The Mathematics That Every Secondary School Math Teacher Needs to Know

572 Chapter 12 Trigonometry Now, the x and y components of the net force Fx and Fy are, respectively, Fx ¼ ðF1Þx þ ðF2Þx þ ðF3Þx ¼ 14:772 þ 17:678 þ 15 ¼ 47:45 and Fy ¼ ðF1Þy þ ðF2Þy þ ðF3Þy ¼ 2:6047 þ 17:678 þ 25:981 ¼ 46:264: pffiffiffiffiSffiffiffioffiffiffiffiffiffiFffiffiffiffiffi=ffiffiffiffiffiffi(ffiffiFffiffixffiffi,ffiffiffiffiFffiffiffiy) = (47.45, 46.26) and the magnitude of the net force, F, in pounds is 47:452 þ 46:2642 % 66:271 pounds. The angle, θ, that that the net force makes with the horizon- tal satisfies tan y % Fy % 46: 264 % 0:975 01: So θ = tanÀ1(0.97501) % 44.275. Thus, our body will Fx 47: 45 move at an angle of 44.275 to the horizontal, and will move as if a single net force of 66.271 pounds were pulling it in that direction. Thus far, the formulas 12.1 and 12.2 have worked beautifully and it would be great if we could always apply them. But, there is still a case we haven’t addressed. If we have the force, F, shown in Figure 12.8, y F 170 x Figure 12.8 and we tried to compute the x component of it as we did earlier, we would have to compute Fx = F cos 170. But what would cosine of 170 degrees mean? Thus far, we have only defined the cosine of an angle in a right triangle where there are no angles of 170 degrees. Furthermore, if the component forces acted as if they were sides of a triangle, with hypotenuse F, then the x component, which depends on cos 170, would have to be negative, since our force is represented by an arrow in the second quadrant. Situations like this are what motivated mathematicians and scientists who worked in tandem to start talking about trigonometric functions of angles more than 90 degrees and trying to define them in a way that made practical sense. We follow that train of thought in the next section. We will return to vectors again in the last section of this chapter after we have developed more needed concepts. Suffice it to say that everything works as above regardless of the angles. Student Learning Opportunities 1* (C) Solve the launch problem. 2* (C) Your students are fascinated to find out that basic trigonometry can be used to find the dis- tance from the earth to the moon and are eager to find out some details. They turn to you for guidance. Use Figure 12.9 and the following suggestion to show how it can be done. Suggestion: It is known that, if when the moon is at its zenith at point B, it is observed to be at the horizon at point A, which is 6155 miles from B along the surface of the earth. Knowing that

12.2 Typical Applications Using Angles and Basic Trigonometric Functions 573 the radius of the earth is roughly 3960 miles, we can find the angle θ in the figure by realizing that the arc AB is 2 pr y , which is the formula for the length of the arc. Use this value of θ to 360 find the distance from point B on the earth to the moon. moon earth B Oq 6155 mi 3960 mi A Figure 12.9 3* (C) After solving the last problem, your students ask if methods similar to that used in question 2 can be used to find things like the distance stars are from the earth, the distance from the earth to the sun, the maximum distance planets are from the sun, and so on. You say, “Yes,” and offer them one more example, that of calculating the distance Venus is from the sun. You give them the following facts and picture: Venus’s orbit around the sun is almost per- fectly circular and the same is true for the earth’s orbit. So we can draw two concentric circles representing their orbits with the sun, S, as center. (See Figure 12.10.) S V E Figure 12.10 When the angle between the sun, earth, and Venus (angle SEV) is maximum, the angle between the sun, Venus, and earth, (angle SVE) is 90. The maximum angle SEV is about 46.3 degrees. Now, using this information calculate the approximate distance from Venus, (V), to the sun, (S), considering them as points. Note that the distance from the earth to the sun is approxi- mately 93,000,000 miles. Compare the answer you got with the latest results scientists have found. How close were you? 4* Find the components of the net force and the direction of the net force acting on a body (assumed to be at the origin) if the individual forces are: (a) F1 = (À2, 3), F2 = (À3, 4), and F3 = (7, 5). (b) F1 = (9, 6), F2 = (2,À5), and F3 = (6, 14).

574 Chapter 12 Trigonometry 5 Find the approximate net force on a body (assumed to be at the origin) if there are two forces, F1 and F2 acting on the body. Here we want the size of the net force as well as the angle it makes with the positive x-axis. (a)* F1 is a force of 30 pounds acting at an angle of 34 degrees with the horizontal and F2 is a force of 50 pounds acting at an angle of 80 degrees with the x-axis. (b) F1 is a force of 40 pounds acting at an angle of 10 degrees with the positive x-axis and F2 is a force of 30 pounds acting at an angle of 70 degrees with the positive x-axis. 6* A woman is standing on a hill 150 feet from a flagpole. She has with her a surveying tool that measures angles. She finds with this tool, that looking to the top of the flagpole, the angle is 45, while looking towards the bottom, the angle is 24. (See Figure 12.11.) What is the height of the flagpole? flagpole 45° 150 feet 24° ground Figure 12.11 7* Let us change the previous problem. Suppose, that instead of assuming that the woman is standing 150 feet from the flagpole, assume that the flagpole is 150 feet high. Approximately how far is the surveying tool from the flagpole? 8 In Figure 12.12 we see an object being suspended by two cables. If the weight of the object is 100 pounds, find the tension in the cables. (The tension is essentially how much force each cable is exerting to do its part to keep the weight from crashing down.) [Hint: Find the horizontal com- ponents of the tension in each cable. Since they must balance each other out, they must be equal. The sum of the vertical components of the forces must be equal to the weight.] 40° 60° T1 T2 100 Figure 12.12

