual, so that we don t forget to distribute the negative. [ (x + h) (x + h) ] ( x x ) = [ (x + h) ( x + xh + h )] ( x x ) Substitute = [ x + h x xh h ] ( x x ) Distribute = x + h x xh h x + x Distribute = x x + h x + x xh h Rearrange terms = h xh h Combine like terms We find no like terms in h xh h so we are finished. We conclude our discussion of polynomial multiplication by showcasing two special products which happen often enough they should be committed to memory. Theorem A.9. Special Products: Let a and b be real numbers: Perfect Square: (a + b) = a + ab + b and (a b) = a ab + b Difference of Two Squares: (a b)(a + b) = a b The formulas in Theorem A.9 can be verified by working through the multiplication. See the remarks following the Properties of Exponents on 87. These are both special cases of F.O.I.L.

370 6 Algebra Review A.8. Polynomial Long Division. We now turn our attention to polynomial long division. Dividing two polynomials follows the same algorithm, in principle, as dividing two natural numbers so we review that process first. Suppose we wished to divide 585 by 79. The standard division tableau is given below In this case, 79 is called the divisor, 585 is called the dividend, is called the quotient and 57 is called the remainder. We can check our answer by showing: 57 dividend = (divisor)(quotient) + remainder or in this case, 585 = (79)()+57. We hope that the long division tableau evokes warm, fuzzy memories of your formative years as opposed to feelings of hopelessness and frustration. If you experience the latter, keep in mind that the Division Algorithm essentially is a two-step process, iterated over and over again. First, we guess the number of times the divisor goes into the dividend and then we subtract off our guess. We repeat those steps with what s left over until what s left over (the remainder) is less than what we started with (the divisor). That s all there is to it! The division algorithm for polynomials has the same basic two steps but when we subtract polynomials, we must take care to subtract like terms only. As a transition to polynomial division, let s write out our previous division tableau in expanded form ( ) ( ) Written this way, we see that when we line up the digits we are really lining up the coefficients of the corresponding powers of 10 - much like how we ll have to keep the powers of x lined up in the same columns. The big difference between polynomial division and the division of natural numbers is that the value of x is an unknown quantity. So unlike using the known value of 10, when we subtract there can be no regrouping of coefficients as in our previous example. (The subtraction requires us to regroup or borrow from the tens digit, then the hundreds digit.) This actually makes polynomial division easier. 4 4 In our opinion - you can judge for yourself.

371 A.8 Polynomial Arithmetic 6 Before we dive into examples, we first state a theorem telling us when we can divide two polynomials, and what to expect when we do so. Theorem A.10. Polynomial Division: Let d and p be nonzero polynomials where the degree of p is greater than or equal to the degree of d. There exist two unique polynomials, q and r, such that p = d q + r, where either r = 0 or the degree of r is strictly less than the degree of d. Essentially, Theorem A.10 tells us that we can divide polynomials whenever the degree of the divisor is less than or equal to the degree of the dividend. We know we re done with the division when the polynomial left over (the remainder) has a degree strictly less than the divisor. It s time to walk through a few examples to refresh your memory. Perform the indicated division. Check your answer by showing dividend = (divisor)(quotient) + remainder 1. ( x + 4x 5x 14 ) (x ). (t + 7) (t 4). ( 6y 1 ) (y + 5) 4. ( w ) ( w ). Solution. 1. To begin ( x + 4x 5x 14 ) (x ), we divide the first term in the dividend, namely x, by the first term in the divisor, namely x, and get x x = x. This then becomes the first term in the quotient. We proceed as in regular long division at this point: we multiply the entire divisor, x, by this first term in the quotient to get x (x ) = x x. We then subtract this result from the dividend. x x x + 4x 5x 14 ( x x ) 6x 5x Now we bring down the next term of the quotient, namely 5x, and repeat the process. We divide 6x = 6x, and add this to the quotient polynomial, multiply it by the divisor (which yields 6x(x ) = x 6x 1x) and subtract. x + 6x x x + 4x 5x 14 ( x x ) 6x 5x ( 6x 1x) 7x 14 Finally, we bring down the last term of the dividend, namely 14, and repeat the process. We divide 7x = 7, add this to the quotient, multiply it by the divisor (which yields 7(x ) = 7x 14) and x

372 64 Algebra Review subtract. x + 6x + 7 x x + 4x 5x 14 ( x x ) 6x 5x ( 6x 1x) 7x 14 (7x 14) 0 In this case, we get a quotient of x + 6x + 7 with a remainder of 0. To check our answer, we compute (x ) ( x + 6x + 7 ) + 0 = x + 6x + 7x x 1x 14 = x + 4x 5x 14. To compute (t + 7) (t 4), we start as before. We find t =, so that becomes the first (and t only) term in the quotient. We multiply the divisor (t 4) by and get t 8. We subtract this from the divided and get 9. t 4 ( t + 7 t 8 ) 9 Our answer is 9 with a remainder of. To check our answer, we compute ( ) (t 4) + 9 = t = t + 1 = t + 7. When we set-up the tableau for ( 6y 1 ) (y + 5), we must first issue a placeholder for the missing y-term in the dividend, 6y 1 = 6y + 0y 1. We then proceed as before. Since 6y y = y, y is the first term in our quotient. We multiply (y + 5) times y and subtract it from the dividend.

373 A.8 Polynomial Arithmetic 65 We bring down the 1, and repeat. y 15 y+5 6y + 0y 1 ( 6y + 15y) ( 15y 1 15y 75 ) 7 Our answer is y 15 7 with a remainder of. To check our answer, we compute: ( (y + 5) y 15 ) + 7 = 6y 15y + 15y = 6y 1 4. For our last example, we need placeholders for both the divisor w = w + 0w and the dividend w = w + 0w + 0w + 0. The first term in the quotient is w = w, and when we multiply and subtract this from the dividend, we re left with just 0w + w + 0 = w. w w +0w w +0w + 0w +0 ( w +0w w ) w 0w + w +0 Since the degree of w (which is 1) is less than the degree of the divisor (which is ), we are done. 5 Our answer is w with a remainder of w. To check, we compute: ( w ) w + w = w w + w = w 5 Since 0w w = 0, we could proceed, write our quotient as w + 0, and move on... but even pedants have limits.

374 66 Algebra Review A.8. Exercises In Exercises 1-15, perform the indicated operations and simplify. 1. (4 x) + (x + x + 7). t + 4t ( t). q(00 q) (5q + 500) ( 4. (y 1)(y + 1) 5. x ) (x + 5) 6. (4t + )(t ) 7. w(w 5)(w + 5) 8. (5a )(5a a + 9) 9. (x x + )(x + x + ) 10. ( 7 z)( 7 + z) 11. (x 5) 1. (x 5)(x + x 5 + 5) 1. (w ) (w + 9) 14. (x +h) (x +h) (x x) 15. (x [ + 5])(x [ 5]) In Exercises 16-7, perform the indicated division. Check your answer by showing dividend = (divisor)(quotient) + remainder 16. (5x x + 1) (x + 1) 17. (y + 6y 7) (y ) 18. (6w ) (w + 5) 19. (x + 1) (x 4) 0. (t 4) (t + 1) 1. (w 8) (5w 10). (x x + 1) (x + 1). (4y 4 + y + 1) (y y + 1) 4. w 4 (w ) 5. (5t t + 1) (t + 4) 6. (t 4) (t 4) 7. (x x 1) (x [1 ]) In Exercises 8 - verify the given formula by showing the left hand side of the equation simplifies to the right hand side of the equation. 8. Perfect Cube: (a + b) = a + a b + ab + b 9. Difference of Cubes: (a b)(a + ab + b ) = a b 0. Sum of Cubes: (a + b)(a ab + b ) = a + b 1. Perfect Quartic: (a + b) 4 = a 4 + 4a b + 6a b + 4ab + b 4. Difference of Quartics: (a b)(a + b)(a + b ) = a 4 b 4. Sum of Quartics: (a + ab + b )(a ab + b ) = a 4 + b 4 4. With help from your classmates, determine under what conditions (a + b) = a + b. What about (a + b) = a + b? In general, when does (a + b) n = a n + b n for a natural number n?

375 A.8 Polynomial Arithmetic 67 A.8.4 Answers 1. x x t + 6t 6. q + 195q y + y 1 5. x + 7 x t t + 8t w 7 50w 8. 15a x 4 + x z 11. x x 5 + x x w 14. h + xh h 15. x 4x quotient: 5x 8, remainder: quotient: y + 15, remainder: quotient:, remainder: quotient: 11, remainder: 0. quotient: t 1 4, remainder: quotient: w 5 + w , remainder: 0. quotient:, remainder: x + 1. quotient: y + y + 1, remainder: 0 4. quotient: w, remainder: w 5. quotient: 5t, remainder: 1t quotient: 6 t + t 4 +, remainder: 0 7. quotient: x 1, remainder: 0 6 Note: 16 =.

376 68 Algebra Review A.9 Basic Factoring Techniques Now that we have reviewed the basics of polynomial arithmetic it s time to review the basic techniques of factoring polynomial expressions. Our goal is to apply these techniques to help us solve certain specialized classes of non-linear equations. Given that factoring literally means to resolve a product into its factors, it is, in the purest sense, undoing multiplication. If this sounds like division to you then you ve been paying attention. Let s start with a numerical example. Suppose we are asked to factor 167. We could write 167 = 167 1, and while this is technically a factorization of 167, it s probably not an answer the poser of the question would accept. Usually, when we re asked to factor a natural number, we are being asked to resolve it into to a product of socalled prime numbers. 1 Recall that prime numbers are defined as natural numbers whose only (natural number) factors are themselves and 1. They are, in essence, the building blocks of natural numbers as far as multiplication is concerned. Said differently, we can build - via multiplication - any natural number given enough primes. So how do we find the prime factors of 167? We start by dividing each of the primes:,, 5, 7, etc., into 167 until we get a remainder of 0. Eventually, we find that = 961 with a remainder of 0, which means 167 = So factoring and division are indeed closely related - factors of a number are precisely the divisors of that number which produce a zero remainder. We continue our efforts to see if 961 can be factored down further, and we find that 961 = 1 1. Hence, 167 can be completely factored as (This factorization is called the prime factorization of 167.) In factoring natural numbers, our building blocks are prime numbers, so to be completely factored means that every number used in the factorization of a given number is prime. One of the challenges when it comes to factoring polynomial expressions is to explain what it means to be completely factored. In this section, our building blocks for factoring polynomials are irreducible polynomials as defined below. A polynomial is said to be irreducible if it cannot be written as the product of polynomials of lower degree. While Definition A.9 seems straightforward enough, sometimes a greater level of specificity is required. For example, x = (x )(x + ). While x and x + are perfectly fine polynomials, factoring which requires irrational numbers is usually saved for a more advanced treatment of factoring. For now, we will restrict ourselves to factoring using rational coefficients. So, while the polynomial x can be factored using irrational numbers, it is called irreducible over the rationals, since there are no polynomials with rational coefficients of smaller degree which can be used to factor it. 4 Since polynomials involve terms, the first step in any factoring strategy involves pulling out factors which are common to all of the terms. For example, in the polynomial 18x y 54x y 1xy, each coefficient is a multiple of 6 so we can begin the factorization as 6(x y 9x y xy ). The remaining coefficients:, 9 and, have no common factors so 6 was the greatest common factor. What about the variables? Each 1 As mentioned in Section A., this is possible, in only one way, thanks to the Fundamental Theorem of Arithmetic. We ll refer back to this when we get to Section??. See Section??. 4 If this isn t immediately obvious, don t worry - in some sense, it shouldn t be. We ll talk more about this later.

377 A.9 Basic Factoring Techniques 69 term contains an x, so we can factor an x from each term. When we do this, we are effectively dividing each term by x which means the exponent on x in each term is reduced by 1: 6x(xy 9x y y ). Next, we see that each term has a factor of y in it. In fact, each term has at least two factors of y in it, since the lowest exponent on y in each term is. This means that we can factor y from each term. Again, factoring out y from each term is tantamount to dividing each term by y so the exponent on y in each term is reduced by two: 6xy (xy 9x ). Just like we checked our division by multiplication in the previous section, we can check our factoring here by multiplication, too. 6xy (xy 9x ) = (6xy )(xy) (6xy )(9x ) (6xy )() = 18x y 54x y 1xy. We summarize how to find the Greatest Common Factor (G.C.F.) of a polynomial expression below. Finding the G.C.F. of a Polynomial Expression If the coefficients are integers, find the G.C.F. of the coefficients. NOTE 1: If all of the coefficients are negative, consider the negative as part of the G.C.F.. NOTE : If the coefficients involve fractions, get a common denominator, combine numerators, reduce to lowest terms and apply this step to the polynomial in the numerator. If a variable is common to all of the terms, the G.C.F. contains that variable to the smallest exponent which appears among the terms. For example, to factor 5 z 6z, we would first get a common denominator and factor as: 5 z 6z = z 0z 5 = z (z + 10) 5 = z (z + 10) 5 = 5 z (z + 10) We now list some common factoring formulas, each of which can be verified by multiplying out the right side of the equation. While they all should look familiar - this is a review section after all - some should look more familiar than others since they appeared as special product formulas in the previous section. Common Factoring Formulas Perfect Square Trinomials: a + ab + b = (a + b) and a ab + b = (a b) Difference of Two Squares: a b = (a b)(a + b) NOTE: In general, the sum of squares, a + b is irreducible over the rationals. Sum of Two Cubes: a + b = (a + b)(a ab + b ) NOTE: In general, a ab + b is irreducible over the rationals. Difference of Two Cubes: a b = (a b)(a + ab + b ) NOTE: In general, a + ab + b is irreducible over the rationals. The example on the next page gives us practice with these formulas.

378 70 Algebra Review Factor the following polynomials completely over the rationals. That is, write each polynomial as a product polynomials of lowest degree which are irreducible over the rationals x 48x +. 64y 1. 75t 4 + 0t + t 4. w 4 z wz t 4 6. x 6 64 Solution. 1. Our first step is to factor out the G.C.F. which in this case is. To match what is left with one of the special forms, we rewrite 9x = (x) and 16 = 4. Since the middle term is 4x = (4)(x), we see that we have a perfect square trinomial. 18x 48x + = (9x 4x + 16) Factor out G.C.F. = ((x) (4)(x) + (4) ) = (x 4) Perfect Square Trinomial: a = x, b = 4 Our final answer is (x 4). To check our work, we multiply out (x 4) to show that it equals 18x 48x +.. For 64y 1, we note that the G.C.F. of the terms is just 1, so there is nothing (of substance) to factor out of both terms. Since 64y 1 is the difference of two terms, one of which is a square, we look to the Difference of Squares Formula for inspiration. Seeing 64y = (8y) and 1 = 1, we get 64y 1 = (8y) 1 = (8y 1)(8y + 1) Difference of Squares, a = 8y, b = 1 As before, we can check our answer by multiplying out (8y 1)(8y +1) to show that it equals 64y 1.. The G.C.F. of the terms in 75t 4 + 0t + t is t, so we factor that out first. We identify what remains as a perfect square trinomial: 75t 4 + 0t + t = t (5t + 10t + 1) Factor out G.C.F. = t ((5t) + (1)(5t) + 1 ) = t (5t + 1) Perfect Square Trinomial, a = 5t, b = 1 Our final answer is t (5t + 1), which the reader is invited to check. 4. For w 4 z wz 4, we identify the G.C.F. as wz and once we factor it out a difference of cubes is revealed: w 4 z wz 4 = wz(w z ) Factor out G.C.F. = wz(w z)(w + wz + z ) Difference of Cubes, a = w, b = z Our final answer is wz(w z)(w + wz + z ). The reader is strongly encouraged to multiply this out to see that it reduces to w 4 z wz 4.

379 A.9 Basic Factoring Techniques The G.C.F. of the terms in 81 16t 4 is just 1 so there is nothing of substance to factor out from both terms. With just a difference of two terms, we are limited to fitting this polynomial into either the Difference of Two Squares or Difference of Two Cubes formula. Since the variable here is t 4, and 4 is a multiple of, we can think of t 4 = (t ). This means that we can write 16t 4 = (4t ) which is a perfect square. (Since 4 is not a multiple of, we cannot write t 4 as a perfect cube of a polynomial.) Identifying 81 = 9 and 16t 4 = (4t ), we apply the Difference of Squares Formula to get: 81 16t 4 = 9 (4t ) = (9 4t )(9 + 4t ) Difference of Squares, a = 9, b = 4t At this point, we have an opportunity to proceed further. Identifying 9 = and 4t = (t), we see that we have another difference of squares in the first quantity, which we can reduce. (The sum of two squares in the second quantity cannot be factored over the rationals.) 81 16t 4 = (9 4t )(9 + 4t ) = ( (t) )(9 + 4t ) = ( t)( + t)(9 + 4t ) Difference of Squares, a =, b = t As always, the reader is encouraged to multiply out ( t)( + t)(9 + 4t ) to check the result. 6. With a G.C.F. of 1 and just two terms, x 6 64 is a candidate for both the Difference of Squares and the Difference of Cubes formulas. Notice that we can identify x 6 = (x ) and 64 = 8 (both perfect squares), but also x 6 = (x ) and 64 = 4 (both perfect cubes). If we follow the Difference of Squares approach, we get: x 6 64 = (x ) 8 = (x 8)(x + 8) Difference of Squares, a = x and b = 8 At this point, we have an opportunity to use both the Difference and Sum of Cubes formulas: x 6 64 = (x )(x + ) = (x )(x + x + )(x + )(x x + ) Sum / Difference of Cubes, a = x, b = = (x )(x + )(x x + 4)(x + x + 4) Rearrange factors From this approach, our final answer is (x )(x + )(x x + 4)(x + x + 4). Following the Difference of Cubes Formula approach, we get x 6 64 = (x ) 4 = (x 4)((x ) + 4x + 4 ) Difference of Cubes, a = x, b = 4 = (x 4)(x 4 + 4x + 16) At this point, we recognize x 4 as a difference of two squares: x 6 64 = (x )(x 4 + 4x + 16) = (x )(x + )(x 4 + 4x + 16) Difference of Squares, a = x, b =

380 7 Algebra Review Unfortunately, the remaining factor x 4 + 4x + 16 is not a perfect square trinomial - the middle term would have to be 8x for this to work - so our final answer using this approach is (x )(x + )(x 4 + 4x + 16). This isn t as factored as our result from the Difference of Squares approach which was (x )(x + )(x x + 4)(x + x + 4). While it is true that x 4 + 4x + 16 = (x x + 4)(x + x + 4), there is no intuitive way to motivate this factorization at this point. 5 The moral of the story? When given the option between using the Difference of Squares and Difference of Cubes, start with the Difference of Squares. Our final answer to this problem is (x )(x + )(x x + 4)(x + x + 4). The reader is strongly encouraged to show that this reduces down to x 6 64 after performing all of the multiplication. The formulas on page 71, while useful, can only take us so far. Thus we need to review some additional factoring strategies which should be good friends from back in the day! Additional Factoring Formulas un-f.o.i.l.ing : Given a trinomial Ax + Bx + C, try to reverse the F.O.I.L. process. That is, find a, b, c and d such that Ax + Bx + C = (ax + b)(cx + d). NOTE: This means ac = A, bd = C and B = ad + bc. Factor by Grouping: If the expression contains four terms with no common factors among the four terms, try factor by grouping : ac + bc + ad + bd = (a + b)c + (a + b)d = (a + b)(c + d) The techniques of un-f.o.i.l.ing and factoring by grouping are difficult to describe in general but should make sense to you with enough practice. Be forewarned - like all Rules of Thumb, these strategies work just often enough to be useful, but you can be sure there are exceptions which will defy any advice given here and will require some inspiration to solve. 6 Even though Chapter?? will give us more powerful factoring methods, we ll find that, in the end, there is no single algorithm for factoring which works for every polynomial. In other words, there will be times when you just have to try something and see what happens. Factor the following polynomials completely over the integers x x 6. t 11t y 1y 4. 18xy 54xy 180x 5. t 10t + t x 4 + 4x Of course, this begs the question, How do we know x x + 4 and x + x + 4 are irreducible? (We were told so on page 71, but no reason was given.) Stay tuned! We ll get back to this in due course. 6 Jeff will be sure to pepper the Exercises with these. 7 This means that all of the coefficients in the factors will be integers. In a rare departure from form, Carl decided to avoid fractions in this set of examples. Don t get complacent, though, because fractions will return with a vengeance soon enough.

381 A.9 Basic Factoring Techniques 7 Solution. 1. The G.C.F. of the terms x x 6 is 1 and x x 6 isn t a perfect square trinomial (Think about why not.) so we try to reverse the F.O.I.L. process and look for integers a, b, c and d such that (ax + b)(cx + d) = x x 6. To get started, we note that ac = 1. Since a and c are meant to be integers, that leaves us with either a and c both being 1, or a and c both being 1. We ll go with a = c = 1, since we can factor 8 the negatives into our choices for b and d. This yields (x + b)(x + d) = x x 6. Next, we use the fact that bd = 6. The product is negative so we know that one of b or d is positive and the other is negative. Since b and d are integers, one of b or d is ±1 and the other is 6 OR one of b or d is ± and the other is. After some guessing and checking, 9 we find that x x 6 = (x + )(x ).. As with the previous example, we check the G.C.F. of the terms in t 11t + 5, determine it to be 1 and see that the polynomial doesn t fit the pattern for a perfect square trinomial. We now try to find integers a, b, c and d such that (at + b)(ct + d) = t 11t + 5. Since ac =, we have that one of a or c is, and the other is 1. (Once again, we ignore the negative options.) At this stage, there is nothing really distinguishing a from c so we choose a = and c = 1. Now we look for b and d so that (t + b)(t + d) = t 11t + 5. We know bd = 5 so one of b or d is ±1 and the other ±5. Given that bd is positive, b and d must have the same sign. The negative middle term 11t guides us to guess b = 1 and d = 5 so that we get (t 1)(t 5) = t 11t + 5. We verify our answer by multiplying. 10. Once again, we check for a nontrivial G.C.F. and see if 6 11y 1y fits the pattern of a perfect square. Twice disappointed, we rewrite 6 11y 1y = 1y 11y + 6 for notational convenience. We now look for integers a, b, c and d such that 1y 11y + 6 = (ay + b)(cy + d). Since ac = 1, we know that one of a or c is ±1 and the other ±1 OR one of them is ± and the other is ±6 OR one of them is ± while the other is ±4. Since their product is 1, however, we know one of them is positive, while the other is negative. To make matters worse, the constant term 6 has its fair share of factors, too. Our answers for b and d lie among the pairs ±1 and ±6, ± and ±18, ±4 and ±9, or ±6. Since we know one of a or c will be negative, we can simplify our choices for b and d and just look at the positive possibilities. After some guessing and checking, 11 we find ( y + 4)(4y + 9) = 1y 11y Since the G.C.F. of the terms in 18xy 54xy 180x is 18x, we begin the problem by factoring it out first: 18xy 54xy 180x = 18x(y y 10). We now focus our attention on y y 10. We can take a and c to both be 1 which yields (y + b)(y + d) = y y 10. Our choices for b and d are among the factor pairs of 10: ±1 and ±10 or ± and ±5, where one of b or d is positive and the other is negative. We find (y 5)(y + ) = y y 10. Our final answer is 18xy 54xy 180x = 18x(y 5)(y + ). 8 Pun intended! 9 The authors have seen some strange gimmicks that allegedly help students with this step. We don t like them so we re sticking with good old-fashioned guessing and checking. 10 That s the checking part of guessing and checking. 11 Some of these guesses can be more educated than others. Since the middle term is relatively small, we don t expect the extreme factors of 6 and 1 to appear, for instance.

382 74 Algebra Review 5. Since t 10t t + 15 has four terms, we are pretty much resigned to factoring by grouping. The strategy here is to factor out the G.C.F. from two pairs of terms, and see if this reveals a common factor. If we group the first two terms, we can factor out a t to get t 10t = t (t 5). We now try to factor something out of the last two terms that will leave us with a factor of (t 5). Sure enough, we can factor out a from both: t + 15 = (t 5). Hence, we get t 10t t + 15 = t (t 5) (t 5) = (t )(t 5) Now the question becomes can we factor t over the integers? This would require integers a, b, c and d such that (at + b)(ct + d) = t. Since ab = and cd =, we aren t left with many options - in fact, we really have only four choices: (t 1)(t + ), (t + 1)(t ), (t )(t + 1) and (t + )(t 1). None of these produces t - which means it s irreducible over the integers - thus our final answer is (t )(t 5). 6. Our last example, x 4 + 4x + 16, is our old friend from Example A.9. As noted there, it is not a perfect square trinomial, so we could try to reverse the F.O.I.L. process. This is complicated by the fact that our highest degree term is x 4, so we would have to look at factorizations of the form (x + b)(x + d) as well as (x + b)(x + d). We leave it to the reader to show that neither of those work. This is an example of where trying something pays off. Even though we ve stated that it is not a perfect square trinomial, it s pretty close. Identifying x 4 = (x ) and 16 = 4, we d have (x + 4) = x 4 + 8x + 16, but instead of 8x as our middle term, we only have 4x. We could add in the extra 4x we need, but to keep the balance, we d have to subtract it off. Doing so produces an unexpected opportunity: x 4 + 4x + 16 = x 4 + 4x (4x 4x ) Adding and subtracting the same term = x 4 + 8x x Rearranging terms = (x + 4) (x) Factoring perfect square trinomial = [(x + 4) x][(x + 4) + x] Difference of Squares: a = (x + 4), b = x = (x x + 4)(x + x + 4) Rearraging terms We leave it to the reader to check that neither x x + 4 nor x + x + 4 factor over the integers, so we are done. A.9.1 Solving Equations by Factoring Many students wonder why they are forced to learn how to factor. Simply put, factoring is our main tool for solving the non-linear equations which arise in many of the applications of Mathematics. 1 We use factoring in conjunction with the Zero Product Property of Real Numbers which was first stated on page 80 and is given here again for reference. The Zero Product Property of Real Numbers: If a and b are real numbers with ab = 0 then either a = 0 or b = 0 or both. 1 Also known as story problems or real-world examples.

383 A.9 Basic Factoring Techniques 75 Consider the equation 6x + 11x = 10. To see how the Zero Product Property is used to help us solve this equation, we first set the equation equal to zero and then apply the techniques from Example A.9: 6x + 11x = 10 6x + 11x 10 = 0 Subtract 10 from both sides (x + 5)(x ) = 0 Factor x + 5 = 0 or x = 0 Zero Product Property x = 5 or x = a = x + 5, b = x The reader should check that both of these solutions satisfy the original equation. It is critical that you see the importance of setting the expression equal to 0 before factoring. Otherwise, we d get something silly like: 6x + 11x = 10 x(6x + 11) = 10 Factor What we cannot deduce from this equation is that x = 10 or 6x + 11 = 10 or that x = and 6x + 11 = 5. (It s wrong and you should feel bad if you do it.) It is precisely because 0 plays such a special role in the arithmetic of real numbers (as the Additive Identity) that we can assume a factor is 0 when the product is 0. No other real number has that ability. We summarize the correct equation solving strategy below. Strategy for Solving Non-linear Equations 1. Put all of the nonzero terms on one side of the equation so that the other side is 0.. Factor.. Use the Zero Product Property of Real Numbers and set each factor equal to Solve each of the resulting equations. Let s finish the section with a collection of examples in which we use this strategy. Solve the following equations. 1. x = 5 16x. t = 1 + 4t 4. (y 1) = (y 1) 4. w 4 = 8w 1 1 w z(z(18z + 9) 50) = 5 6. x 4 8x 9 = 0

384 76 Algebra Review Solution. 1. We begin by gathering all of the nonzero terms to one side getting 0 on the other and then we proceed to factor and apply the Zero Product Property. x = 5 16x x + 16x 5 = 0 Add 16x, subtract 5 (x 5)(x + 7) = 0 Factor x 5 = 0 or x + 7 = 0 Zero Product Property x = 5 or x = 7 We check our answers by substituting each of them into the original equation. Plugging in x = 5 yields 5 on both sides while x = 7 gives 147 on both sides.. To solve t = 1+4t 4, we first clear fractions then move all of the nonzero terms to one side of the equation, factor and apply the Zero Product Property. t = 1 + 4t 4 4t = 1 + 4t Clear fractions (multiply by 4) 0 = 1 + 4t 4t Subtract 4 0 = 4t 4t + 1 Rearrange terms 0 = (t 1) Factor (Perfect Square Trinomial) At this point, we get (t 1) = (t 1)(t 1) = 0, so, the Zero Product Property gives us t 1 = 0 in both cases. 1 Our final answer is t = 1, which we invite the reader to check.. Following the strategy outlined above, the first step to solving (y 1) = (y 1) is to gather the nonzero terms on one side of the equation with 0 on the other side and factor. (y 1) = (y 1) (y 1) (y 1) = 0 Subtract (y 1) (y 1)[(y 1) ] = 0 Factor out G.C.F. (y 1)(y ) = 0 Simplify y 1 = 0 or y = 0 y = 1 or y = Both of these answers are easily checked by substituting them into the original equation. An alternative method to solving this equation is to begin by dividing both sides by (y 1) to simplify things outright. As we saw in Example A.4.1, however, whenever we divide by a variable quantity, we make the explicit assumption that this quantity is nonzero. Thus we must stipulate that y 1 0. (y 1) (y 1) = (y 1) (y 1) y 1 = y = 1 More generally, given a positive power p, the only solution to x p = 0 is x = 0. Divide by (y 1) - this assumes (y 1) 0

385 A.9 Basic Factoring Techniques 77 Note that in this approach, we obtain the y = solution, but we lose the y = 1 solution. How did that happen? Assuming y 1 0 is equivalent to assuming y 1. This is an issue because y = 1 is a solution to the original equation and it was divided out too early. The moral of the story? If you decide to divide by a variable expression, double check that you aren t excluding any solutions Proceeding as before, we clear fractions, gather the nonzero terms on one side of the equation, have 0 on the other and factor. w 4 = 8w 1 w 4 ( ) ( 1 4 w 4 8w 1 1 = 1 w ) 4 Multiply by w 4 = (8w 1) (w 4) Distribute 4w 4 = 8w 1 w + 1 Distribute 0 = 8w 1 w + 1 4w 4 Subtract 4w 4 0 = 8w w 4w 4 Gather like terms 0 = w (8w 4w ) Factor out G.C.F. At this point, we apply the Zero Product Property to deduce that w = 0 or 8w 4w = 0. From w = 0, we get w = 0. To solve 8w 4w = 0, we rearrange terms and factor: 4w + 8w = (w 1)( w + ) = 0. Applying the Zero Product Property again, we get w 1 = 0 (which gives w = 1 ), or w + = 0 (which gives w = ). Our final answers are w = 0, w = 1 and w =. The reader is encouraged to check each of these answers in the original equation. (You need the practice with fractions!) 5. For our next example, we begin by subtracting the 5 from both sides then work out the indicated operations before factoring by grouping. z(z(18z + 9) 50) = 5 z(z(18z + 9) 50) 5 = 0 Subtract 5 z(18z + 9z 50) 5 = 0 Distribute 18z + 9z 50z 5 = 0 Distribute 9z (z + 1) 5(z + 1) = 0 Factor (9z 5)(z + 1) = 0 Factor At this point, we use the Zero Product Property and get 9z 5 = 0 or z + 1 = 0. The latter gives z = 1 whereas the former factors as (z 5)(z + 5) = 0. Applying the Zero Product Property again gives z 5 = 0 (so z = 5 ) or z + 5 = 0 (so z = 5.) Our final answers are z = 1, z = 5 and z = 5, each of which is good fun to check. 6. The nonzero terms of the equation x 4 8x 9 = 0 are already on one side of the equation so we proceed to factor. This trinomial doesn t fit the pattern of a perfect square so we attempt to reverse 14 You will see other examples throughout this text where dividing by a variable quantity does more harm than good. Keep this basic one in mind as you move on in your studies - it s a good cautionary tale.

386 78 Algebra Review the F.O.I.L.ing process. With an x 4 term, we have two possible forms to try: (ax + b)(cx + d) and (ax + b)(cx + d). We leave it to you to show that (ax + b)(cx + d) does not work and we show that (ax + b)(cx + d) does. Since the coefficient of x 4 is 1, we take a = c = 1. The constant term is 9 so we know b and d have opposite signs and our choices are limited to two options: either b and d come from ±1 and ±9 OR one is while the other is. After some trial and error, we get x 4 8x 9 = (x 9)(x + 1). Hence x 4 8x 9 = 0 reduces to (x 9)(x + 1) = 0. The Zero Product Property tells us that either x 9 = 0 or x + 1 = 0. To solve the former, we factor: (x )(x + ) = 0, so x = 0 (hence, x = ) or x + = 0 (hence, x = ). The equation x + 1 = 0 has no (real) solution, since for any real number x, x is always 0 or greater. Thus x + 1 is always positive. Our final answers are x = and x =. As always, the reader is invited to check both answers in the original equation.