12.3 Extending Notions of Trigonometric Functions 575 12.3 Extending Notions of Trigonometric Functions LAUNCH 1 Find, without the use of a calculator, the sine, cosine, and tangent of 0. Explain how you got your answers. 2 Find, without the use of a calculator, the sine, cosine, and tangent of 90. Explain how you got your answers. 3 Make conjectures about the values of the sine, cosine, and tangent of 270. Explain your reasoning. In middle school you most likely learned the mnemonic, SOHCAHTOA, which was a way of remembering how to compute the sine, cosine, and tangent of an angle. But, as you have seen from the launch question, this mnemonic is not very helpful for special angles of 0 and 90 or for angles that are larger than 90. After reading this section, you will learn how to expand the def- initions of the trigonometric relationships so that these cases will be easily dealt with. 12.3.1 Trigonometric Functions of Angles More Than 90 Degrees In the last section we spoke of how practical considerations led to the need to consider angles of more than 90 degrees, which we will now examine in more detail. Suppose that we have an angle θ. As we know, every angle has two sides, an initial side and a terminal side. The angle is always measured as a rotation from the initial side to the terminal side. (See Figure 12.13.) terminal side initial side q Figure 12.13 An angle is said to be positive if the direction from the initial side to the terminal side is counter- clockwise, and negative if the direction from the initial side to the terminal side is clockwise. (This is just a convention. Mathematicians could have done it the opposite way.) Thus, with the labeling in the figure, θ is a positive angle. An angle of 360 represents a complete counterclockwise revolution. Angles can be more than 360. An angle of 720 represents two complete revolutions, while an angle of 800 represents two complete counterclockwise revolutions and another 80. (See Figure 12.14.)

576 Chapter 12 Trigonometry 800 Figure 12.14 An angle of 800° (An angle of À800 represents the same angle, only the rotation is in the clockwise direction.) In what follows, we will assume that the initial side of any angle we discuss is along the positive x-axis. (In Figure 12.15 we see an angle, θ, whose initial side is the x-axis.) We pick an arbitrary point, (x, y), on the terminal side, and suppose that it is at a distance r from the origin, where r is greater than 0. y-axis P (x, y) r y x-axis q x Figure 12.15 Then the horizontal leg of the triangle is x, the vertical leg is y, and the hypotenuse is r, as indicated in Figure 12.15. Accordingly, we have cos y ¼ x ; r sin y ¼ y ; ð12:3Þ r tan y ¼ y : x It is this basic definition that leads us to analogous definitions of functions more than 90 degrees. For any angle, we define the sine, cosine, and tangent of θ as we did in display (12.3). Namely, we pick an arbitrary point (x, y) on the terminal side of the angle at a distance r from the origin and use the definitions given in display (12.3). It is important to note that x and y in the definitions given in display (12.3) do not always refer to sides of a triangle, rather to coordinates of points

12.3 Extending Notions of Trigonometric Functions 577 on the terminal side of the angle. (This will play a part when we discuss things such as sine of 90.) The reciprocals of the functions sin θ, cos θ, and tan θ are known as cosecant θ, secant θ, and cotan- gent θ (abbreviated csc θ, sec θ, and cot θ), respectively. These will be developed more in the Student Learning Opportunities, since most of you are familiar with these terms. The definitions given in display (12.3) for sine, cosine, and tangent immediately imply that it is possible for the sine or cosine of an angle to be negative. Specifically, in the first quadrant, accord- ing to display (12.3), all of the functions, sine, cosine, and tangent are positive, since x, y, are pos- itive in that quadrant and r is always positive. (Recall that the quadrants are numbered 1–4 counterclockwise, where quadrant 1 is where x > 0 and y > 0.) In the second quadrant, the sine is positive, since y and r are positive in that quadrant, while cosine and tangent are negative, since x is negative. In the third quadrant, the tangent is positive and the sine and cosine are neg- ative. In the fourth quadrant, the cosine is positive and the sine and tangent are negative. You should make sure you see why before you proceed. Figure 12.16 illustrates these facts: sine and its all trig functions reciprocal are are positive positive tangent and cosine and its its reciprocal reciprocal are are positive positive Figure 12.16 A way to remember the quadrants in which the trigonometric functions are positive is A-S-T-C. All Students Talk Constantly! Here is a typical secondary school problem. Example 12.5 The point P(À8, 15) is on the terminal side of an angle θ. Find the sine, cosine, and tangent of θ. Solution: We draw the picture first as shown in Figure 12.17. y (−8, 15) x Figure 12.17