387 A.9 Basic Factoring Techniques 79 A.9. Exercises In Exercises 1-0, factor completely over the integers. Check your answer by multiplication. 1. x 10x. 1t 5 8t. 16xy 1x y 4. 5(m + ) 4(m + ) 5. (x 1)(x + ) 4(x 1) 6. t (t 5) + t 5 7. w t 9. 81t z 64y (y + ) 4y 1. (x + h) (x + h) 1. y 4y t + 10t x 6x + 7x 16. m m x 18. t 6 + t 19. x 5x y 1y t + 16t x x m m 4. 7w w 5. m + 9m 1m 6. x 4 + x (t 1) + (t 1) x 5x 9x t + t t y 4 + 5y + 9 In Exercises 1-45, find all rational number solutions. Check your answers. 1. (7x + )(x 5) = 0. (t 1) (t + 4) = 0. (y + 4)(y + y 10) = t = t 5. y + = y 6. 6x = 8x x 4 = 9x 8. w(6w + 11) = w + 5w + = (w + 1) 40. x (x ) = 16(x ) 41. (t + 1) = (t + 1) 4. a = 6 a 4. 8t = t x + x = x y 4 y = (y + ) 46. With help from your classmates, factor 4x 4 + 8x With help from your classmates, find an equation which has, 1, and 117 as solutions. 15 y 4 + 5y + 9 = (y 4 + 6y + 9) y

388 80 Algebra Review A.9. Answers 1. x(1 5x). 4t (t ). 4xy(4y x) 4. (m + ) (4m + 7) 5. (x 1)(x 1) 6. (t 5)(t + 1) 7. (w 11)(w + 11) 8. (7 t)(7 + t) 9. (t )(t + )(9t + 4) 10. (z 8y )(z + 8y ) 11. (y )(y + 1) 1. (x + h)(x + h 1)(x + h + 1) 1. (y 1) 14. (5t + 1) 15. x(x ) 16. (m + 5) 17. ( x)(9 + 6x + 4x ) 18. t (t + 1)(t t + 1) 19. (x 7)(x + ) 0. (y 9)(y ) 1. (t + 1)(t + 5). (x 5)(x 4). (7 m)(5 + m) 4. ( w + 1)(w ) 5. m(m 1)(m + 4) 6. (x )(x + )(x + 5) 7. (t )(t + )(t + 1) 8. (x )(x + )(x 5) 9. (t )(1 t)(1 + t) 0. (y y + )(y + y + ) 1. x = 7 or x = 5. t = 1 or t = 4. y = 5 or y = 4. t = 0 or t = 4 5. y = 1 or y = 6. x = or x = x = 0 or x = ± 4 8. w = 5 or w = 9. w = 5 or w = x = or x = ±4 41. t = 1, t = 1, or t = 0 4. a = ±1 4. t = 4 or t = 44. x = 45. y = ±

389 A.10 Quadratic Equations 81 A.10 Quadratic Equations In Section A.9.1, we reviewed how to solve basic non-linear equations by factoring. The astute reader should have noticed that all of the equations in that section were carefully constructed so that the polynomials could be factored using the integers. To demonstrate just how contrived the equations had to be, we can solve x + 5x = 0 by factoring, (x 1)(x + ) = 0, from which we obtain x = 1 and x =. If we change the 5 to a 6 and try to solve x + 6x = 0, however, we find that this polynomial doesn t factor over the integers and we are stuck. It turns out that there are two real number solutions to this equation, but they are irrational numbers, and the goal of this section is to review the techniques which allow us to find these solutions. 1 In this section, we focus our attention on quadratic equations. An equation is said to be quadratic in a variable x if it can be written in the form ax + bx + c = 0 where a, b and c are expressions which do not involve x and a 0. Think of quadratic equations as equations that are one degree up from linear equations - instead of the highest power of x being just x = x 1, it s x. The simplest class of quadratic equations to solve are the ones in which b = 0. In that case, we have the following. Solving Quadratic Equations by Extracting Square Roots If c is a real number with c 0, the solutions to x = c are x = ± c. Note: If c < 0, x = c has no real number solutions. There are a couple different ways to see why Extracting Square Roots works, both of which are demonstrated by solving the equation x =. If we follow the procedure outlined in the previous section, we subtract from both sides to get x = 0 and we now try to factor x. As mentioned in the remarks following Definition A.9, we could think of x = x ( ) and apply the Difference of Squares formula to factor x = (x )(x + ). We solve (x )(x + ) = 0 by using the Zero Product Property as before by setting each factor equal to zero: x = 0 and x + 0. We get the answers x = ±. In general, if c 0, then c is a real number, so x c = x ( c) = (x c)(x + c). Replacing the with c in the above discussion gives the general result. Another way to view this result is to visualize taking the square root of both sides: since x = c, x = c. How do we simplify x? We have to exercise a bit of caution here. Note that (5) and ( 5) both simplify to 5 = 5. In both cases, x returned a positive number, since the negative in 5 was squared away before we took the square root. In other words, x is x if x is positive, or, if x is negative, we make x positive - that is, x = x, the absolute value of x. So from x =, we take the square root of both sides of the equation to get x =. This simplifies to x =, which by Theorem A.5 is equivalent to x = or x =. Replacing the in the previous argument with c, gives the general result. As you might expect, Extracting Square Roots can be applied to more complicated equations. Consider the equation below. We can solve it by Extracting Square Roots provided we first isolate the quantity that 1 While our discussion in this section departs from factoring, we ll see in Chapter?? that the same correspondence between factoring and solving equations holds whether or not the polynomial factors over the integers.

390 8 Algebra Review is being squared : ( x + ) 15 = 0 ( x + ) = 15 ( x + ) = 15 4 x + 15 = ± 4 x + 15 = ± x = 15 ± x = ± 15 Add 15 Divide by Extract Square Roots Property of Radicals Subtract Add fractions Let s return to the equation x + 6x = 0 from the beginning of the section. We leave it to the reader to expand the left side and show that ( x + ) 15 = x + 6x. In other words, we can solve x + 6x = 0 by transforming into an equivalent equation. This process, you may recall, is called Completing the Square. We ll revisit Completing the Square in Section?? in more generality and for a different purpose but for now we revisit the steps needed to complete the square to solve a quadratic equation. Solving Quadratic Equations: Completing the Square To solve a quadratic equation ax + bx + c = 0 by Completing the Square: 1. Subtract the constant c from both sides.. Divide both sides by a, the coefficient of x. (Remember: a 0.). Add ( b a) to both sides of the equation. (That s half the coefficient of x, squared.) 4. Factor the left hand side of the equation as ( x + b a). 5. Extract Square Roots. 6. Subtract b from both sides. a

391 A.10 Quadratic Equations 8 To refresh our memories, we apply this method to solve x 4x + 5 = 0: x 4x + 5 = 0 x 4x = 5 Subtract c = 55 x 8x = 5 Divide by a = x 8x + 16 = Add ( b a) = ( 4) = 16 (x 4) = 4 4 x 4 = ± 4 x = 4 ± Factor: Perfect Square Trinomial Extract Square Roots Add 4 At this point, we use properties of fractions and radicals to rationalize the denominator: = = = 9 We can now get a common (integer) denominator which yields: 4 19 x = 4 ± = 4 ± = 1 ± 19 The key to Completing the Square is that the procedure always produces a perfect square trinomial. To see why this works every single time, we start with ax + bx + c = 0 and follow the procedure: ax + bx + c = 0 ax + bx = c Subtract c x + bx a = c a Divide by a 0 x + bx a + ( b a ) = c a + ( ) ( b b Add a a (Hold onto the line above for a moment.) Here s the heart of the method - we need to show that x + bx a + ( b a) = ( x + b a ) ) To show this, we start with the right side of the equation and apply the Perfect Square Formula from Theorem A.9 ( x + a) b ( ) ( ) b b = x + x + = x + bx ( ) b a a a + a Recall that this means we want to get a denominator with rational (more specifically, integer) numbers.

392 84 Algebra Review With just a few more steps we can solve the general equation ax + bx + c = 0 so let s pick up the story where we left off. (The line on the previous page we told you to hold on to.) x + bx ( ) b a + = c ( ) b a a + a ( x + b ) = c a a + b 4a ( x + b ) = 4ac a 4a + b 4a ( x + b ) = b 4ac a 4a x + b b 4ac = ± a 4a x + b b 4ac = ± a a x = b a ± b 4ac a x = b ± b 4ac a Factor: Perfect Square Trinomial Get a common denominator Add fractions Extract Square Roots Properties of Radicals b Subtract a Add fractions. Lo and behold, we have derived the legendary Quadratic Formula! Theorem A.11. Quadratic Formula: The solution(s) to ax + bx + c = 0 with a 0 is/are: x = b ± b 4ac a We can check our earlier solutions to x + 6x = 0 and x 4x + 5 = 0 using the Quadratic Formula. For x + 6x = 0, we identify a =, b = 6 and c =. The quadratic formula gives: x = 6 ± 6 4()( ) () 6 ± = 6 ± 60 4 Using properties of radicals ( 60 = 15), this reduces to ( ± 15) 4 = ± 15. We leave it to the reader to show these two answers are the same as ± 15, as required. For x 4x + 5 = 0, we identify a =, b = 4 and c = 5. Here, we get: x = ( 4) ± ( 4) 4()(5) () = 4 ± Since 516 = 19, this reduces to x = 1± 19. Think about what ( ± 15) is really telling you.

393 A.10 Quadratic Equations 85 It is worth noting that the Quadratic Formula applies to all quadratic equations - even ones we could solve using other techniques. For example, to solve x + 5x = 0 we identify a =, b = 5 and c =. Plugging those into the Quadratic Formula yields: At this point, we have x = = 1 x = 5 ± 5 4()( ) () = 5 ± 49 4 = 5 ± 7 4 and x = = 1 4 = - the same two answers we obtained factoring. We can also use it to solve x =, if we wanted to. From x = 0, we have a = 1, b = 0 and c =. The Quadratic Formula produces x = 0 ± 0 4(1)() (1) = ± 1 = ± = ± As this last example illustrates, while the Quadratic Formula can be used to solve every quadratic equation, that doesn t mean it should be used. Many times other methods are more efficient. We now provide a more comprehensive approach to solving Quadratic Equations. Strategies for Solving Quadratic Equations If the variable appears in the squared term only, isolate it and Extract Square Roots. Otherwise, put the nonzero terms on one side of the equation so that the other side is 0. Try factoring. If the expression doesn t factor easily, use the Quadratic Formula. The reader is encouraged to pause for a moment to think about why Completing the Square doesn t appear in our list of strategies despite the fact that we ve spent the majority of the section so far talking about it. 4 Let s get some practice solving quadratic equations, shall we? Find all real number solutions to the following equations. 1. (w 1) = 0. 5x x(x ) = 7. (y 1) = y (5 1x) = x t + 10t + = 0 6. x = x 4 6 Solution. 1. Since (w 1) = 0 contains a perfect square, we isolate it first then extract square roots: (w 1) = 0 = (w 1) Add (w 1) ± = w 1 Extract Square Roots 1 ± = w Add 1 1 ± = w Divide by 4 Unacceptable answers include Jeff and Carl are mean and It was one of Carl s Pedantic Rants.

394 86 Algebra Review We find our two answers w = 1±. The reader is encouraged to check both answers by substituting each into the original equation. 5. To solve 5x x(x ) = 7, we perform the indicated operations and set one side equal to 0. 5x x(x ) = 7 5x x + x = 7 Distribute x + 8x = 7 Gather like terms x + 8x 7 = 0 Subtract 7 At this point, we attempt to factor and find x + 8x 7 = (x 1)( x + 7). Using the Zero Product Property, we get x 1 = 0 or x + 7 = 0. Our answers are x = 1 or x = 7, which are easily verified.. Even though we have a perfect square in (y 1) = y+, Extracting Square Roots won t help matters since we have a y on the other side of the equation. Our strategy here is to perform the indicated operations (and clear the fraction for good measure) and get 0 on one side of the equation. (y 1) = y + y y + 1 = y + Perfect Square Trinomial ( (y y + 1) = y + ) Multiply by ( ) y + y 6y + = 6 Distribute y 6y + = 6 (y + ) y 6y (y + ) = 0 Subtract 6, Add (y + ) y 5y 1 = 0 A cursory attempt at factoring bears no fruit, so we run this through the Quadratic Formula with a =, b = 5 and c = 1. y = ( 5) ± ( 5) 4()( 1) () y = 5 ± y = 5 ± 7 6 Since 7 is prime, we have no way to reduce 7. Thus, our final answers are y = 5± 7 6. The reader is encouraged to supply the details of the challenging verification of the answers. 5 It s excellent practice working with radicals and fractions so we really, really want you to take the time to do it.

395 A.10 Quadratic Equations We proceed as before; our goal is to gather the nonzero terms on one side of the equation. 5(5 1x) = x x = x Distribute ( ) 59 4(15 105x) = 4 4 5x Multiply by x = x Distribute x x = 0 Subtract 59, Add 100x 100x 40x = 0 Gather like terms With highly composite numbers like 100 and 441, factoring seems inefficient at best, 6 so we apply the Quadratic Formula with a = 100, b = 40 and c = 441: x = ( 40) ± ( 40) 4(100)(441) (100) = 40 ± = 40 ± 0 00 = 40 ± 0 00 = = 1 10 To our surprise and delight we obtain just one answer, x = Our next equation 4.9t + 10t + = 0, already has 0 on one side of the equation, but with coefficients like 4.9 and 10, factoring with integers is not an option. We could make things a bit easier by clearing the decimal (by multiplying through by 10) to get 49t + 100t + 0 = 0 but we simply cannot rid ourselves of the irrational number. The Quadratic Formula is our only recourse. With a = 49, b = 100 and c = 0 we get: 6 This is actually the Perfect Square Trinomial (10x 1).

396 88 Algebra Review t = 100 ± (100 ) 4( 49)(0) ( 49) = 100 ± = 100 ± = 100 ± = ( 50 ± 4 50) ( 49) = 50 ± = ( 50 ± 4 50) 49 = Reduce Properties of Negatives Distribute You ll note that when we distributed the negative in the last step, we changed the ± to a. While this is technically correct, at the end of the day both symbols mean plus or minus, 7 so we can write our answers as t = 50 ± Checking these answers are a true test of arithmetic mettle. 6. At first glance, the equation x = x 4 6 seems misplaced. The highest power of the variable x here is 4, not, so this equation isn t a quadratic equation - at least not in terms of the variable x. It is, however, an example of an equation that is Quadratic in Disguise. 8 We introduce a new variable u to help us see the pattern - specifically we let u = x. Thus u = (x ) = x 4. So in terms of the variable u, the equation x = x 4 6 is u = u 6. The latter is a quadratic equation, which we can solve using the usual techniques: u = u 6 0 = u u 6 Subtract u After a few attempts at factoring, we resort to the Quadratic Formula with a =, b = and c = 6 7 There are instances where we need both symbols, however. For example, the Sum and Difference of Cubes Formulas (page 71) can be written as a single formula: a ± b = (a ± b)(a ab + b ). In this case, all of the top symbols are read to give the sum formula; the bottom symbols give the difference formula. 8 More formally, quadratic in form. Carl likes Quadratics in Disguise since it reminds him of the tagline of one of his beloved childhood cartoons and toy lines.

397 A.10 Quadratic Equations 89 to get the following: u = ( ) ± ( ) 4()( 6) () = ± = ± 76 6 = ± = ± 19 6 = (1 ± 19) () = 1 ± 19 Properties of Radicals Factor Reduce We ve solved the equation for u, but what we still need to solve the original equation 9 - which means we need to find the corresponding values of x. Since u = x, we have two equations: x = or x = We can solve the first equation by extracting square roots to get x = ±. The second equa- tion, however, has no real number solutions because 1 19 is a negative number. For our final answers we can rationalize the denominator 10 to get: x = ± = ± = ± + 19 As with the previous exercise, the very challenging check is left to the reader. Our last example above, the Quadratic in Disguise, hints that the Quadratic Formula is applicable to a wider class of equations than those which are strictly quadratic. We give some general guidelines to recognizing these beasts in the wild on the next page. 9 Or, you ve solved the equation for you (u), now you have to solve it for your instructor (x). 10 We ll say more about this technique in Section A.1.

398 90 Algebra Review Identifying Quadratics in Disguise An equation is a Quadratic in Disguise if it can be written in the form: ax m + bx m + c = 0. In other words: There are exactly three terms, two with variables and one constant term. The exponent on the variable in one term is exactly twice the variable on the other term. To transform a Quadratic in Disguise to a quadratic equation, let u = x m so u = (x m ) = x m. This transforms the equation into au + bu + c = 0. For example, x 6 x + 1 = 0 is a Quadratic in Disguise, since 6 =. If we let u = x, we get u = (x ) = x 6, so the equation becomes u u + 1 = 0. However, x 6 x + 1 = 0 is not a Quadratic in Disguise, since 6. The substitution u = x yields u = (x ) = x 4, not x 6 as required. We ll see more instances of Quadratics in Disguise in later sections. We close this section with a review of the discriminant of a quadratic equation as defined below. The Discriminant: Given a quadratic equation ax + bx + c = 0, the quantity b 4ac is called the discriminant of the equation. The discriminant is the radicand of the square root in the quadratic formula: x = b ± b 4ac a It discriminates between the nature and number of solutions we get from a quadratic equation. The results are summarized below. Theorem A.1. Discriminant Theorem: Given a Quadratic Equation ax + bx + c = 0, let D = b 4ac be the discriminant. If D > 0, there are two distinct real number solutions to the equation. If D = 0, there is one repeated real number solution. Note: Repeated here comes from the fact that both solutions b±0 a reduce to b a. If D < 0, there are no real solutions. For example, the equation x + x 1 = 0 has two real number solutions since the discriminant works out to be (1) 4(1)( 1) = 5 > 0. This results in a ± 5 in the Quadratic Formula which then generates two different answers. On the other hand, x + x + 1 = 0 has no real solutions since here, the discriminant is (1) 4(1)(1) = < 0 which generates a ± in the Quadratic Formula. The equation x + x + 1 = 0 has discriminant () 4(1)(1) = 0 so in the Quadratic Formula we get a ± 0 = 0 thereby generating just one solution. More can be said as well. For example, the discriminant of 6x x 40 = 0 is 961. This is a perfect square, 961 = 1, which means our solutions are rational numbers. When our solutions are

399 A.10 Quadratic Equations 91 rational numbers, the quadratic actually factors nicely. In our example 6x x 40 = (x + 5)(x 8). Admittedly, if you ve already computed the discriminant, you re most of the way done with the problem and probably wouldn t take the time to experiment with factoring the quadratic at this point but we ll see another use for this analysis of the discriminant in Example A.1.

400 9 Algebra Review A.10.1 Exercises In Exercises 1-1, find all real solutions. Check your answers, as directed by your instructor. ( 1. x 1 ) = (5t + ) =. (y ) = x + x 1 = 0 5. w = w 6. y(y + 4) = 1 7. z = 4z v + 0.v = x = x t = (t + 1) 11. (x ) = x (y 1)(y + 1) = 5y 1. w 4 + w 1 = x 4 + x = 15. ( y) 4 = ( y) x 4 + 6x = 15x 17. 6p + = p + p v = 7v v y 8y = 18y 1 0. x = x = v + 1 v In Exercises - 7, find all real solutions and use a calculator to approximate your answers, rounded to two decimal places b = 6. πr = = 8r + πr t + 100t = x = 1.65( x) 7. (0.5 + A) = 0.7(0.1 A) In Exercises 8-0, use Theorem A.5 along with the techniques in this section to find all real solutions to the following. 8. x x = 9. x x = x 1 0. x x + = 4 x 1. Prove that for every nonzero number p, x + xp + p = 0 has no real solutions.. Solve for t: 1 gt + vt + h = 0. Assume g > 0, v 0 and h 0.

401 A.10 Quadratic Equations 9 A.10. Answers 1. x = ± x = 1 ± 5. t = 4 5, 5 5. w = 1,. y = ±1, ± 5 6. y = ± 5 7. z = 1 ± t = 5 ± w = ± 16. x = 0, 5 ± v =, 1 9. No real solution. 11. x = 0 1. y = ± x = ±1 15. y = 4 ± p = 1, ± 18. v = 0, ±, ± y = 5 ± 46 ± x = 1. v = b = ± ±.0. r = ± 50 π ±.4 4. r = 4 ± 54π + π 6. x = 99 ± x = 1,, ± 17,, r 6.,.7 5. t = 500 ± , t 5.68, , x 1.69, A = 107 ± 7 70, A 0.50, x = ±1, ± 0. x = 1, 1, 7 1. The discriminant is: D = p 4p = p < 0. Since D < 0, there are no real solutions.. t = v ± v + gh g

402 94 Algebra Review A.11 Complex Numbers The results of Section A.10 tell us that the equation x + 1 = 0 has no real number solutions. However, it would have solutions if we could make sense of 1. The Complex Numbers do just that - they give us a mechanism for working with 1. As such, the set of complex numbers fill in an algebraic gap left by the set of real numbers. Here s the basic plan. There is no real number x with x = 1, since for any real number x 0. However, we could formally extract square roots and write x = ± 1. We build the complex numbers by relabeling the quantity 1 as i, the unfortunately misnamed imaginary unit. 1 The number i, while not a real number, is defined so that it plays along well with real numbers and acts very much like any other radical expression. For instance, (i) = 6i, 7i i = 4i, ( 7i)+(+4i) = 5 i, and so forth. The key properties which distinguish i from the real numbers are listed below. The imaginary unit i satisfies the two following properties: 1. i = 1. If c is a real number with c 0 then c = i c Property 1 in Definition A.11 establishes that i does act as a square root of 1, and property establishes what we mean by the principal square root of a negative real number. In property, it is important to remember the restriction on c. For example, it is perfectly acceptable to say 4 = i 4 = i() = i. However, ( 4) i 4, otherwise, we d get = 4 = ( 4) = i 4 = i(i) = i = ( 1) =, which is unacceptable. The moral of this story is that the general properties of radicals do not apply for even roots of negative quantities. With Definition A.11 in place, we can define the set of complex numbers. A complex number is a number of the form a+bi, where a and b are real numbers and i is the imaginary unit. The set of complex numbers is denoted C. Complex numbers include things you d normally expect, like + i and 5 i. However, don t forget that a or b could be zero, which means numbers like i and 6 are also complex numbers. In other words, don t forget that the complex numbers include the real numbers, so 0 and π 1 are both considered complex numbers. The arithmetic of complex numbers is as you would expect. The only things you need to remember are the two properties in Definition A.11. The next example should help recall how these animals behave. 1 Some Technical Mathematics textbooks label it j. While it carries the adjective imaginary, these numbers have essential real-world implications. For example, every electronic device owes its existence to the study of imaginary numbers. Note the use of the indefinite article a. Whatever beast is chosen to be i, i is the other square root of 1. To use the language of Section A.1., R C.

403 A.11 Complex Numbers 95 Perform the indicated operations. 1. (1 i) ( + 4i). (1 i)( + 4i). 1 i 4i ( )( 1) 6. (x [1 + i])(x [1 i]) Solution. 1. As mentioned earlier, we treat expressions involving i as we would any other radical. We distribute and combine like terms: (1 i) ( + 4i) = 1 i 4i Distribute = 6i Gather like terms Technically, we d have to rewrite our answer 6i as ( ) + ( 6)i to be (in the strictest sense) in the form a + bi. That being said, even pedants have their limits, so 6i is good enough.. Using the Distributive Property (a.k.a. F.O.I.L.), we get (1 i)( + 4i) = (1)() + (1)(4i) (i)() (i)(4i) F.O.I.L. = + 4i 6i 8i = i 8( 1) i = 1 = i + 8 = 11 i. How in the world are we supposed to simplify 1 i? Well, we deal with the denominator 4i as 4i we would any other denominator containing two terms, one of which is a square root. 4 We multiply both numerator and denominator by + 4i, the (complex) conjugate of 4i. Doing so produces 1 i 4i = (1 i)( + 4i) ( 4i)( + 4i) Equivalent Fractions = + 4i 6i 8i 9 16i F.O.I.L. = i 8( 1) 9 16( 1) i = 1 = 11 i 5 = i 4. We use property of Definition A.11 first, then apply the rules of radicals applicable to real numbers to get 1 = ( i ) ( i 1 ) = i 1 = 6 = 6. 4 See subsection A.1.1 for a more thorough treatment of this type of maneuver.

404 96 Algebra Review 5. We adhere to the order of operations here and perform the multiplication before the radical to get ( )( 1) = 6 = We brute force multiply using the distributive property and find that (x [1 + i])(x [1 i]) = x x[1 i] x[1 + i] + [1 i][1 + i] F.O.I.L. This type of factoring will be revisited in Section??. = x x + ix x ix + 1 i + i 4i Distribute = x x + 1 4( 1) Gather like terms = x x + 5 i = 1 In the previous example, we used the conjugate idea from Section A.1 to divide two complex numbers. More generally, the complex conjugate of a complex number a + bi is the number a bi. The notation commonly used for complex conjugation is a bar : a + bi = a bi. For example, + i = i and i = + i. To find 6, we note that 6 = 6 + 0i = 6 0i = 6, so 6 = 6. Similarly, 4i = 4i, since 4i = 0 + 4i = 0 4i = 4i. Note that + 5 = + 5, not 5, since + 5 = i = + 5 0i = + 5. Here, the conjugation specified by the bar notation involves reversing the sign before i = 1, not before 5. The properties of the conjugate are summarized in the following theorem. Theorem A.1. Properties of the Complex Conjugate: Let z and w be complex numbers. z = z z + w = z + w zw = z w z n = (z) n, for any natural number n z is a real number if and only if z = z. Theorem A.1 says in part that complex conjugation works well with addition, multiplication and powers. The proofs of these properties can best be achieved by writing out z = a+bi and w = c +di for real numbers a, b, c and d. Next, we compute the left and right sides of each equation and verify that they are the same. The proof of the first property is a very quick exercise. 5 To prove the second property, we compare z + w with z + w. We have z + w = a + bi + c + di = a bi + c di. To find z + w, we first compute z + w = (a + bi) + (c + di) = (a + c) + (b + d)i so z + w = (a + c) + (b + d)i = (a + c) (b + d)i = a + c bi di = a bi + c di = z + w As such, we have established z + w = z + w. The proof for multiplication works similarly. The proof that the conjugate works well with powers can be viewed as a repeated application of the product rule, and is best 5 Trust us on this.

405 A.11 Complex Numbers 97 proved using a technique called Mathematical Induction. 6 The last property is a characterization of real numbers. If z is real, then z = a + 0i, so z = a 0i = a = z. On the other hand, if z = z, then a + bi = a bi which means b = b so b = 0. Hence, z = a + 0i = a and is real. We now return to the business of solving quadratic equations. Consider x x + 5 = 0. The discriminant b 4ac = 16 is negative, so we know by Theorem A.1 there are no real solutions, since the Quadratic Formula would involve the term 16. Complex numbers, however, are built just for such situations, so we can go ahead and apply the Quadratic Formula to get: x = ( ) ± ( ) 4(1)(5) (1) = ± 16 = ± 4i = 1 ± i. Find the complex solutions to the following equations Solution. x x + 1 = x +. t 4 = 9t + 5. z + 1 = 0 1. Clearing fractions yields a quadratic equation so we then proceed as in Section A.10. x x + 1 = x + x = (x + )(x + 1) Multiply by (x + 1) to clear denominators x = x + x + x + F.O.I.L. x = x + 4x + Gather like terms 0 = x + x + Subtract x From here, we apply the Quadratic Formula x = ± 4(1)() (1) = ± 8 = ± i 8 = ± i ( 1 = ± i ) = 1 ± i Quadratic Formula Simplify Definition of i Product Rule for Radicals Factor and reduce We get two answers: x = 1 + i and its conjugate x = 1 i. Checking both of these answers reviews all of the salient points about complex number arithmetic and is therefore strongly encouraged. 6 See Section??. 7 Remember, all real numbers are complex numbers, so complex solutions means both real and non-real answers.

406 98 Algebra Review. Since we have three terms, and the exponent on one term ( 4 on t 4 ) is exactly twice the exponent on the other ( on t ), we have a Quadratic in Disguise. We proceed accordingly. t 4 = 9t + 5 t 4 9t 5 = 0 Subtract 9t and 5 (t + 1)(t 5) = 0 Factor t + 1 = 0 or t = 5 Zero Product Property From t + 1 = 0 we get t = 1, or t = 1. We extract square roots as follows: t = ± = ±i = ±i 1 = ±i = ± i, where we have rationalized the denominator per convention. From t = 5, we get t = ± 5. In total, we have four complex solutions - two real: t = ± 5 and two non-real: t = ± i.. To find the real solutions to z + 1 = 0, we can subtract the 1 from both sides and extract cube roots: z = 1, so z = 1 = 1. It turns out there are two more non-real complex number solutions to this equation. To get at these, we factor: z + 1 = 0 (z + 1)(z z + 1) = 0 Factor (Sum of Two Cubes) z + 1 = 0 or z z + 1 = 0 From z + 1 = 0, we get our real solution z = 1. From z z + 1 = 0, we apply the Quadratic Formula to get: z = ( 1) ± ( 1) 4(1)(1) (1) = 1 ± = 1 ± i Thus we get three solutions to z + 1 = 0 - one real: z = 1 and two non-real: z = 1±i. As always, the reader is encouraged to test their algebraic mettle and check these solutions. It is no coincidence that the non-real solutions to the equations in Example A.11 appear in complex conjugate pairs. Any time we use the Quadratic Formula to solve an equation with real coefficients, the answers will form a complex conjugate pair owing to the ± in the Quadratic Formula. This leads us to a generalization of Theorem A.1 which we state below. Theorem A.14. Discriminant Theorem: Given a Quadratic Equation ax + bx + c = 0, where a, b and c are real numbers, let D = b 4ac be the discriminant. If D > 0, there are two distinct real number solutions to the equation. If D = 0, there is one (repeated) real number solution. Note: Repeated here comes from the fact that both solutions b±0 a reduce to b a. If D < 0, there are two non-real solutions which form a complex conjugate pair. We will have much more to say about complex solutions to equations in Section?? and we will revisit Theorem A.14 then.

407 A.11 Complex Numbers 99 A.11.1 Exercises In Exercises 1-10, use the given complex numbers z and w to find and simplify the following. z + w zw z 1 z z w w z z zz (z) 1. z = + i, w = 4i. z = 1 + i, w = i. z = i, w = 1 + i 4. z = 4i, w = i 5. z = 5i, w = + 7i 6. z = 5 + i, w = 4 + i 7. z = i, w = + i 8. z = 1 i, w = 1 i 9. z = 1 + i, w = 1 + i 10. z = + i, w = i In Exercises 11-18, simplify the quantity ( 5)( 4) ( 9)( 16) 17. ( 9) 18. ( 9) We know that i = 1 which means i = i i = ( 1) i = i and i 4 = i i = ( 1)( 1) = 1. In Exercises 19-6, use this information to simplify the given power of i. 19. i 5 0. i 6 1. i 7. i 8. i i 6 5. i i 04 In Exercises 7-5, find all complex solutions. 7. x + 6 = 4x 8. 15t + t + 5 = t(t + 1) 9. y + 4 = y w = w 1. y y = y. x x 1 = x. x = 5 x Multiply and simplify: ( x [ i ] ) ( x [ + i ] ) 5y y 1 = y 5. z 4 = 16

408 400 Algebra Review A.11. Answers 1. For z = + i and w = 4i z + w = + 7i zw = 1 + 8i z = 5 + 1i 1 z = 1 1 i z w = 4 1 i w z = i z = i zz = 1 (z) = 5 1i. For z = 1 + i and w = i z + w = 1 zw = 1 i z = i 1 z = 1 1 i z w = 1 + i w z = 1 1 i z = 1 i zz = (z) = i. For z = i and w = 1 + i z + w = 1 + i zw = i z = 1 1 z = i z w = i w z = + i z = i zz = 1 (z) = 1 4. For z = 4i and w = i z + w = + i zw = 8 + 8i z = 16 1 z = 1 4 i z w = 1 + i w z = 1 1 i z = 4i zz = 16 (z) = For z = 5i and w = + 7i z + w = 5 + i zw = i z = 16 0i 1 z = i z w = i w z = i z = + 5i zz = 4 (z) = i

409 A.11 Complex Numbers For z = 5 + i and w = 4 + i z + w = 1 + i zw = 6i z = 4 10i 1 z = i z w = i w z = i z = 5 i zz = 6 (z) = i 7. For z = i and w = + i z + w = zw = 4 z = 4i 1 z = i z w = i w z = i z = + i zz = 4 (z) = 4i 8. For z = 1 i and w = 1 i z + w = i zw = 4 z = i 1 z = i z w = 1 + i w z = 1 i z = 1 + i zz = 4 (z) = + i 9. For z = 1 + i and w = 1 + i z + w = i zw = 1 z = 1 + i 1 z = 1 i z w = 1 i w z = 1 + i z = 1 i zz = 1 (z) = 1 i 10. For z = + i and w = i z + w = zw = 1 z = i 1 z = i z w = i w z = i z = i zz = 1 (z) = i 11. 7i 1. i

410 40 Algebra Review i 19. i 5 = i 4 i = 1 i = i 0. i 6 = i 4 i = 1 ( 1) = 1 1. i 7 = i 4 i = 1 ( i) = i. i 8 = i 4 i 4 = ( i 4) = (1) = 1. i 15 = ( i 4) i = 1 ( i) = i 4. i 6 = ( i 4)6 i = 1 ( 1) = 1 5. i 117 = ( i 4) 9 i = 1 i = i 6. i 04 = ( i 4) 76 = 1 76 = 1 7. x = ± i w = 1 ± i 7 5 ± i. x = 6. x 6x + 8. t = 5, ± i 1. y = ± i 4. y = ±i, ± i 9. y = ±, ±i. x = 0, 1 ± i 5. z = ±, ±i

411 A.1 Rational Expressions and Equations 40 A.1 Rational Expressions and Equations We now turn our attention to rational expressions - that is, algebraic fractions - and equations which contain them. The reader is encouraged to keep in mind the properties of fractions listed on page 8 because we will need them along the way. Before we launch into reviewing the basic arithmetic operations of rational expressions, we take a moment to review how to simplify them properly. As with numeric fractions, we cancel common factors, not common terms. That is, in order to simplify rational expressions, we first factor the numerator and denominator. For example: x 4 + 5x x 5x x x x 5x but, rather x 4 + 5x x 5x = x (x + 5) x(x 5) Factor G.C.F. = x (x + 5) x(x 5)(x + 5) Difference of Squares = x x (x + 5) x(x 5) (x + 5) Cancel common factors = x x 5 This equivalence holds provided the factors being canceled aren t 0. Since a factor of x and a factor of x + 5 were canceled, x 0 and x + 5 0, so x 5. We usually stipulate this as: x 4 + 5x x 5x = x x 5, provided x 0, x 5 While we re talking about common mistakes, please notice that 5 x x Just like their numeric counterparts, you don t add algebraic fractions by adding denominators of fractions with common numerators - it s the other way around: 1 x = x It s time to review the basic arithmetic operations with rational expressions. 1 One of the most common errors students make on college Mathematics placement tests is that they forget how to add algebraic fractions correctly. This places many students into remedial classes even though they are probably ready for collegelevel Math. We urge you to really study this section with great care so that you don t fall into that trap.

412 404 Algebra Review Perform the indicated operations and simplify. 1.. x 5x x 4 4 x x x 5 + x. y 8y y y y 4. 5 w 9 w + w 9 4 (x + h) 4 x h 5. t (t) 6. 10x(x ) 1 + 5x ( 1)(x ) Solution. 1. As with numeric fractions, we divide rational expressions by inverting and multiplying. Before we get too carried away however, we factor to see what, if any, factors cancel. x 5x x 4 4 x x x 5 + x = x 5x x 4 4 x 5 + x x x = (x 5x )(x 5 + x ) (x 4 4)(x x ) = = = (x + 1)(x )x (x + ) (x )(x + )(x )(x + 1) (x + 1) (x )x (x + ) (x ) (x + ) (x )(x + 1) x (x + 1) (x + 1)(x ) Invert and multiply Multiply fractions Factor Cancel common factors Provided x The x is mentioned since a factor of (x ) was canceled as we reduced the expression. We also canceled a factor of (x + ). Why is there no stipulation as a result of canceling this factor? Because x + 0 for all real x. (See Section A.11 for details.) At this point, we could go ahead and multiply out the numerator and denominator to get x (x + 1) (x + 1)(x ) = x 4 + x x + x x but for most of the applications where this kind of algebra is needed (solving equations, for instance), it is best to leave things factored. Your instructor will let you know whether to leave your answer in factored form or not. Speaking of factoring, do you remember why x can t be factored over the integers?