578 Chapter 12 Trigonometry qTffiffihffiffiffieffiffiffiffixffiffiffifficffiffioffiffiffioffiffirffiffidffiffiffiinate of P is À8, the y coordinate is 15, and the distance this point is from the origin is ðÀ8Þ2 þ 152 ¼ 17. Thus, cos y ¼ À8 ; 17 sin y ¼ 15 ; 17 tan y ¼ 15 : À8 So, we do see that the sine is positive and the cosine and tangents are negative, as they should be in the second quadrant. An angle whose terminal side is not in the quadrant but on one of the axes is known as a qua- drantal angle. Thus, the quadrantal angles are 0, 90, 180, 270, 360, and so on. Example 12.6 Find, without the use of a calculator, the sine, cosine, and tangent of each of the fol- lowing quadrantal angles (a) 90 (b) 180 (c) 270. Solution: (a) The terminal side of a 90 degree angle is on the y-axis. Pick the point (0, 1) on the terminal side of θ = 90. Here x = 0, y = 1 and r = 1. So sin 90 ¼ y ¼ 1 ¼ 1; r 1 cos 90 ¼ x ¼ 0 ¼ 0; and r 1 tan 90 ¼ y ¼ 1 ; which is not defined: x 0 (b) Pick the point (À1, 0) on the terminal side of θ = 180 (which is along the negative x-axis). Then x = À1, y = 0, and r = 1. Thus, sin 180 ¼ y ¼ 0 ¼ 0; r 1 cos 180 ¼ x ¼ À1 ¼ À1; and r 1 tan 180 ¼ y ¼ 0 ¼ 0: x À1 (c) Pick the point (0,À1) on the terminal side of θ = 270. Then r = 1 and sin 270 ¼ y ¼ À1 ¼ À1; r 1 cos 270 ¼ x ¼ 0 ¼ 0; and r 1 tan 270 ¼ y ¼ À1 ; which is not defined: x 0 We have, in this example given a non-calculator solution to parts of the launch. Using Fig- ure 12.18,

12.3 Extending Notions of Trigonometric Functions 579 y-axis P (x, y) ry q x-axis x Figure 12.18 we see that x2 þ y2 ¼ r2: Dividing this by r2 we get x2 þ y2 ¼ 1: r2 r2 This can be rewritten as x2 þ y2 ¼ 1: rr Using the definitions of sin θ and cos θ from equation (12.3), this becomes sin2 y þ cos2 y ¼ 1: If we took a similar picture with the terminal side in another quadrant, we would have the exact same result. Thus, regardless of what θ is, sin2 θ + cos2 θ = 1. This you have undoubtedly seen and is called an identity. (You can see a visual representation of this identity by graphing the equation Y = sin2 θ + cos2 θ on your calculator. What do you see? Is this what you expected?) We will develop the other Pythagorean identities related to the reciprocal functions in the Student Learning Opportunities. 12.3.2 Some Useful Trigonometric Relationships In Figure 12.19 you see an angle of Àθ and θ for θ in the first quadrant and points (x, y) and (x,Ày) on the terminal sides of θ and Àθ. y (x, y) r q x −q (x, −y) Figure 12.19

580 Chapter 12 Trigonometry There is a similar picture for θ in the third quadrant, or any other quadrant. θ and Àθ are always reflections of each other about the x-axis. It follows from this immediately that cos ðÀyÞ ¼ cos ðyÞ ð12:4Þ ð12:5Þ  x  since r according to the diagram they are both and that sin ðÀyÞ ¼ Àsin ðyÞ since sin (Àθ) is Ày and sin θ is yr . r In Figure 12.20, we see both θ and 180 À θ. You will find a similar picture regardless of which quadrant θ is in. (You should try some numerical values of θ to convince yourself of this.) y (−x, y) (x, y) rr 180−q q x Figure 12.20 We notice right away when we express everything in terms of x, y, and r, that sinð180 À yÞ ¼ sin y ð12:6Þ  both are y  and that since r , cosð180 À yÞ ¼ Àcos y ð12:7Þ  one is x and the other is Àx  since r r . We will make good use of these facts soon, but what we want to point out now is how the proofs that we gave in Chapter 5 for the Laws of Sines and Cosines can be modified so that they are now valid even if θ is obtuse. You may recall that the proofs we gave in Chapter 5 required the triangles to be acute. (We had you do the obtuse case in Student Learning Opportunity 4 Section 2.1 Chapter 5 asking you to accept facts given in equations (12.6) and (12.7).) So, for the sake of completeness we will now give the proof of the Law of Cosines for obtuse triangles. We begin with the triangle ABC as shown in Figure 12.21 and draw altitude with length h to side AB extended. C a hb Dx q c B A Figure 12.21