413 A.1 Rational Expressions and Equations 405. As with numeric fractions we need common denominators in order to subtract. This is already the case here so we proceed by subtracting the numerators. 5 w 9 w + w 9 = 5 (w + ) w 9 = 5 w w 9 = w w 9 Subtract fractions Distribute Combine like terms At this point, we need to see if we can reduce this expression so we proceed to factor. It first appears as if we have no common factors among the numerator and denominator until we recall the property of factoring negatives from Page 81: w = (w ). This yields: w w 9 = (w ) (w )(w + ) (w ) = (w )(w + ) = 1 w + Factor Cancel common factors Provided w The stipulation w comes from the cancellation of the (w ) factor.. In this next example, we are asked to add two rational expressions with different denominators. As with numeric fractions, we must first find a common denominator. To do so, we start by factoring each of the denominators. y 8y y y y = = (y 4) + y + 1 y(16 y ) (y 4) + y + 1 y(4 y)(4 + y) Factor Factor some more To find the common denominator, we examine the factors in the first denominator and note that we need a factor of (y 4). We now look at the second denominator to see what other factors we need. We need a factor of y and (4 + y) = (y + 4). What about (4 y)? As mentioned in the last example, we can factor this as: (4 y) = (y 4). Using properties of negatives, we migrate this negative out to the front of the fraction, turning the addition into subtraction. We find the (least) common denominator to be (y 4) y(y + 4). We can now proceed to multiply the numerator and denominator of each fraction by whatever factors are missing from their respective denominators to

414 406 Algebra Review produce equivalent expressions with common denominators. (y 4) + y + 1 y(4 y)(4 + y) = = = = (y 4) + y + 1 y( (y 4))(y + 4) (y 4) y + 1 y(y 4)(y + 4) y(y + 4) (y 4) y(y + 4) y + 1 y(y 4)(y + 4) y(y + 4) (y 4) y(y + 4) (y + 1)(y 4) y(y 4) (y + 4) (y 4) (y 4) Equivalent Fractions Multiply Fractions At this stage, we can subtract numerators and simplify. We ll keep the denominator factored (in case we can reduce down later), but in the numerator, since there are no common factors, we proceed to perform the indicated multiplication and combine like terms. y(y + 4) (y 4) y(y + 4) (y + 1)(y 4) y(y 4) (y + 4) = y(y + 4) (y + 1)(y 4) (y 4) y(y + 4) = y + 1y (y y 4) (y 4) y(y + 4) = y + 1y y + y + 4 (y 4) y(y + 4) = y + 15y + 4 y(y + 4)(y 4) Subtract numerators Distribute Distribute Gather like terms We would like to factor the numerator and cancel factors it has in common with the denominator. After a few attempts, it appears as if the numerator doesn t factor, at least over the integers. As a check, we compute the discriminant of y + 15y + 4 and get 15 4()(4) = 19. This isn t a perfect square so we know that the quadratic equation y + 15y + 4 = 0 has irrational solutions. This means y + 15y + 4 can t factor over the integers so we are done. 4. In this example, we have a compound fraction, and we proceed to simplify it as we did its numeric counterparts in Example A.. Specifically, we start by multiplying the numerator and denominator of the big fraction by the least common denominator of the little fractions inside of it - in this case we need to use (4 (x + h))(4 x) - to remove the compound nature of the big fraction. Once we have See the remarks following Theorem A.1.

415 A.1 Rational Expressions and Equations 407 a more normal looking fraction, we can proceed as we have in the previous examples. 4 (x + h) ( 4 x 4 (x + h) ) 4 x (4 (x + h))(4 x) = h h (4 (x + h))(4 x) = ( 4 (x + h) 4 x ) (4 (x + h))(4 x) h(4 (x + h))(4 x) Equivalent fractions Multiply = = = (4 (x + h))(4 x) (4 (x + h))(4 x) 4 (x + h) 4 x h(4 (x + h))(4 x) (4 (x + h))(4 x) (4 (x + h)) (4 x) (4 (x + h)) (4 x) h(4 (x + h))(4 x) (4 x) (4 (x + h)) h(4 (x + h))(4 x) Distribute Reduce Now we can clean up and factor the numerator to see if anything cancels. (This why we kept the denominator factored.) (4 x) (4 (x + h)) h(4 (x + h))(4 x) = = = = [(4 x) (4 (x + h))] h(4 (x + h))(4 x) [4 x 4 + (x + h)] h(4 (x + h))(4 x) [4 4 x + x + h] h(4 (x + h))(4 x) h h(4 (x + h))(4 x) Factor out G.C.F. Distribute Rearrange terms Gather like terms = = h h(4 (x + h))(4 x) (4 (x + h))(4 x) Reduce Provided h 0 Your instructor will let you know if you are to expand the denominator or not At first glance, it doesn t seem as if there is anything that can be done with t (t) because the exponents on the variables are different. However, since the exponents are negative, these are actually rational expressions. In the first term, the exponent applies to the t only but in the second 4 We ll keep it factored because in Calculus it needs to be factored.

416 408 Algebra Review term, the exponent applies to both the and the t, as indicated by the parentheses. One way to proceed is as follows: t (t) = t 1 (t) = t 1 9t We see that we are being asked to subtract two rational expressions with different denominators, so we need to find a common denominator. The first fraction contributes a t to the denominator, while the second contributes a factor of 9. Thus our common denominator is 9t, so we are missing a factor of 9 in the first denominator and a factor of t in the second. t 1 9t = = t t t t 18 9t t 9t = 18 t 9t Equivalent Fractions Multiply Subtract We find no common factors among the numerator and denominator so we are done. A second way to approach this problem is by factoring. We can extend the concept of the Polynomial G.C.F. to these types of expressions and we can follow the same guidelines as set forth on page 71 to factor out the G.C.F. of these two terms. The key ideas to remember are that we take out each factor with the smallest exponent and that factoring is the same as dividing. We first note that t (t) = t t and we see that the smallest power on t is. Thus we want to factor out t from both terms. It s clear that this will leave in the first term, but what about the second term? Since factoring is the same as dividing, we would be dividing the second term by t which thanks to the properties of exponents is the same as multiplying by 1 = t. The same holds t for. Even though there are no factors of in the first term, we can factor out by multiplying it by 1 = = 9. We put these ideas together below. t (t) = t t Properties of Exponents = t (() t 1 ) Factor 1 1 = (18 t) t Rewrite = 18 t 9t Multiply While both ways are valid, one may be more of a natural fit than the other depending on the circumstances and temperament of the student. 6. As with the previous example, we show two different yet equivalent ways to approach simplifying 10x(x ) 1 + 5x ( 1)(x ). First up is what we ll call the common denominator approach where we rewrite the negative exponents as fractions and proceed from there.

417 A.1 Rational Expressions and Equations 409 Common Denominator Approach: 10x(x ) 1 + 5x ( 1)(x ) = = = = = = = 10x x + 5x ( 1) (x ) 10x x x x 5x (x ) Equivalent Fractions 10x(x ) (x ) 5x (x ) Multiply 10x(x ) 5x (x ) Subtract 5x((x ) x) (x ) Factor out G.C.F. 5x(x 6 x) (x ) Distribute 5x(x 6) (x ) Combine like terms Both the numerator and the denominator are completely factored with no common factors so we are done. Factoring Approach : In this case, the G.C.F. is 5x(x ). Factoring this out of both terms gives: 10x(x ) 1 + 5x ( 1)(x ) = 5x(x ) ((x ) 1 x) Factor = = 5x (x 6 x) Rewrite, distribute (x ) 5x(x 6) (x ) Multiply As expected, we got the same reduced fraction as before. Next, we review the solving of equations which involve rational expressions. As with equations involving numeric fractions, our first step in solving equations with algebraic fractions is to clear denominators. In doing so, we run the risk of introducing what are known as extraneous solutions - answers which don t satisfy the original equation. As we illustrate the techniques used to solve these basic equations, see if you can find the step which creates the problem for us.

418 410 Algebra Review Solve the following equations x = x. t t + 1 t 1 = 1 t 1. 1 w 1 w + 5 = 0 4. (x + 4) 1 + x( 1)(x + 4) (x) = 0 5. Solve x = y + 1 y for y. 6. Solve 1 f = 1 S S for S 1. Solution. 1. Our first step is to clear the fractions by multiplying both sides of the equation by x. In doing so, we are implicitly assuming x 0; otherwise, we would have no guarantee that the resulting equation is equivalent to our original equation = x ( x ) x = (x)x Provided x 0 x 1(x) + 1 x (x) = x Distribute x + x x = x Multiply x + 1 = x 0 = x x 1 Subtract x, subtract 1 x = ( 1) ± ( 1) 4(1)( 1) (1) x = 1 ± 5 Quadratic Formula Simplify We obtain two answers, x = 1± 5. Neither of these are 0 thus neither contradicts our assumption that x 0. The reader is invited to check both of these solutions. 6 5 See page The check relies on being able to rationalize the denominator - a skill we haven t reviewed yet. (Come back after you ve read Section A.1.1 if you want to!) Additionally, the positive solution to this equation is the famous Golden Ratio.

419 A.1 Rational Expressions and Equations 411. To solve the equation, we clear denominators. Here, we need to assume t 1 0, or t 1. t t = t 1 t 1 ( ) ( ) t t (t 1) = t 1 t 1 (t 1) Provided t 1 (t t + 1)( (t 1)) 1 (t = 1) t( (t 1)) 1((t 1)) Multiply, distribute (t t + 1) = t t t + Distribute t 4t + = t t + Distribute, combine like terms t t t = 0 Subtract t, add t, subtract t(t t 1) = 0 Factor t = 0 or t t 1 = 0 Zero Product Property t = 0 or (t + 1)(t 1) = 0 Factor t = 0 or t + 1 = 0 or t 1 = 0 t = 0, 1 or 1 We assumed that t 1 in order to clear denominators. Sure enough, the candidate t = 1 doesn t check in the original equation since it causes division by 0. In this case, we call t = 1 an extraneous solution. Note that t = 1 does work in every equation after we clear denominators. In general, multiplying by variable expressions can produce these extra solutions, which is why checking our answers is always encouraged. 7 The other two candidates, t = 0 and t = 1, are solutions.. As before, we begin by clearing denominators. Here, we assume 1 w 0 (so w 1 ) and w (so w 5 ). 1 w 1 ( w + 5 = 0 1 w 1 ) (1 w )(w + 5) = 0(1 w )(w + 5) w 1, 5 w + 5 (1 w )(w + 5) (1 w ) 1(1 w ) (w + 5) (w + 5) (w + 5) (1 w ) = 0 = 0 Distribute The result is a linear equation in w so we gather the terms with w on one side of the equation and 7 Contrast this with what happened in Example A.9.1 when we divided by a variable and lost a solution.

420 41 Algebra Review put everything else on the other. We factor out w and divide by its coefficient. (w + 5) (1 w ) = 0 6w w = 0 Distribute 6w + w = 14 Subtract 14 (6 + )w = 14 Factor w = Divide by This solution is different than our excluded values, and 5 14, so we keep w = answer. The reader is invited to check this in the original equation. 6+ as our final 4. To solve our next equation, we have two approaches to choose from: we could rewrite the quantities with negative exponents as fractions and clear denominators, or we can factor. We showcase each technique below. Clearing Denominators Approach: We rewrite the negative exponents as fractions and clear denominators. In this case, we multiply both sides of the equation by (x + 4), which is never 0. (Think about that for a moment.) As a result, we need not exclude any x values from our solution set. (x + 4) 1 + x( 1)(x + 4) (x) = 0 x x( 1)(x) (x + 4) = 0 Rewrite ( x + 4 6x ) (x + 4) (x + 4) = 0(x + 4) Multiply (x +4) (x + 4) (x 6x (x + 4) + 4) (x + 4) = 0 Distribute (x + 4) 6x = 0 x + 1 6x = 0 Distribute x = 1 Combine like terms, subtract 1 x = 4 Divide by x = ± 4 = ± Extract square roots We leave it to the reader to show that both x = and x = satisfy the original equation. Factoring Approach: Since the equation is already set equal to 0, we re ready to factor. Following the guidelines presented in Example A.1, we factor out (x + 4) from both terms and

421 A.1 Rational Expressions and Equations 41 look to see if more factoring can be done: (x + 4) 1 + x( 1)(x + 4) (x) = 0 (x + 4) ((x + 4) 1 + x( 1)(x)) = 0 Factor (x + 4) (x + 4 x ) = 0 (x + 4) (4 x ) = 0 Gather like terms (x + 4) = 0 or 4 x = 0 Zero Product Property x + 4 = 0 or 4 = x The first equation yields no solutions (Think about this for a moment.) while the second gives us x = ± 4 = ± as before. 5. We are asked to solve this equation for y so we begin by clearing fractions with the stipulation that y 0 or y. We are left with a linear equation in the variable y. To solve this, we gather the terms containing y on one side of the equation and everything else on the other. Next, we factor out the y and divide by its coefficient, which in this case turns out to be x. In order to divide by x, we stipulate x 0 or, said differently, x. x = y + 1 y ( ) y + 1 x(y ) = (y ) Provided y y xy x = (y + 1) (y ) (y Distribute, multiply ) xy x = y + 1 xy y = x + 1 Add x, subtract y y(x ) = x + 1 Factor y = x + 1 Divide provided x x We highly encourage the reader to check the answer algebraically to see where the restrictions on x and y come into play Our last example comes from physics and the world of photography. 9 We take a moment here to note that while superscripts in Mathematics indicate exponents (powers), subscripts are used primarily to distinguish one or more variables. In this case, S 1 and S are two different variables (much like x and y) and we treat them as such. Our first step is to clear denominators by multiplying both sides by fs 1 S - provided each is nonzero. We end up with an equation which is linear in S 1 so we proceed 8 It involves simplifying a compound fraction! 9 See this article on focal length.

422 414 Algebra Review as in the previous example. 1 = f ( ) 1 (fs 1 S ) = f S 1 S ( ) (fs 1 S ) Provided f 0, S 1 0, S 0 S 1 S fs 1 S f f S 1 S f = fs 1S S 1 + fs 1S S = f S 1 S + fs 1 S S 1 S Multiply, distribute Cancel S 1 S = fs + fs 1 S 1 S fs 1 = fs Subtract fs 1 S 1 (S f ) = fs Factor S 1 = fs S f Divide provided S f As always, the reader is highly encouraged to check the answer and see what the restriction S f means in terms of focusing a camera!

423 A.1 Rational Expressions and Equations 415 A.1.1 Exercises In Exercises 1-18, perform the indicated operations and simplify. 1. x 9 x x x x 6. t t t + 1 (t t 8). 4y y y + 1 y 16 y 5y 4. x x 1 1 x x 1 7. b x x 1 5. w 1 w + 1 w 1 1 b 8. x x 4 1 x h h y y 1 y y m m m 1 x + h 1 x h + y 1 y 1. w 1 (w) y 1 + ( y) 15. (x ) 1 x(x ) 16. t 1 + t t 17. ( + h) () h 18. (7 x h) 1 (7 x) 1 h In Exercises 19-7, find all real solutions. Be sure to check for extraneous solutions. 19. x 5x + 4 = 0. y 1 y + 1 = w w = w w 9. x + 17 x + 1 = x + 5. t t + 1 t + t t = 1 5. w + = w w w 6. x 1 1 = x y + 4y y 9 7. = 4y x (1 + x ) = In Exercises 8-0, use Theorem A.5 along with the techniques in this section to find all real solutions. 8. n n 1 = 9. x x 1 = 0. t 4 t = t In Exercises 1 -, find all real solutions and use a calculator to approximate your answers, rounded to two decimal places = πR. x = (.1 x) c = 1 4

424 416 Algebra Review In Exercises 4-9, solve the given equation for the indicated variable. 4. Solve for y: 1 y y + = x 5. Solve for y: x = 1 y Solve for T : V 1 T 1 = V T 7. Solve for t 0 : 8. Solve for x: = 5 9. Solve for R: P = x v r x + v r t 0 1 t 0 t 1 = 5R (R + 4) 11 Recall: subscripts on variables have no intrinsic mathematical meaning; they re just used to distinguish one variable from another. In other words, treat quantities like V 1 and V as two different variables as you would x and y.

425 A.1 Rational Expressions and Equations 417 A.1. Answers (x + ) x(x + ), x. t ) (t + 4)(t, t. y(y + 1) y + 4, y 1,, 4 x 1 x 1 b 5b + 7 b 5. w 1, w x + x + 4 (x 4)(x + 1) 9. y, y 0 m + 1 m +, m 10. x, x h, h x(x + h), h w 14. (y 7y + 9) y(y ) (x ) 16. t (h + 6) + t, t (h + ), h (7 x)(7 x h), h x = y = 1, 1. w = 1. x = 6,. No solution. 4. y = 0, ± 5. w =, 1 6. x =, 7. x =, 4 8. n = 1 9. x = 1 ± 5, 1 ± 5 0. t = R = ± ±0.05. x = , x = 9.64π c = ± = ± (You actually didn t need a calculator for this!) 4. y = 1 x x 1, y, x 5. y = x + x, y 1, x 6. T = V T 1 V 1, T 1 0, T 0, V t 0 = 8. x = 1 ± 5v r + 1, x ±v r. 5 t 1 + 1, t R = (8P 5) ± (8P 5) 64P P = (5 8P) ± P, P 0, R 4 P

426 418 Algebra Review A.1 Radical Equations In this section we review simplifying expressions and solving equations involving radicals. In addition to the product, quotient and power rules stated in Theorem A.1 in Section A., we present the following result which states that n th roots and n th powers more or less undo each other. 1 Theorem A.15. Simplifying n th powers of n th roots and n th roots of n th powers: Suppose n is a natural number, a is a real number and n a is a real number. Then ( n a) n = a if n is odd, n a n = a; if n is even, n a n = a. Since n a is defined so that ( n a) n = a, the first claim in the theorem is just a re-wording of Definition A.. The second part of the theorem breaks down along odd/even exponent lines due to how exponents affect negatives. To see this, consider the specific cases of ( ) and 4 ( ) 4. In the first case, ( ) = 8 =, so we have an instance of when n a n = a. The reason that the cube root undoes the third power in ( ) = is because the negative is preserved when raised to the third (odd) power. In 4 ( ) 4, the negative goes away when raised to the fourth (even) power: 4 ( ) 4 = According to Definition A., the fourth root is defined to give only non-negative numbers, so 4 16 =. Here we have a case where 4 ( ) 4 = =, not. In general, we need the absolute values to simplify n a n only when n is even because a negative to an even power is always positive. In particular, x = x, not just x (unless we know x 0.) We practice these formulas in the following example. Perform the indicated operations and simplify. 1. x + 1. t 10t x x 4 + ( 1 ( x 4) ) (x) 6. 48x πr 4 4 L 8 ( 18y 8y) + ( 0 80) Solution. 1. We told you back on page 9 that roots do not distribute across addition and since x + 1 cannot be factored over the real numbers, x + 1 cannot be simplified. It may seem silly to start with this example but it is extremely important that you understand what maneuvers are legal and which ones are not. 1 See Sections?? and?? for a more precise understanding of what we mean here. This discussion should sound familiar - see the discussion following Definition A. and the discussion following Extracting the Square Root on page 8. You really do need to understand this otherwise horrible evil will plague your future studies in Math. If you say something totally wrong like x + 1 = x + 1 then you may never pass Calculus. PLEASE be careful!

427 A.1 Radical Equations 419. Again we note that t 10t + 5 t 10t + 5, since radicals do not distribute across addition and subtraction. 4 In this case, however, we can factor the radicand and simplify as t 10t + 5 = (t 5) = t 5 Without knowing more about the value of t, we have no idea if t 5 is positive or negative so t 5 is our final answer. 5. To simplify 48x 14, we need to look for perfect cubes in the radicand. For the cofficient, we have 48 = 8 6 = 6. To find the largest perfect cube factor in x 14, we divide 14 (the exponent on x) by (since we are looking for a perfect cube). We get 4 with a remainder of. This means 14 = 4 +, so x 14 = x 4 + = x 4 x = (x 4 ) x. Putting this altogether gives: 48x 14 = = 6 (x 4 ) x (x 4 ) 6x = x 4 6x Factor out perfect cubes Rearrange factors, Product Rule of Radicals 4. In this example, we are looking for perfect fourth powers in the radicand. In the numerator r 4 is clearly a perfect fourth power. For the denominator, we take the power on the L, namely 1, and divide by 4 to get. This means L 8 = L 4 = (L ) 4. We get πr 4 4 L 1 = = = 4 πr 4 4 L π r 4 4 (L ) 4 4 π r L Quotient Rule of Radicals Product Rule of Radicals Simplify Without more information about r, we cannot simplify r any further. However, we can simplify L. Regardless of the choice of L, L 0. Actually, L > 0 because L is in the denominator which means L 0. Hence, L = L. Our answer simplifies to: 4 π r L = r 4 π L 5. After a quick cancellation (two of the s in the second term) we need to obtain a common denominator. Since we can view the first term as having a denominator of 1, the common denominator is precisely the denominator of the second term, namely ( x 4). With common denominators, 4 Let t = 1 and see what happens to t 10t + 5 versus t 10t In general, t 5 = t 5 and t 5 t + 5 so watch what you re doing!

428 40 Algebra Review we proceed to add the two fractions. Our last step is to factor the numerator to see if there are any cancellation opportunities with the denominator. x ( ) x ( (x) = x ( ) x x 4) ( (x) x 4) Reduce = x x x 4 + ( x 4) Mutiply = (x x 4) ( x 4) ( x 4) + x ( x 4) Equivalent fractions = x( x 4) x ( + x 4) ( Multiply x 4) = x(x 4) ( x 4) + x ( Simplify x 4) = x(x 4) + x ( x 4) Add = x(x 4 + 1) ( x 4) Factor = x(x ) ( x 4) We cannot reduce this any further because x is irreducible over the rational numbers. 6. We begin by working inside each set of parentheses, using the product rule for radicals and combining like terms. ( 18y 8y) + ( 0 80) = = = = = ( 9 y 4 y) + ( ) ( 9 y 4 y) + ( ) ( y y) + ( 5 4 5) ( y) + ( 5) y + ( ) ( 5) = y = y + 0 To see if this simplifies any further, we factor the radicand: perfect square factors, we are done. y + 0 = (y + 10). Finding no

429 A.1 Radical Equations 41 Theorem A.15 allows us to generalize the process of Extracting Square Roots to Extracting n th Roots which in turn allows us to solve equations 6 of the form X n = c. Extracting n th roots: If c is a real number and n is odd then the real number solution to X n = c is X = n c. If c 0 and n is even then the real number solutions to X n = c are X = ± n c. Note: If c < 0 and n is even then X n = c has no real number solutions. n Essentially, we solve X n = c by taking the n th root of both sides: X n = n c. Simplifying the left side gives us just X if n is odd or X if n is even. In the first case, X = n c, and in the second, X = ± n c. Putting this together with the other part of Theorem A.15, namely ( n a) n = a, gives us a strategy for solving equations which involve n th powers and n th roots. Strategies for Solving Power and Radical Equations If the equation involves an n th power and the variable appears in only one term, isolate the term with the n th power and extract n th roots. If the equation involves an n th root and the variable appears in that n th root, isolate the n th root and raise both sides of the equation to the n th power. Note: When raising both sides of an equation to an even power, be sure to check for extraneous solutions. The note about extraneous solutions can be demonstrated by the basic equation: x =. This equation has no solution since, by definition, x 0 for all real numbers x. However, if we square both sides of this equation, we get ( x) = ( ) or x = 4. However, x = 4 doesn t check in the original equation, since 4 =, not. Once again, the root 7 of all of our problems lies in the fact that a negative number to an even power results in a positive number. In other words, raising both sides of an equation to an even power does not produce an equivalent equation, but rather, an equation which may possess more solutions than the original. Hence the cautionary remark above about extraneous solutions. Solve the following equations. 1. (5x + ) 4 (5 w) = = 9. t + t + = 6 4. y + 1 = x x = n + + n = 0 For the remaining problems, assume that all of the variables represent positive real numbers. 8 6 Well, not entirely. The equation x 7 = 1 has seven answers: x = 1 and six complex number solutions which we ll find using techniques in Section??. 7 Pun intended! 8 That is, you needn t worry that you re multiplying or dividing by 0 or that you re forgetting absolute value symbols.

430 4 Algebra Review 7. Solve for r: V = 4π (R r ). 8. Solve for M 1 : r 1 r = 9. Solve for v: m = Solution. m 0 1 v c M. Again, assume that no arithmetic rules are violated. M 1 1. In our first equation, the quantity containing x is already isolated, so we extract fourth roots. The exponent is even, so when the roots are extracted we need both the positive and negative roots. (5x + ) 4 = 16 5x + = ± 4 16 Extract fourth roots 5x + = ± 5x + = or 5x + = x = 1 5 or x = 1 We leave it to the reader to verify that both of these solutions satisfy the original equation.. In this example, we first need to isolate the quantity containing the variable w. Here, third (cube) roots are required and since the exponent (index) is odd, we do not need the ±: (5 w) 1 7 = 9 (5 w) 7 = 8 Subtract 1 (5 w) = 56 Multiply by 7 5 w = 56 Extract cube root 5 w = ( 8)(7) 5 w = w = 7 Product Rule w = 5 7 Subtract 5 w = 5 7 w = Divide by Properties of Negatives The reader should check the answer because it provides a hearty review of arithmetic.. To solve t + t + = 6, we first isolate the square root, then proceed to square both sides of the equation. In doing so, we run the risk of introducing extraneous solutions so checking our answers

431 A.1 Radical Equations 4 here is a necessity. t + t + = 6 t + = 6 t Subtract t ( t + ) = (6 t) Square both sides t + = 6 1t + t F.O.I.L. / Perfect Square Trinomial 0 = t 14t + Subtract t and 0 = (t )(t 11) Factor From the Zero Product Property, we know either t = 0 (which gives t = ) or t 11 = 0 (which gives t = 11). When checking our answers, we find t = satisfies the original equation, but t = 11 does not. 9 So our final answer is t = only. 4. In our next example, we locate the variable (in this case y) beneath a cube root, so we first isolate that root and cube both sides. y + 1 = 0 y + 1 = Subtract y + 1 = Divide by y + 1 = Properties of Negatives ( ) ( y + 1) = Cube both sides y + 1 = ( ) y + 1 = 7 y = 7 1 Subtract 1 y = y = 7 7 y = 7 54 Common denominators Subtract fractions Divide by ( multiply by 1 Since we raised both sides to an odd power, we don t need to worry about extraneous solutions but we encourage the reader to check the solution just for the fun of it. 9 It is worth noting that when t = 11 is substituted into the original equation, we get = 6. If the + 5 were 5, the solution would check. Once again, when squaring both sides of an equation, we lose track of ±, which is what lets extraneous solutions in the door. )

432 44 Algebra Review 5. In the equation 4x x = 1, we have not one but two square roots. We begin by isolating one of the square roots and squaring both sides. 4x x = 1 4x 1 = 1 1 x Subtract 1 x from both sides ( 4x 1) = (1 1 x) Square both sides 4x 1 = x + ( 1 x) F.O.I.L. / Perfect Square Trinomial 4x 1 = x + 4(1 x) 4x 1 = x + 4 8x Distribute 4x 1 = 5 8x 4 1 x Gather like terms At this point, we have just one square root so we proceed to isolate it and square both sides a second time. 10 4x 1 = 5 8x 4 1 x 1x 6 = 4 1 x Subtract 5, add 8x (1x 6) = ( 4 1 x) Square both sides 144x 144x + 6 = 16(1 x) 144x 144x + 6 = 16 x 144x 11x + 0 = 0 Subtract 16, add x 4(6x 8x + 5) = 0 Factor 4(x 1)(18x 5) = 0 Factor some more From the Zero Product Property, we know either x 1 = 0 or 18x 5 = 0. The former gives x = 1 while the latter gives us x = 5. Since we squared both sides of the equation (twice!), we need to 18 check for extraneous solutions. We find x = 5 18 to be extraneous, so our only solution is x = As usual, our first step in solving 4 n + + n = 0 is to isolate the radical. We then proceed to raise both sides to the fourth power to eliminate the fourth root: 4 n + + n = 0 4 n + = n Subtract n ( 4 n + ) 4 = ( n) 4 Raise both sides to the 4 th power n + = n 4 Properties of Negatives 0 = n 4 n Subtract n and 0 = (n )(n + 1) Factor - this is a Quadratic in Disguise At this point, the Zero Product Property gives either n = 0 or n + 1 = 0. From n = 0, we get n =, so n = ±. From n + 1 = 0, we get n = 1, which gives no real solutions. 11 Since we raised both sides to an even (the fourth) power, we need to check for extraneous solutions. We find that n = works but n = is extraneous. 10 To avoid complications with fractions, we ll forego dividing by the coefficient of 1 x, namely 4. This is perfectly fine so long as we don t forget to square it when we square both sides of the equation. 11 Why is that again?

433 A.1 Radical Equations In this problem, we are asked to solve for r. While there are a lot of letters in this equation 1, r appears in only one term: r. Our strategy is to isolate r then extract the cube root. V = 4π (R r ) V = 4π(R r ) Multiply by to clear fractions V = 4πR 4πr Distribute V 4πR = 4πr Subtract 4πR V 4πR 4π 4πR V 4π 4πR V 4π = r Divide by 4π = r Properties of Negatives = r Extract the cube root The check is, as always, left to the reader and highly encouraged. 8. The equation we are asked to solve in this example is from the world of Chemistry and is none other than Graham s Law of Effusion. As was mentioned in Example A.1, subscripts in Mathematics are used to distinguish between variables and have no arithmetic significance. In this example, r 1, r, M 1 and M are as different as x, y, z and 117. Since we are asked to solve for M 1, we locate M 1 and see it is in the denominator of a fraction which is inside of a square root. We eliminate the square root by squaring both sides and proceed from there. (r1 r 1 = r ) = M M 1 ( M ) Square both sides r M 1 r 1 r = M M 1 r M 1 1 = M r Multiply by r M 1 to clear fractions, assume r, M 1 0 M 1 = M r r 1 Divide by r 1, assume r 1 0 As the reader may expect, checking the answer amounts to a good exercise in simplifying rational and radical expressions. The fact that we are assuming all of the variables represent positive real numbers comes in to play, as well. 9. Our last equation to solve comes from Einstein s Special Theory of Relativity and relates the mass of an object to its velocity as it moves. 1 We are asked to solve for v which is located in just one term, 1 including a Greek letter, no less! 1 See this article on the Lorentz Factor.

434 46 Algebra Review namely v, which happens to lie in a fraction underneath a square root which is itself a denominator. We have quite a lot of work ahead of us! m = m 0 1 v c m 1 v c = m 0 Multiply by 1 v ( m 1 v c ( m 1 v ) c to clear fractions c ) = m 0 Square both sides = m 0 Properties of Exponents m m v c = m 0 Distribute m v c = m 0 m Subtract m m v = c (m 0 m ) Multiply by c (c 0) m v = c m 0 + c m Distribute v = c m c m 0 m Rearrange terms, divide by m (m 0) v = c m c m 0 m Extract Square Roots, v > 0 so no ± v = c (m m ) 0 m Properties of Radicals, factor v = c m m 0 m v = c m m 0 m c > 0 and m > 0 so c = c and m = m Checking the answer algebraically would earn the reader great honor and respect on the Algebra battlefield so it is highly recommended. A.1.1 Rationalizing Denominators and Numerators In Section A.10, there were a few instances where we needed to rationalize a denominator - that is, take a fraction with radical in the denominator and re-write it as an equivalent fraction without a radical in the denominator. There are various reasons for wanting to do this, 14 but the most pressing reason is 14 Before the advent of the handheld calculator, rationalizing denominators made it easier to get decimal approximations to fractions containing radicals. However, some (admittedly more abstract) applications remain today one of which we ll explore in Section A.11; one you ll see in Calculus.

435 A.1 Radical Equations 47 that rationalizing denominators - and numerators as well - gives us an opportunity for more practice with fractions and radicals. To refresh your memory, we rationalize a denominator and a numerator below: 1 = = 4 = and 7 4 = 7 4 = 7 8 = 7 = 14 In general, if the fraction contains either a single term numerator or denominator with an undesirable n th root, we multiply the numerator and denominator by whatever is required to obtain a perfect n th power in the radicand that we want to eliminate. If the fraction contains two terms the situation is somewhat more complicated. To see why, consider the fraction 4. Suppose we wanted to rid the denominator of the 5 5 term. We could try as above and multiply numerator and denominator by 5 but that just yields: 4 5 = 5 (4 5) 5 = = We haven t removed 5 from the denominator - we ve just shuffled it over to the other term in the denominator. As you may recall, the strategy here is to multiply both the numerator and the denominator by what s called the conjugate. Congugate of a Square Root Expression: If a, b and c are real numbers with c > 0 then the quantities (a + b c) and (a b c) are conjugates of one another. a Conjugates multiply according to the Difference of Squares Formula: (a + b c)(a b c) = a (b c) = a b c a As are (b c a) and (b c + a) because (b c a)(b c + a) = b c a. That is, to get the conjugate of a two-term expression involving a square root, you change the to a +, or vice-versa. For example, the conjugate of 4 5 is 4 + 5, and when we multiply these two factors together, we get (4 5)(4 + 5) = 4 ( 5) = 16 5 = 11. Hence, to eliminate the 5 from the denominator of our original fraction, we multiply both the numerator and the denominator by the conjugate of 4 5 to get: 4 5 = (4 + 5) (4 5)(4 + 5) = (4 + 5) 4 ( 5) = (4 + 5) 16 5 = What if we had 5 instead of 5? We could try multiplying 4 5 by to get (4 5)(4 + 5) = 4 ( 5) = 16 5, which leaves us with a cube root. What we need to undo the cube root is a perfect cube, which means we look to the Difference of Cubes Formula for inspiration: a b = (a b)(a + ab + b ). If we take a = 4 and b = 5, we multiply (4 5)( ( 5) ) = ( 5) ( 5) = 64 5 = 59

436 48 Algebra Review So if we were charged with rationalizing the denominator of 4, we d have: = ( ( 5) ) (4 5)( ( 5) ) = 59 This sort of thing extends to n th roots since (a b) is a factor of a n b n for all natural numbers n, but in practice, we ll stick with square roots with just a few cube roots thrown in for a challenge. 15 Rationalize the indicated numerator or denominator: 1. Rationalize the denominator: 5 4x. Rationalize the numerator: 9 + h h Solution. 1. We are asked to rationalize the denominator, which in this case contains a fifth root. That means we need to work to create fifth powers of each of the factors of the radicand. To do so, we first factor the radicand: 4x = 8 x = x. To obtain fifth powers, we need to multiply by 4 x inside the radical. 5 4x = 5 x = = 5 4 x 5 x 5 4 x 5 4 x 5 x 4 x Equivalent Fractions Product Rule = 5 4 x x 5 = 5 4 x x 5 = 5 4 x x = x x 5 4x = x Property of Exponents Product Rule Product Rule Reduce Simplify. Here, we are asked to rationalize the numerator. Since it is a two term numerator involving a square root, we multiply both numerator and denominator by the conjugate of 9 + h, namely 9 + h+. 15 To see what to do about fourth roots, use long division to find (a 4 b 4 ) (a b), and apply this to

437 A.1 Radical Equations 49 After simplifying, we find an opportunity to reduce the fraction: 9 + h h = ( 9 + h )( 9 + h + ) h( 9 + h + ) = ( 9 + h) h( 9 + h + ) = = = = (9 + h) 9 h( 9 + h + ) h h( 9 + h + ) 1 h h( 9 + h + ) h + Equivalent Fractions Difference of Squares Simplify Simplify Reduce We close this section with an awesome example from Calculus. 1 1 (x + h) + 1 x + 1 Simplify the compound fraction then rationalize the numerator of the result. h Solution. We start by multiplying the top and bottom of the big fraction by x + h + 1 x (x + h) + 1 x + 1 h = = = 1 1 x + h + 1 x + 1 h ( ) 1 1 x + h + 1 x + 1 x + h + 1 x + 1 x + h + 1 x + 1 x + h + 1 h x + h + 1 x + 1 h x + h + 1 x + 1 x + h + 1 x + 1 x + 1 = x + 1 x + h + 1 h x + h + 1 x + 1 Next, we multiply the numerator and denominator by the conjugate of x + 1 x + h + 1, namely

438 40 Algebra Review x x + h + 1, simplify and reduce: x + 1 x + h + 1 h x + h + 1 x + 1 = ( x + 1 x + h + 1)( x x + h + 1) h x + h + 1 x + 1( x x + h + 1) = = = = = ( x + 1) ( x + h + 1) h x + h + 1 x + 1( x x + h + 1) (x + 1) (x + h + 1) h x + h + 1 x + 1( x x + h + 1) x + 1 x h 1 h x + h + 1 x + 1( x x + h + 1) h h x + h + 1 x + 1( x x + h + 1) x + h + 1 x + 1( x x + h + 1) While the denominator is quite a bit more complicated than what we started with, we have done what was asked of us. In the interest of full disclosure, the reason we did all of this was to cancel the original h from the denominator. That s an awful lot of effort to get rid of just one little h, but you ll see the significance of this in Calculus.