12.3 Extending Notions of Trigonometric Functions 581 In right triangle BDC, we have BC2 ¼ BD2 þ DC2 or just a2 ¼ ðc þ xÞ2 þ h2: This last equation expands to a2 ¼ c2 þ 2 cx þ x2 þ h2: ð12:8Þ From right triangle ADC, we have x2 + h2 = b2. Substituting for x2 + h2 in equation (12.8) we get that a2 ¼ c2 þ 2 cx þ b2: ð12:9Þ From triangle ADC we have  ¼ x which tells us that x = b cos(∡CAD). Substituting cos ∡CADÞ b, this value for x into equation (12.9) we get a2 ¼ c2 þ 2 cb cosð∡CADÞ þ b2: ð12:10Þ But ∡CAD = 180 À θ and cos(∡CAD) = cos(180 À θ) = Àcos θ by equation (12.7). Replacing cos (∡CAD) by Àcos θ we get that a2 ¼ c2 À 2cb cos y þ b2: Rewriting this last equation we get a2 ¼ b2 þ c2 À 2cb cos A; which is the usual form of the Law of Cosines. So, what does this tell us? In any type of triangle, if we know two sides and one included angle, we can find the third side. Or, if we know three sides, we can find all of the angles. Also, notice how similar this formula is to the Pythagorean Theorem. The size of a2 increases or decreases based on the size of angle A. Notice if angle A is 90, cos A = 0 and we get the Pythagorean Theorem. Since we now have meanings for sines and cosines of angles more than 90 it is natural to ask if the Law of Sines, presented earlier in Chapter 5, is also valid when one of the angles of the triangle is more than 90. The answer is “Yes,” and we give a partial proof here when angle A is obtuse. We know by equation (12.6) that sin(A) = sin(180ÀA). Thus, a ¼ a À AÞ : ð12:11Þ sinðAÞ sinð180 Using Figure 12.22, C a hb 180−A A B Figure 12.22

582 Chapter 12 Trigonometry we see that sinð180 À AÞ ¼ h : Thus, equation (12.11) becomes b a ¼ a À AÞ ¼ a ¼ ab : ð12:12Þ sinðAÞ sinð180 h h ð12:13Þ a Similarly, from the picture, sin B ¼ h : Thus, a b B ¼ b ¼ ab : sin h h a Since by equations (12.12) and (12.13) a and b are both ahb, we see that sinðAÞ sin B a ¼ b B : sinðAÞ sin You should finish the proof and show that b ¼ c C, so that sin B sin a ¼ b B ¼ c C : sinðAÞ sin sin The Law of Sines is more important than you might think. Other than the logical applications to architecture, construction, and astronomy, where triangles abound, it is a key tool in surveying. It is often used to estimate distances over hilly terrain and was used in the Great Survey of India, which took about 100 years to complete. It uses a method called triangulation, where only one distance is measured to start with and then the region you are interested in surveying is broken into adjacent triangles. Using an instrument called a theodolite, which measures angles, all the other distances can be calculated from that one distance by using the Laws of Sines and/or Cosines repeatedly. We illustrate that with a scaled down version of this type of problem. Example 12.7 In Figure 12.23, B 120° D C 130° 10° 15° A Figure 12.23 we see a schematic of a certain part of an area consisting of rocks and difficult to navigate terrain. The distance AD has been measured to be 500 feet. Using the theodolite, all other angles have been deter- mined and are, as shown in the picture. Find all the remaining lengths. Solution: We won’t solve the whole problem. We will only give you an outline of the solution process. Since we know AD, we can use the Law of Sines in triangle ABD to find both AB and BD. Similarly, we can find AC and DC using the Law of Sines in triangle ADC. Now that we know BD and DC, and angle BDC (which is 110) we can find BC using the Law of Cosines. It is this kind of process that is done on a grand scale when surveying land.

12.3 Extending Notions of Trigonometric Functions 583 The proof that we gave in Chapter 5 for sin(A + B) using Ptolemy’s Theorem was given for acute angles A and B. (See Example 5.20.) The proof can be modified with equations (12.6) and (12.7) to prove this theorem in the case when A or B is obtuse. Are the usual rules for sin(A + B) and cos(A + B) true even if the angles are more than 180 degrees? The answer is “Yes” to both questions and we turn to a proof of that now. The proof is somewhat algebraic but the speed with which we get the result in all cases makes the work worth- while. In many ways, this is an elegant proof. Theorem 12.8 For any angles θ1 and θ2, cos(θ1 + θ2) = cos θ1 cos θ2 À sin θ1 sin θ2. Proof. The beauty of this proof is how it ties the notion of distance from Cartesian coordinates to trigonometry. The proof is somewhat surprising. Since when dealing with trigonometric functions of any angle we are able to take any point on the terminal side, we take all our points at a distance r from the origin, which means they are all on a circle with radius r. Place angles θ1, θ2, and Àθ2 as shown in Figure 12.24. (For clarification, θ2 is the angle from OP to OQ where O is the origin and Àθ2 is the angle of the same size as θ2 drawn from the positive x-axis, but drawn clockwise.) x QP r Rx q2 q1 −q 2 S Figure 12.24 We list the coordinates of P, R, Q, and S. P ¼ ðr cos y1; r sin y1Þ R ¼ ðr; 0Þ Q ¼ ðr cosðy1 þ y2Þ; r sinðy1 þ y2ÞÞ S ¼ ðr cosðÀy2Þ; r sinðÀy2ÞÞ ¼ ðr cosðy2Þ; Àr sinðy2ÞÞ These follow immediately from equation (12.3) since x = r cos θ and y = r sin θ always. Notice when finding the coordinates of S, we also used equations (12.4) and (12.5). Now the arc RQ is intercepted by a central angle of θ1 + θ2, as is arc PS. (For arc PS, although the angle is Àθ2, its size is θ2. The negative sign simply tells us the direction from the initial side to the terminal side.) Thus, the chords RQ and PS have the same length, from the well-known result in geometry which says that in a circle, equal arcs have equal chords. Since RQ = PS we have, using the distance formula ð12:14Þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi |fflfflfflðfflfflrfflfflfflcfflfflofflfflfflsfflfflðfflfflyfflffl1fflfflfflþfflfflfflfflfflyfflffl2fflfflÞfflfflfflÀfflfflfflfflfflrfflfflÞffl2{zþfflfflfflfflðfflrfflfflfflsfflfflifflnfflfflfflfflðfflyfflfflffl1fflfflfflþfflfflfflfflfflyfflffl2fflfflÞfflfflfflÀfflfflfflfflffl0fflfflÞffl }2 ¼ |fflfflfflðfflfflrfflfflfflcfflfflofflfflfflsfflfflfflyfflffl2fflfflfflÀfflfflfflfflfflrfflfflcfflfflfflofflfflfflsfflfflyfflfflffl1fflfflÞffl2fflfflfflþ{zðfflfflðfflfflÀfflfflfflfflrfflfflfflsfflifflfflnfflfflfflfflyfflffl2fflfflÞfflfflfflÀfflfflfflfflfflrfflfflfflsfflfflifflnfflfflfflfflyfflffl1fflfflÞffl}2 PQ RS The rest is algebra.

584 Chapter 12 Trigonometry Square both sides in equation (12.14) to eliminate the square roots and then expand what is left to get r2 cos2ðy1 þ y2Þ À 2r2 cosðy1 þ y2Þ þ r2 þ r2 sin2ðy1 þ y2Þ ¼ r2 cos2 y2 À 2r2 cos y1 cos y2 þ r2 cos2 y1 þ r2 sin2 y2 þ 2r2 sin y1 sin y2 þ r2 sin2 y1: Divide this last equation by r2 to get cos2ðy1 þ y2Þ À 2 cosðy1 þ y2Þ þ 1 þ sin2ðy1 þ y2Þ ð12:15Þ ¼ cos2 y2 À 2 cos y1 cos y2 þ cos2 y1 þ sin2 y2 þ 2 sin y1 sin y2 þ sin2 y1: Now on the right side of equation (12.15) we have cos2 θ2 + sin2 θ2 and cos2 θ1 + sin2 θ1 both of which are 1, and on the left side we have cos2(θ1 + θ2) + sin2(θ1 + θ2) which is also 1. So equation (12.15) simplifies to 2 À 2 cosðy1 þ y2Þ ¼ 2 À 2 cos y1 cos y2 þ 2 sin y1 sin y2: ð12:16Þ Subtracting 2 from both sides of equation (12.16) and dividing by À2 we get cosðy1 þ y2Þ ¼ cos y1 cos y2 À sin y1 sin y2 and we are done. How nice! & Corollary 12.9 For any angles θ1 and θ2, cos(θ1 À θ2) = cos θ1 cos θ2 + sin θ1 sin θ2. Proof. Since the formula cos(θ1 + θ2) = cos θ1 cos θ2 À sin θ1 sin θ2 is true for all angles θ1 and θ2 we can replace θ2 by Àθ2 to get cosðy1 À y2Þ ¼ cos y1 cosðÀy2Þ À sin y1 sinðÀy2Þ: Now we use equations (12.4) and (12.5) to immediately get that cosðy1 À y2Þ ¼ cos y1 cos y2 þ sin y1 sin y2: & Corollary 12.10 cos(90 À θ) = sin θ regardless of what θ is. (θ is in degrees.) Proof. cos(90 À θ) = cos 90 cos(θ) + sin90 sin(θ) = 0 cos(θ) + 1 sin(θ) = sin θ. & Corollary 12.11 sin(90 À θ) = cos θ regardless of what θ is. (θ is in degrees.) Proof. From the previous corollary which is true for all angles θ, we get, replacing θ by 90 À θ: cosð90 À ð90 À yÞÞ ¼ sinð90 À yÞ or just, cosðyÞ ¼ sinð90 À yÞ which is what we were trying to prove. &

12.3 Extending Notions of Trigonometric Functions 585 Student Learning Opportunities 1 (C) Solve part (a) of the launch problem. 2 (C) Your students want to know why a rotation in the counterclockwise direction is called pos- itive, rather than negative, which would make more sense to them. According to the text, what is the answer? 3 Find the approximate net force on a body (assumed to be at the origin) if there are two forces, F1, and F2 acting on the body. Here we want the size of the net force as well as the angle it makes with the positive x-axis. (a) F1 a force of 30 pounds acting at an angle of 34 degrees with the positive x-axis. F2 is a force of 50 pounds acting at an angle of 150 degrees with the positive x-axis. (b) F1 a force of 40 pounds acting at an angle of 10 degrees with the positive x-axis and F2 is a force of 30 pounds acting at an angle of 110degrees with the positive x-axis. 4 (C) Your students are confused about how to find the sine, cosine, and tangent of À90 without the use of a calculator. What do you tell them? 5 Following the convention in the book, we define the secant of θ to be xr , the cosecant of θ to be r x y and the cotangent of θ to be y The secant, cosecant, and cotangent of θ are abbreviated as sec θ, csc θ, and cot θ respectively. (a)* Find, without a calculator: sec 90, csc 90, cot 90. (b) Find, without a calculator: sec 180, csc 180, cot 180. (c) Find, WITH a calculator: sec 70, csc 70, cot 70. 6 Prove that for any angle where cos θ is not zero, tan y ¼ sin y : cos y 7 Is it always true that cot y ¼ 1 y ? tan 8 Prove that for any angle θ for which tan θ and sec θ are defined, that 1 + tan2 θ = sec2 θ. 9 Prove that for any angle θ for which cot θ and csc θ are defined, that 1 + cot2 θ = csc2 θ. 10 (C) (a) Your students make the continual error of saying that sin(A + B) = sin A + sin B. They are thinking “Distributive Law.” How can you convince them that there is no distributive law for trigonometric functions? (b) How can you prove by using corollaries (12.10) and (12.11), that sin(A + B) = sin A cos B + cos A sin B? [Hint: Write sin(A + B) = cos(90 À (A + B)) = cos((90 À A) À B). Now use Theorem 12.8.] (c) Derive a formula for sin(A À B) using your answer from part (b). 11* (C) A very common error that students make is to say that sin 2A = 2 sin A. Show, by example or by graphing y = sin 2A and y = 2 sin A on the same set of axes, that this is not true. Then show how you would guide your students in using the formula for sin(A + B) to help them see that sin 2A = 2 sin A cos A. Finally, find specific angles when sin 2A does equal 2 sin A. pffiffiffi pffiffiffi 1 2 23, 12* Using the known values of sin 30 ¼ 2 and sin 45 ¼ cos 45 ¼ 2 ; cos 30 ¼ find the exact values of sin 75 and sin 15.