439 A.1 Radical Equations 41 A.1. Exercises In Exercises 1-1, perform the indicated operations and simplify. 1. 9x. 8t. 50y t + 4t w 16w ( 1x x) + 1 c v 4πr 5 7. c 8. L 4 πε 8 9. ρ x x + 1 ( ) t 1 + t ( t) x 1 t ( ) ( ) z + z ( 1 z ) 1 ( 1) 1. + (x) x 1 ( x 1 ) () 4 In Exercises 14-5, find all real solutions. 14. (x + 1) + 8 = (1 y) 4 = t = x + 1 = t + 1 = x + 1 = x y + y + 10 = 1. t + 6 9t =. x 1 = x +. w = 4 1 w 4. x + x 5 = 5. x + 1 = + 4 x In Exercises 6-9, solve each equation for the indicated variable. Assume all quantities represent positive real numbers. 6. Solve for h: I = bh L 8. Solve for g: T = π g Solve for a: I 0 = 5 a Solve for v: L = L 0 In Exercises 0-5, rationalize the numerator or denominator, and simplify. 1 v c x 7. x c x c. x + h + 1 x + 1 h 4. x + 1 x 7 5. x + h x h

440 4 Algebra Review A.1. Answers 1. x. t. 5 y 4. t w 8 6. x c v c 8. r πr L 9. ε 4 π ρ x 11. 6t 1 t z ( 1 z) 1. 4x (x 1) x x = 15. y = 1, 16. t = 17. x = t = ± x = 0. y = 1. t = 1,. x = w = 4. x = 6 5. x = 4 6. h = 1I b 8. g = 4π L T 7. a = 4 I v = c L 0 L L x 6x. 1 x + c. x + h x ( x + 1) + x ( x + h) + x + h x + ( x)

441 A.14 Variation 4 A.14 Variation In many instances in the sciences, equations are encountered as a result of fundamental natural laws which are typically a result of assuming certain basic relationships between variables. These basic relationships are summarized in the definition below. Suppose x, y and z are variable quantities. We say y varies directly with (or is directly proportional to) x if there is a constant k such that y = kx y varies inversely with (or is inversely proportional to) x if there is a constant k such that y = k x z varies jointly with (or is jointly proportional to) x and y if there is a constant k such that z = kxy The constant k in the above definitions is called the constant of proportionality. Translate the following into mathematical equations using Definition A Hooke s Law: The force F exerted on a spring is directly proportional the extension x of the spring.. Boyle s Law: At a constant temperature, the pressure P of an ideal gas is inversely proportional to its volume V. (We explore this one more deeply in Example??.. The volume V of a right circular cone varies jointly with the height h of the cone and the square of the radius r of the base. 4. Ohm s Law: The current I through a conductor between two points is directly proportional to the voltage V between the points and inversely proportional to the resistance R between the points. 5. Newton s Law of Universal Gravitation: Suppose two objects, one of mass m and one of mass M, are positioned so that the distance between their centers of mass is r. The gravitational force F exerted on the two objects varies directly with the product of the two masses and inversely with the square of the distance between their centers of mass. Solution. 1. Applying the definition of direct variation, we get F = kx for some constant k.. Since P and V are inversely proportional, we write P = k V.

442 44 Algebra Review. There is a bit of ambiguity here. It s clear that the volume and the height of the cone are represented by the quantities V and h, respectively, but does r represent the radius of the base or the square of the radius of the base? It is the former. Usually, if an algebraic operation is specified (like squaring), it is meant to be expressed in the formula. We apply Definition A.14 to get V = khr. 4. Even though the problem doesn t use the phrase varies jointly, it is implied by the fact that the current I is related to two different quantities. Since I varies directly with V but inversely with R, we write I = kv R. 5. We write the product of the masses mm and the square of the distance as r. We have that F varies directly with mm and inversely with r, so F = kmm r. A note about units is in order. The formulas given in Example A.14 above all have quantities from the real world and we would disappoint our friends who teach Science if we didn t remind you to pay attention to units when working with these equations. The natural question that arises is What units does k have? The answer is whatever works and by that we mean the units on k will be whatever it takes to make the equation have the same units on both sides. For example, in Hooke s Law we have that F = kx. If F is in newtons and x is in meters then k must be in newton. This can lead to some odd sounding units, such as the units on the constant R in the Ideal Gas meter Law PV = nrt (see Exercise 11) or no units at all (see Exercise 9a). Unit conversions can mess things up as well - see Exercise 9b for a sample of that kind of nonsense! We end this section with an example that first requires us to find the value of k and then use it to solve another problem. Suppose it takes 11 pounds of force to hold a spring inches beyond its natural length. What force is required to hold it 7 inches beyond natural length? Solution. Using Hooke s Law with F = 11 pounds and x = inches we solve 11 = k for k and find k = 5.5 pound. Setting x = 7 in Hooke s Law with k = 5.5 yields F = = 8.5 pounds of force. (Check inch the units to convince yourself that this worked!)

443 A.14 Variation 45 A.14.1 Exercises In Exercises 1-6, translate the following into mathematical equations. 1. At a constant pressure, the temperature T of an ideal gas is directly proportional to its volume V. (This is Charles s Law). The frequency of a wave f is inversely proportional to the wavelength of the wave λ.. The density d of a material is directly proportional to the mass of the object m and inversely proportional to its volume V. 4. The square of the orbital period of a planet P is directly proportional to the cube of the semi-major axis of its orbit a. (This is Kepler s Third Law of Planetary Motion ) 5. The drag of an object traveling through a fluid D varies jointly with the density of the fluid ρ and the square of the velocity of the object ν. 6. Suppose two electric point charges, one with charge q and one with charge Q, are positioned r units apart. The electrostatic force F exerted on the charges varies directly with the product of the two charges and inversely with the square of the distance between the charges. (This is Coulomb s Law) 7. According to this webpage, the frequency f of a vibrating string is given by f = 1 T where T is L µ the tension, µ is the linear mass 1 of the string and L is the length of the vibrating part of the string. Express this relationship using the language of variation. 8. According to the Centers for Disease Control and Prevention a person s Body Mass Index B is directly proportional to his weight W in pounds and inversely proportional to the square of his height h in inches. (a) Express this relationship as a mathematical equation. (b) If a person who was 5 feet, 10 inches tall weighed 5 pounds had a Body Mass Index of.7, what is the value of the constant of proportionality? (c) Rewrite the mathematical equation found in part 8a to include the value of the constant found in part 8b and then find your Body Mass Index. 9. This exercise refers back to the volume of a right circular cone formula found in Example A.14. (a) First assume that V, h and r are all measured using the same unit of length. Work with your classmates to show that in this case, the k needed for the volume formula V = khr has no units on it. 1 Also known as the linear density. It is simply a measure of mass per unit length.

444 46 Algebra Review (b) Now assume that V is measured in milliliters, h is measured in meters and r is measured in yards. Work with your classmates to find the units on k so that the volume formula V = khr makes sense. 10. We know that the circumference of a circle varies directly with its radius with π as the constant of proportionality. (That is, we know C = πr.) With the help of your classmates, compile a list of other basic geometric relationships which can be seen as variations. 11. Research the Ideal Gas Law PV = nrt to see what sorts of units are used for the constant R. What other formulations of this law did you find in your research?

445 A.14 Variation 47 A.14. Answers 1. T = kv. f = k λ. d = km V 4. P = ka 5. D = kρν 6. 4 F = kqq 7. Rewriting f = 1 T 1 T L µ as f = L we see that the frequency f varies directly with the square root µ of the tension and varies inversely with the length and the square root of the linear mass. 8. (a) B = kw h (b) 5 k = (c) B = 70.68W h r The character λ is the lower case Greek letter lambda. The characters ρ and ν are the lower case Greek letters rho and nu, respectively. 4 Note the similarity to this formula and Newton s Law of Universal Gravitation as discussed in Example 5. 5 The CDC uses 70.

446 48 Algebra Review

447 Appendix B Geometry Review The authors really wanted the Trigonometry portion of Precalculus, Episode IV to start with the definitions of the circular functions so one purpose of this Geometry Review Appendix is to find a home for the material that is prerequisite to those definitions. Another reason for this Appendix is to further support a co-requisite approach to teaching a Precalculus 1 class. As is the case with the Algebra Review Appendix, this chapter is not designed for students who have never seen this material before. In fact, our treatment of Geometry is even more brief than that of Algebra because we assume a student who is taking a stand alone college-level Trigonometry class is already proficient in College Algebra, and those learning the Trigonometry portion of a full Precalculus class have ostensibly survived the College Algebra portion. Thus we review only some very basic concepts covered in a typical high school Geometry course. Where appropriate, we have referenced specific sections of the main body of the Precalculus text in an effort to assist faculty who would like to assign the Appendix as just in time review reading to their students. This Appendix contains two sections which are briefly described below: Section B.1 (Angles in Degrees) is a brief review of some of the terminology and concepts from a typical high school Geometry course. Radian measure is deferred until Chapter 4. Section B. (Basic Right Triangle Trigonometry) defines the trigonometric functions in the context of a right triangle using angles measured in degrees. Basic applications are discussed and a proof of the Pythagorean Theorem is given but trigonometric identities are deferred until Chapter 4. 1 Remember how we define Precalculus - to us, Precalculus = College Algebra + College Trigonometry without formal limits. In order to fully support a co-requisite approach to a class that has Trigonometry in it, we felt it necessary to provide some material to assist students who have gaps in their Geometry background. The careful reader will note that all of this material was in the main body of our third edition so it can be included nearly seemlessly into a regular Trigonometry class.

448 440 Geometry Review B.1 Angles in Degrees This section serves as a review of the concept of angle and the use of the degree system to measure angles. Recall that a ray is usually described as a half-line and can be thought of as a line segment in which one of the two endpoints is pushed off infinitely distant from the other, as pictured below. The point from which the ray originates is called the initial point of the ray. P A ray with initial point P. When two rays share a common initial point they form an angle and the common initial point is called the vertex of the angle. Two examples of what are commonly thought of as angles are Q P An angle with vertex P. An angle with vertex Q. However, the two figures below also depict angles - albeit these are, in some sense, extreme cases. In the first case, the two rays are directly opposite each other forming what is known as a straight angle; in the second, the rays are identical so the angle is indistinguishable from the ray itself. P A straight angle. Q The measure of an angle is a number which indicates the amount of rotation that separates the rays of the angle. There is one immediate problem with this, as pictured below. Which amount of rotation are we attempting to quantify? What we have just discovered is that we have at least two angles described by this diagram. 1 Clearly these two angles have different measures because one appears to represent a larger rotation than the other, so we must label them differently. In this book, 1 The phrase at least will be justified in short order.

449 B.1 Angles in Degrees 441 we use lower case Greek letters such as α (alpha), β (beta), γ (gamma) and θ (theta) to label angles. So, for instance, we have β α One system to measure angles is degree measure. Quantities measured in degrees are denoted by the symbol. One complete revolution as shown below is 60, and parts of a revolution are measured proportionately. Thus half of a revolution (a straight angle) measures 1 (60 ) = 180, a quarter of a revolution (a right angle) measures 1 4 (60 ) = 90 and so on. One revolution Note that in the above figure, we have used the small square to denote a right angle, as is commonplace in Geometry. Recall that if an angle measures strictly between 0 and 90 it is called an acute angle and if it measures strictly between 90 and 180 it is called an obtuse angle. It is important to note that, theoretically, we can know the measure of any angle as long as we know the proportion it represents of entire revolution. For instance, the measure of an angle which represents a rotation of of a revolution would measure (60 ) = 40, the measure of an angle which constitutes only 1 of a revolution measures (60 ) = 0 and an angle which indicates no rotation at all is measured as 0. The choice of 60 is most often attributed to the Babylonians. This is how a protractor is graded.

450 44 Geometry Review Using our definition of degree measure, we have that 1 represents the measure of an angle which constitutes 1 of a revolution. Even though it may be hard to draw, it is nonetheless not difficult to imagine 60 an angle with measure smaller than 1. There are two ways to subdivide degrees. The first, and most familiar, is decimal degrees. For example, an angle with a measure of 0.5 would represent a rotation halfway between 0 and 1, or equivalently, = 61 of a full rotation. This can be taken to the 70 limit using Calculus so that measures like make sense. 4 The second way to divide degrees is the Degree - Minute - Second (DMS) system. In this system, one degree is divided equally into sixty minutes, and in turn, each minute is divided equally into sixty seconds. 5 In symbols, we write 1 = 60 and 1 = 60, from which it follows that 1 = 600. To convert a measure of 4.15 to the DMS system, we start by noting ( that ) 4.15 = Converting the partial amount of degrees to minutes, we find = 7.5 = Converting the partial amount of minutes to seconds gives ( ) = 0. Putting it all together yields 4.15 = = = = = On the other hand, to convert to decimal degrees, we first compute 15 ( ) 1 60 = 1 4 and 45 ( ) = Then we find = = = = Recall that two acute angles are called complementary angles if their measures add to 90. Two angles, either a pair of right angles or one acute angle and one obtuse angle, are called supplementary angles 4 Awesome math pun aside, this is the same idea behind defining irrational exponents in Section??. 5 Does this kind of system seem familiar?

451 B.1 Angles in Degrees 44 if their measures add to 180. In the diagram below, the angles α and β are supplementary angles while the pair γ and θ are complementary angles. β θ α γ Supplementary Angles Complementary Angles In practice, the distinction between the angle itself and its measure is blurred so that the sentence α is an angle measuring 4 is often abbreviated as α = 4. It is now time for an example. Let α = and β = Convert α to the DMS system. Round your answer to the nearest second.. Convert β to decimal degrees. Round your answer to the nearest thousandth of a degree.. Sketch α and β. 4. Find a supplementary angle for α. 5. Find a complementary angle for β. Solution. 1. To convert ( ) α to the DMS system, we start with = 111 ( ). Next we convert =.6. Writing.6 = + 0.6, we convert = Hence, = Rounding to seconds, we obtain α = = = = To convert β to decimal degrees, we convert 8 ( ) 1 60 = 7 15 and 17 ( ) = Putting it all together, we have = = =

452 444 Geometry Review. To sketch α, we first note that 90 < α < 180. Dividing this range in half, we get 90 < α < 15, and once more, we have 90 < α < This gives us a pretty good estimate for α, as shown below. 6 Proceeding similarly for β, we find 0 < β < 90, then 0 < β < 45,.5 < β < 45, and lastly,.75 < β < 45. Angle α Angle β 4. To find a supplementary angle for α, we seek an angle θ so that α +θ = 180. We get θ = 180 α = = To find a complementary angle for β, we seek an angle γ so that β + γ = 90. We get γ = 90 β = While we could reach for the calculator to obtain an approximate answer, we choose instead to do a bit of sexagesimal 7 arithmetic. We first rewrite 90 = = = In essence, we are borrowing 1 = 60 from the degree place, and then borrowing 1 = 60 from the minutes place. 8 This yields, γ = = = Up to this point, we have discussed only angles which measure between 0 and 60, inclusive. Ultimately, we want to use the arsenal of Algebra which we have stockpiled in Chapters?? through?? to not only solve geometric problems involving angles, but also to extend their applicability to other real-world phenomena. A first step in this direction is to extend our notion of angle from merely measuring an extent of rotation to quantities which indicate an amount of rotation along with a direction. To that end, we introduce the concept of an oriented angle. As its name suggests, in an oriented angle, the direction of the rotation is important. We imagine the angle being swept out starting from an initial side and ending at a terminal side, as shown below. When the rotation is counter-clockwise 9 from initial side to terminal side, we say that the angle is positive; when the rotation is clockwise, we say that the angle is negative. Initial Side Terminal Side Terminal Side Initial Side A positive angle, 45 A negative angle, 45 6 If this process seems hauntingly familiar, it should. Compare this method to the Bisection Method introduced in Section??. 7 Like latus rectum, this is also a real math term. 8 This is the exact same kind of borrowing you used to do in Elementary School when trying to find Back then, you were working in a base ten system; here, it is base sixty. 9 widdershins

453 B.1 Angles in Degrees 445 At this point, we also extend our allowable rotations to include angles which encompass more than one revolution. For example, to sketch an angle with measure 450 we start with an initial side, rotate counterclockwise one complete revolution (to take care of the first 60 ) then continue with an additional 90 counter-clockwise rotation, as seen below. 450 To further connect angles with the Algebra which has come before, we shall often overlay an angle diagram on the coordinate plane. An angle is said to be in standard position if its vertex is the origin and its initial side coincides with the positive horizontal (usually labeled as the x-) axis. Angles in standard position are classified according to where their terminal side lies. For instance, an angle in standard position whose terminal side lies in Quadrant I is called a Quadrant I angle. If the terminal side of an angle lies on one of the coordinate axes, it is called a quadrantal angle. Two angles in standard position are called coterminal if they share the same terminal side. 10 In the figure below, α = 10 and β = 40 are two coterminal Quadrant II angles drawn in standard position. Note that α = β + 60, or equivalently, β = α 60. We leave it as an exercise to the reader to verify that coterminal angles always differ by a multiple of More precisely, if α and β are coterminal angles, then β = α + 60 k where k is an integer. 1 4 y α = β = 40 x 4 Two coterminal angles, α = 10 and β = 40, in standard position. 10 Note that by being in standard position they automatically share the same initial side which is the positive x-axis. 11 It is worth noting that all of the pathologies of Analytic Trigonometry result from this innocuous fact. 1 Recall that this means k = 0, ±1, ±,...