586 Chapter 12 Trigonometry 13 (C) Your students use the following formulas in secondary school when proving identities and also in calculus when finding certain trigonometric integrals: cos2y ¼ cos 2 y À sin 2 y ¼ 2cos 2 y À 1 ¼ 1 À 2sin 2 y: They need help in trying to prove them. How can you help them understand how to go about proving these identities? 14 Using the identities from the previous exercise, show that sin2 y ¼ 1 À cos2y and that cos2 y ¼ 2 1 þ cos2y : (These identities are used to evaluate certain trigonometric integrals in calculus.) 2 15 A nice proof of the Law of Cosines can be given using coordinate geometry. In any triangle c2 = a2 + b2 À 2ab cos C. Prove this using the following hint: We start with a triangle and place it on a coordinate plane as shown in Figure 12.25 y A(b cos C, b sin C ) x B (a, 0) c b C (0, 0) a Figure 12.25 and call the coordinates of A, (x, y). Since cos C ¼ x ; x ¼ r cos C, which equals b cos C since r = b. Similarly since sin C ¼ y ; y ¼ r b b sin C. Now the distance from A to B is c, and this can be computed by the distance formula, qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi c ¼ ðb cos C À aÞ2 þ ðb sin CÞ2: Finish it. 16* A surveyor sees a building across the river. Standing at point A he measures the angle of ele- vation from the ground to the top of the building to be 30 degrees. He steps back 100 feet and again measures the angle of elevation and finds it to be 15. (See Figure 12.26.) Assuming that it makes a 90-degree angle with the floor, approximately how tall is the building? building B 15° 30° 100 A river Figure 12.26

12.3 Extending Notions of Trigonometric Functions 587 17 The method used in the previous problem can be generalized. Suppose we wish to find the height, h, of some object. We take two points A and B on the ground as shown in Fig- ure 12.27 and measure the distance between them. h B A d Figure 12.27 (a) Show that h ¼ cot B d cot A : ð12:17Þ À ð12:18Þ (b) Show that h can also be calculated by h ¼ d sin A sin B : sinðA À BÞ (c) Setting equations (12.17) and (12.18) equal to each other and dividing both sides by d we get that cot B 1 cot A ¼ sin A sin B : À sinðA À BÞ Assuming that none of the denominators in this display are zero, does this equation rep- resent an identity in general? Why or why not? 18* A cruise ship starts out from point P and travels 30 miles at a bearing of 50 degrees northwest to a point Q. It then turns and travels a distance of 48 miles, at a bearing of 20 southwest to a point R as shown in Figure 12.28. How far is it from its original debarkation point? 48 mi Q 30 mi 20° 50° P R Figure 12.28

588 Chapter 12 Trigonometry 19 (C) In your calculus class your students have just learned the series for sin x and cos x. They are: sin x ¼ x À x3 þ x5 À x7 þ . . . 3! 5! 7! and cos x ¼ 1 À x2 þ x4 À x6 þ . . . 2! 4! 6! One of your students notices that these relationships can be used to explain why cos(Àx) = cos (x) and why sin(Àx) = Àsin x for all angles x. Without reference to any pictures, explain how. 12.4 Radian Measure 12.4.1 Conversion While degree measure seems to be the most natural way to measure angles, when calculus came on the scene, a different kind of measure was needed that would be more useful. So mathematicians created radian measure. Suppose that we have a circle of radius r and we mark off a distance of r along the circumference of the circle, the central angle formed is called an angle of 1 radian. (See Figure 12.29.) Since the circumference of a circle is 2πr % 6.28r, we can mark off the radius of the circle 2π or r 1 radian r r Figure 12.29 approximately 6.28 times. Since the central angle formed by marking off the radius 2π times along the circumference is 360 degrees, we have that 2p radians ¼ 360 degrees: ð12:19Þ Dividing both sides of this equation by 2π, we get that 1 radian ¼ 180 degrees % 57:296 degrees: ð12:20Þ p Dividing both sides of equation (12.19) by 360, we get that p radians ¼ 1 degree: ð12:21Þ 180 Equations (12.19) and (12.21) are the basis of the conversion formulas that one uses in high school. p Since 1 degree equals p radians, the number of degrees in θ radians is 180 y. Thus, to convert from 180 degrees to radians, we multiply the number of degrees by p . To illustrate: to convert 90 to 180

12.4 Radian Measure 589 radians, we multiply by p to get 180 90 ¼ 90  p radians ¼ p radians: 180 2 Similarly, since by equation (12.20) 1 radian is 180 degrees, θ radians is 180 y degrees. That is, to p p convert from radians to degrees, we multiply the number of radians by 180 . To illustrate: p p radians ¼ p  180 degrees ¼ 45: 4 4 p One advantage of radian measure is that it makes certain formulas in calculus much simpler. For example, if we take the derivative, y0, of y = sin x and x is measured in radians, as is always done in calculus, then y0 = cos x. But if we were to measure x in degrees, then the derivative of sin x would not be cos x, rather it would be p cos x, and a similar statement can be made for the other trig- 180 p onometric functions. This extra factor of 180 would really be annoying to work with, especially when taking higher order derivatives. But in radian measure, this factor does not appear, giving radian measure a big advantage over degree measure. There is another more technical reason why radian measure is useful in the sciences. Radian measure is “dimensionless,” which allows sci- entists to divorce it completely from angular measure and apply it to situations not relating to angles, as we shall see. The following conversions occur so often in secondary school that it is good to know them. Number of degrees Number of radians 0 30 0 45 p 60 6 90 p 4 p 3 p 2 We will use both degrees and radians in our discussion depending on what seems easiest in a particular context. 12.4.2 Areas and Arc Length in Terms of Radians How can we represent area and lengths of arcs on a circle using radian measure? We know that the area of a circle is πr2, where r is the radius of the circle. Therefore, if we have a sector of a circle with central angle k as shown in Figure 12.30, A k° B Figure 12.30

590 Chapter 12 Trigonometry then the area of that sector is pr2 k and the length of the arc okft,haendsetchtoerfaisct2oprr36kp0. We observe that the area of the sector can 360   be rewritten as pr2 k ¼ 1 r2 p k is simply 360 2 180 180 θ, the radian measure of the angle k as we observe from the conversion formulas. Substituting into the preceding expression, we get that the area of a sector is A sector ¼ 1 r2 y 2 where θ is the radian measure of an angle. In a similar manner, the length of the arc of the sector, Á Á2 prk ¼ r p k ¼ ry where θ is the radian measure of the angle. Thus, if Lsector represents the 360 180 length of the arc of a sector, we have L sector ¼ ry: ð12:22Þ Let us illustrate these formulas with some examples. The first is one that is often given in sec- ondary school as a “fun type” of problem. Example 12.12 A dog located at point D is attached to a rope 25 feet long which is attached to the corner of an enclosed rectangular plot of grass, with dimensions 10 feet by 30 feet as shown in Figure 12.31. D 10 ft 25 ft 30 ft Figure 12.31 The dog cannot get inside the plot. If the rope is taut, find the area within which the dog can roam. Solution: As the dog roams, it can trace out the area shown in Figure 12.32. 25 ft 15 ft 10 ft 30 ft Figure 12.32

12.4 Radian Measure 591 This area consists of 3 of a circle with radius 25 feet and 1 of a circle with radius 15 feet. The 4 4 central angle associated with the 3 part of the circle is 270 or 3p radians, and with the 1 part of a 4 2 4 circle is 90 or p radians. Using the formula that the area of a sector is given by 1 r2 y, we find that 2 2 the area the dog has to roam is given by:  1 ð25Þ2 3 p þ 1 ð15Þ2 p 2 22 2 % 1649:3 square feet: Example 12.13 Every car has a system of belts and pulleys that makes such things as the air condi- tioner and fan belt run. Consider Figure 12.33, where we have a system of two pulleys and a belt wrapped around them tightly. Assuming the angle where the belt crosses itself is p radians, and the 6 radii of the pulleys are 3 and 5 inches, find the approximate length of the belt. 5 3 p 6 Figure 12.33 p Solution: Draw a line joining the centers of the pulleys which divides the angle into two equal 6 parts of length p . See FG in Figure 12.34: 12 Ap 12 C 5 B3 F G E D Figure 12.34 Now, using triangle ABF, we have that tan1p2 ¼ AF ¼ 5 : AB AB Solving for AB, we get that AB % 5À Á % 18:66 inches. In a similar manner, using triangle BEG tan p 12 we get that tan1p2 ¼ GE ¼ 3 : BE BE