454 446 Geometry Review Graph each of the (oriented) angles below in standard position and classify them according to where their terminal side lies. Find three coterminal angles, at least one of which is positive and one of which is negative. 1. α = 60. β = 5. γ = φ = 750 Solution. 1. To graph α = 60, we draw an angle with its initial side on the positive x-axis and rotate counterclockwise = 1 of a revolution. We see that α is a Quadrant I angle. To find angles which are 6 coterminal, we look for angles θ of the form θ = α + 60 k, for some integer k. When k = 1, we get θ = = 40. Substituting k = 1 gives θ = = 00. Finally, if we let k =, we get θ = = Since β = 5 is negative, we start at the positive x-axis and rotate clockwise 5 60 = 5 8 of a revolution. We see that β is a Quadrant II angle. To find coterminal angles, we proceed as before and compute θ = k for integer values of k. We find 15, 585 and 495 are all coterminal with 5. y y α = x x 4 β = 5 4 α = 60 in standard position. β = 5 in standard position.. Since γ = 540 is positive, we rotate counter-clockwise from the positive x-axis. One full revolution accounts for 60, with 180, or 1 of a revolution remaining. Since the terminal side of γ lies on the negative x-axis, γ is a quadrantal angle. All angles coterminal with γ are of the form θ = k, where k is an integer. Working through the arithmetic, we find three such angles: 180, 180 and The Greek letter φ is pronounced fee or fie and since φ is negative, we begin our rotation clockwise from the positive x-axis. Two full revolutions account for 70, with just 0 or 1 of a revolution to go. 1 We find that φ is a Quadrant IV angle. To find coterminal angles, we compute θ = k for a few integers k and obtain 90, 0 and 0.

455 B.1 Angles in Degrees 447 y y 4 4 γ = x x 4 4 φ = 750 γ = 540 in standard position. φ = 750 in standard position. Note that since there are infinitely many integers, any given angle has infinitely many coterminal angles, and the reader is encouraged to plot the few sets of coterminal angles found in Example B.1 to see this. As we ll see in Section B. and throughout Chapter??, degree measure is very popular for many applications involving geometry and modeling physical forces. In Section??, we ll introduce a different method of measuring angles, radian measure, which is tied directly to arc length and is useful in other applications involving circular motion and periodic phenomenon.

456 448 Geometry Review B.1.1 Exercises In Exercises 1-4, convert the angles into the DMS system. Round each of your answers to the nearest second In Exercises 5-8, convert the angles into decimal degrees. Round each of your answers to three decimal places In Exercises 9-0, graph the oriented angle in standard position. Classify each angle according to where its terminal side lies and then give two coterminal angles, one of which is positive and the other negative With help from your classmates, explain why if (x, y) is a point on the terminal side of an angle α in standard position, then so is (r x, r y) for any number r > 0. What happens if r < 0?

457 B.1 Angles in Degrees 449 B.1. Answers is a Quadrant I angle coterminal with 90 and y is a Quadrant II angle coterminal with 480 and 40 4 y x x is a Quadrant III angle coterminal with 585 and y 1. 0 is a Quadrant IV angle coterminal with 690 and y x x 1. 0 is a Quadrant IV angle coterminal with 0 and y is a Quadrant III angle coterminal with 5 and y x x 4 4

458 450 Geometry Review is a Quadrant II angle coterminal with 10 and y x is a quadrantal angle coterminal with 90 and y x is a Quadrant I angle coterminal with 45 and y is a Quadrant II angle coterminal with 10 and 40 4 y x x is a Quadrant III angle coterminal with 150 and 10 4 y is a quadrantal angle coterminal with 180 and y x x

459 B. Right Triangle Trigonometry 451 B. Right Triangle Trigonometry The word trigonometry literally means measuring triangles, so naturally most students first introduction to trigonometry focuses on triangles. This section focuses on right triangles, triangles in which one angle measures 90. Consider the right triangle below, where, as usual, the small square denotes the right angle, the labels a, b, and c denote the lengths of the sides of the triangle, and α and β represent the (measure of) the non-right angles. As you may recall, the side opposite the right angle is called the hypotenuse of the right triangle. Also note that since the sum of the measures of all angles in a triangle must add to 180, we have α + β + 90 = 180, or α + β = 90. Said differently, the non-right angles in a right triangle are complements. c α b β a We now state and prove the most famous result about right triangles: The Pythagorean Theorem. Theorem B.1. (The Pythagorean Theorem) The square of the length of the hypotenuse of a right triangle is equal to the sums of the squares of the other two sides. More specifically, if c is the length of the hypotenuse of a right triangle and a and b are the lengths of the other two sides, then a + b = c. There are several proofs of the Pythagorean Theorem, 1 but the one we choose to reproduce here showcases a nice interplay between algebra and geometry. Consider taking four copies of the right triangle below on the left and arranging them as seen below on the right. b α β c a β b α c c a β α b a b α c c β a β α a b It should be clear that we have produced a large square with a side length of (a + b). What is also true, but may not be obvious, is that the shaded quadrilateral is also a square. We can readily see the shaded 1 Including one by Mentor, Ohio native President James Garfield.

460 45 Geometry Review quadrilateral has equal sides of length c. Moreover, since α + β = 90, we get the interior angles of the shaded quadrilateral are each 90. Hence, the shaded quadrilateral is indeed a square. We finish the proof by computing the area of the of the large square in two ways. First, we square the length of its side: (a + b). Next, we add up the areas of the four triangles, each having area 1 ab along with the area of the shaded square, c. Equating these to expressions gives: (a + b) = 4 ( 1 ab) + c. Since (a + b) = a + ab + b and 4 ( 1 ab) = ab, we have a + ab + b = ab + c or a + b = c, as required. It should be noted that the converse of the Pythagorean Theorem is also true. That is if a, b, and c are the lengths of sides of a triangle and a + b = c, then c the triangle is a right triangle. A list of integers (a, b, c) which satisfy the relationship a + b = c is called a Pythagorean Triple. Some of the more common triples are: (, 4, 5), (5, 1, 1), (7, 4, 5), and (8, 15, 17). We leave it to the reader to verify these integers satisfy the equation a + b = c and suggest committing these triples to memory. Next, we set about defining characteristic ratios associated with acute angles. Given any acute angle θ, we can imagine θ being an interior angle of a right triangle as seen below. c b θ a Focusing on the arrangement of the sides of the triangle with respect to the angle θ, we make the following definitions: the side with length a is called the side of the triangle which is adjacent to θ and the side with length b is called the side of the triangle opposite θ. As usual, the side labeled c (the side opposite the right angle) is the hypotenuse. Using this diagram, we define three important trigonometric ratios of θ. Suppose θ is an acute angle residing in a right triangle as depicted above. The sine of θ, denoted sin(θ) is defined by the ratio: sin(θ) = b c, or The cosine of θ, denoted cos(θ) is defined by the ratio: cos(θ) = a c, or length of opposite length of hypotenuse. length of adjacent length of hypotenuse. The tangent of θ, denoted tan(θ) is defined by the ratio: tan(θ) = b length of opposite, or a length of adjacent. For example, consider the angle θ indicated in the triangle below on the left. Using Definition B., we get sin(θ) = 4 5, cos(θ) = 5, and tan(θ) = 4. One may well wonder if these trigonometric ratios we ve found for θ change if the triangle containing θ changes. For example, if we scale all the sides of the triangle below on the left by a factor of, we produce the similar triangle below in the middle. Using this triangle to We will prove this in Section?? by generalizing the Pythagorean Theorem to a formula that works for all triangles. That is, a triangle with the same shape - that is, the same angles.

461 B. Right Triangle Trigonometry 45 compute our ratios for θ, we find sin(θ) = 8 10 = 4 5, cos(θ) = 6 10 = 5, and tan(θ) = 8 6 = 4. Note that the scaling factor, here, is common to all sides of the triangle, and, hence, cancels from the numerator and denominator when simplifying each of the ratios r 4r θ θ 6 θ r In general, thanks to the Angle Angle Similarity Postulate, any two right triangles which contain our angle θ are similar which means there is a positive constant r so that the sides of the triangle are r, 4r, and 5r as seen above on the right. Hence, regardless of the right triangle in which we choose to imagine θ, sin(θ) = 4r = 4 r, cos(θ) = = 4r, and tan(θ) = = 4. Generalizing this same argument to any acute angle 5r 5 5r 5 r θ assures us that the ratios as described in Definition B. are independent of the triangle we use. Our next objective is to determine the values of sin(θ), cos(θ), and tan(θ) for some of the more commonly used angles. We begin with 45. In a right triangle, if one of the non-right angles measures 45, then the other measures 45 as well. It follows that the two legs of the triangle must be congruent. Since we may choose any right triangle containing a 45 angle for our computations, we choose the length of one (hence both) of the legs to be 1. The Pythagorean Theorem gives the hypotenuse is: c = =, so c =. (We take only the positive square root here since c represents the length of the hypotenuse here, so, necessarily c > 0.) From this, we obtain the values below, and suggest committing them to memory. c = sin (45 ) = 1 = cos (45 ) = 1 = tan (45 ) = 1 1 = 1 Note that we have rationalized here to avoid the irrational number appearing in the denominator. This is a common convention in trigonometry, and we will adhere to it unless extremely inconvenient. Next, we investigate 60 and 0 angles. Consider the equilateral triangle below each of whose sides measures units. Each of its interior angles is necessarily 60, so if we drop an altitude, we produce two triangles each having a base measuring 1 unit and a hypotenuse of units. Using the Pythagorean Theorem, we can find the height, h of these triangles: 1 + h = so h = or h =. Using these, we can find the values of the trigonometric ratios for both 60 and 0. Again, we recommend committing these values to memory.

462 454 Geometry Review 0 h = sin (60 ) = cos (60 ) = 1 tan (60 ) = 1 = sin (0 ) = 1 cos (0 ) = tan (0 ) = 1 = Since 0 and 60 are complements, the side adjacent to the 60 angle is the side opposite the 0 and the side opposite the 60 angle is the side adjacent to the 0. This sort of swapping is true of all complementary angles and will be generalized in Section??, Theorem??. Note that the values of the trigonometric ratios we have derived for 0, 45, and 60 angles are the exact values of these ratios. For these angles, we can conveniently express the exact values of their sines, cosines, and tangents resorting, at worst, to using square roots. The reader may well wonder if, for instance, we can express the exact value of, say, sin (4 ) in terms of radicals. The answer in this case is yes (see here), but, in general, we will not take the time to pursue such representations. 4 Hence, if a problem requests an exact answer involving sin (4 ), we will leave it written as sin (4 ) and use a calculator to produce a suitable approximation as the situation warrants. Our first example requires the concept of an angle of inclination. The angle of inclination (or angle of elevation) of an object refers to the angle whose initial side is some kind of base-line (say, the ground), and whose terminal side is the line-of-sight to an object above the base-line. Schematically: object θ base line The angle of inclination from the base line to the object is θ 1. The angle of inclination from a point on the ground 0 feet away to the top of Lakeland s Armington Clocktower 5 is 60. Find the height of the Clocktower to the nearest foot. 4 We will do a little of this in Section??. 5 Named in honor of Raymond Q. Armington, Lakeland s Clocktower has been a part of campus since 197.

463 B. Right Triangle Trigonometry 455. The Americans with Disabilities Act (ADA) stipulates the incline on an accessibility ramp be 5. If a ramp is to be built so that it replaces stairs that measure 1 inches tall, how long does the ramp need to be? Round your answer to the nearest inch.. In order to determine the height of a California Redwood tree, two sightings from the ground, one 00 feet directly behind the other, are made. If the angles of inclination were 45 and 0, respectively, how tall is the tree to the nearest foot? Solution. 1. We can represent the problem situation using a right triangle as shown below on the left. If we let h denote the height of the tower, then we have tan (60 ) =. From this we get an exact answer of h = 0 tan (60 ) = 0 feet. Using a calculator, we get the approximation which, when rounded to the nearest foot, gives us our answer of 5 feet.. We diagram the situation below on the left using l to represent the unknown length of the ramp. We have sin (5 ) = 1 1 l so that l = sin( inches. Hence, the ramp is 41 inches long. ) h 0 l in. 1 in. h ft ft. Finding the height of the Clocktower Finding the length of an accessibility ramp.. Sketching the problem situation below, we find ourselves with two unknowns: the height h of the tree and the distance x from the base of the tree to the first observation point. h ft ft. x ft. Finding the height of a California Redwood

464 456 Geometry Review Luckily, we have two right triangles to help us find each unknown, as shown below. From the triangle below on the left, we get tan (45 ) = h x. From the triangle below on the right, we see tan (0 ) = h x+00. h ft. h ft x ft. x + 00 ft. Since tan (45 ) = 1, the first equation gives h = 1, or x = h. Substituting this into the second equation x h gives h+00 = tan (0 ) =. Clearing fractions, we get h = (h + 00). The result is a linear equation for h, so we expand the right hand side and gather all the terms involving h to one side. h = (h + 00) h = h + 00 h h = 00 ( )h = 00 h = Hence, the tree is approximately 7 feet tall. There are three more trigonometric ratios which are commonly used and they are defined in the same manner the ratios in Definition B. are defined. They are listed below. Suppose θ is an acute angle residing in a right triangle as depicted on page 454. The cosecant of θ, denoted csc(θ) is defined by the ratio: csc(θ) = c b length of hypotenuse, or. length of opposite The secant of θ, denoted sec(θ) is defined by the ratio: sec(θ) = c a length of hypotenuse, or. length of adjacent The cotangent of θ, denoted cot(θ) is defined by the ratio: cot(θ) = a b, or length of adjacent length of opposite. We practice these definitions in the following example. Suppose θ is an acute angle with cot(θ) =. Find the values of the remaining five trigonometric ratios: sin(θ), cos(θ), tan(θ), csc(θ), and sec(θ).

465 B. Right Triangle Trigonometry 457 Solution. We are given cot(θ) =. So, to proceed, we construct a right triangle in which the length of the side adjacent to θ and the length of the side opposite of θ has a ratio of =. Note there are infinitely 1 many such right triangles - we have produced two below for reference. We will focus our attention on the triangle below on the left and encourage the reader to work through the details using the triangle below on the right to verify the choice of triangle doesn t matter. c = 10 θ 1 θ 6 From the diagram, we see immediately tan(θ) = 1, but in order to determine the remaining four trigonometric ratios, we need to first find the value of the hypotenuse. The Pythagorean Theorem gives 1 + = c so c = 10 or c = 10. Rationalizing denominators, we find sin(θ) = = 10 10, cos(θ) = 10 = 10 10, csc(θ) = = 10 and sec(θ) = 1 While we learned all about the trigonometric ratios of θ in Example B., the identity of θ remains unknown. Since sin(θ) = is decidedly less than sin (0 ) = 1 = 0.5, it stands to reason that θ < 0. It turns out the calculator can provide for us a decimal approximation of θ by way of the sin 1 (x) function. Here, the 1 exponent denotes an inverse function (see Section??) does not mean reciprocal. ( 6 10 That ) is, sin 1 (x) (read sine-inverse of x ) gives an angle whose sine is x. Hence, we may write θ = sin The functions cos 1 (x) and tan 1 (x) work similarly. Indeed, ( ) ( ) ( ) θ = sin 1 = cos 1 = tan 1, and the reader is encouraged to use a calculator to verify these statements. Please note there is much more to these inverse functions than the angle finder description use here. 7 That being said, we finish this section showcasing a use for the tan 1 (x) function below. 8 The roof on the house below has a 6/1 pitch. This means that when viewed from the side, the roof line has a rise of 6 feet over a run of 1 feet. Find the angle of inclination from the bottom of the roof to the top of the roof. Round your answer to the nearest hundredth of a degree. 10 Front View Side View 6 That is, sin 1 (x) 1. That being said, sin(x) (sin(x)) 1 = 1 = csc(x). 7 sin(x) See Section?? for all of the pedantic details. 8 The authors would like to thank Dan Stitz for this problem and associated graphics.

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