592 Chapter 12 Trigonometry Thus, BE ¼ 3À Á % 11: 196. Since AE = AB + BE, we have that tan p 12 AE % 29: 856 inches: Similarly, CD % 29: 856 inches: To finish finding the length of the belt, we need to find the length of the (larger) arc AD and the (larger) arc EC. Now ∡AFB ¼ 2pÀÀ1∡10A2pBF ¼ 2p11À421p2p.¼T51h2pe and the same is true for angle BFD. Thus, the larger angle ∡AFD ¼ 2 p ¼ same is true for the larger angle CGE. So the length of (the larger)  _ 14 p 12 AD ¼ ry ¼ 5 by equation (12.22), and the length of (the larger)  _ 14 p : EC ¼ 3 12 The length of the belt is therefore _ _ AE þ ðlargerÞ EC þ CD þ ðlargerÞ AD   14 p 14 p % 29:856 þ 3 12 þ 29:856 þ 5 12 % 89:034 inches: Another interesting application of these ideas is mapmaking. Let us quickly review latitude and longitude. The earth, considered as a sphere, is broken into circles parallel to the equator called lat- itude lines. Similarly, all circles on the earth’s surface that pass through both the north and south pole are called longitude lines or meridians. The equator is given a latitude of 0. Any longitude line may be taken as a longitude of 0. But the longitude line that passes through Greenwich, England, by inter- national agreement, is the line taken as the one with longitude 0, and that line is called the prime meridian. The intersection of the prime meridian and the equator is labeled O in Figure 12.35. Each point, P, on the earth’s surface is given a latitude and longitude. These are both angular mea- surements. The longitude measures how many degrees from the prime meridian we have to turn to get to the meridian that P is on (angle OCD in Figure 12.35), while the latitude is how many degrees up or down we must travel to get to the latitude line P is on (angle PCD in Figure 12.35). Greenwich Greenwich P C P C OD OD Angle OCD is the longitude of P Angle DCP is the latitude of P Figure 12.35

12.4 Radian Measure 593 Suppose we are interested in making a map of part of the surface of the earth. There are many ways to do this, each with its advantages. One way to do this is with a method called cylindrical projection. In this method, the earth is considered to be enclosed by a cylinder (Figure 12.36). y y P has P = (L, M ) P P' latitude β Greenwich β M x Cα 0 latitude D OL 0 longitude Figure 12.36 Now each point P on the part of the earth that we want to map is projected onto the cylinder in the following way. We draw a line from the center, C, of the earth to the point P on the earth’s surface and then continue the line until it hits the cylinder at P0 shown in Figure 12.36. Once we do this, we unroll our cylinder as if it is a sheet of paper and we have our map. The equator we take as our x-axis, and the y-axis is the prime meridian through Greenwich England, 0. The question is, how do we get the coordinates (x, y) of any point P0 on the earth’s surface on this planar representation of the map? The answer is, we use trigonometry. If the point P has longitude α and latitude β, then the x coordinate of P0 is the length of the arc L shown in the picture when the cylinder is unwrapped. Furthermore, the y coordinate is the height M shown in the picture. We know that L ¼ 2 pR Á a , where r is the radius of the earth and thus, this is 360 our x coordinate. Again, using trigonometry, we find that M = y = R tan β. Doing this for various points on the surface of the earth, we get our map. Of course, the question arises, how do we wrap the cylinder around the earth? After all, the earth is pretty big, and furthermore the cylinder when unrolled will be quite big. True. That is why our map is a scaled down version of the real thing! Student Learning Opportunities 1* (C) Your students want to know how to draw an angle that measures 1 radian and approxi- mately how many degrees 1 radian represents. What is the answer? 2 (C) Your students know that π is approximately equal to 3.14. Now they are learning that π radians = 180 and this confuses them. What is the source of their confusion concerning the value of π and how can you resolve it?

594 Chapter 12 Trigonometry 3 Convert each of the following from degrees to radians: (a)* 210 (b) À300 (c) 450 4* Convert each of the following from radians to degrees: (a) 7p 6 (b) À 3p 4 (c) 2p 5 5 A pulley system consists of two pulleys having radii R and r. There is a belt tightly placed around the system and the angle where the belt meets itself is θ radians as shown in Figure 12.37. A R B C F q r G E D Figure 12.37 Show that the length of the belt is given by  y L ¼ ðR þ r Þ y þ p þ 2cot 2 : 6* A cow is attached to an L-shaped building surrounded by grass as shown. If the rope attached to the cow’s neck is 60 feet long and the dimensions of the building are all in feet, how much area can the cow graze? The lengths given are the lengths of the solid lines, AB, BC, CD, and DE in Figure 12.38. Cow • A 20 ft B 30 ft 60 ft C D 40 ft 50 ft FE Figure 12.38

12.5 Graphing Trigonometric Curves 595 12.5 Graphing Trigonometric Curves LAUNCH In the following questions, use the graphing calculator to draw the graphs of the given functions. Then sketch these graphs on your graphing paper. 1 Graph the functions: y = sin x, y = 2 sin x, y = 3 sin x. If you graph the equation y = A sin x, how does changing A affect the graph of y = sin x? 2 How are the graphs of y = sin 2x, and y = sin 3x different from the graph of y = sin x? Predict how y = sin Bx will look where B is a positive integer. 3 Graph the functions: y = sin x, y = 3 + sin x, y = À1 + sin x. If you graph the equation y = D + sin x, how does changing D affect the graph of y = sin x?  4p.  À 2p; y sin x 4 Graph the functions: y = sin x, y ¼ sin x ¼ sinðx À pÞ, and y ¼ þ If you graph the equation y = sin(x À C), how does changing C affect the graph of y = sin x? 5 Sketch the graph of y = 2 + 4 sin(x + π). Now graph it on your graphing calculator. Was your sketch accurate? 6 Describe what you have learned about the effects of changing A, B, C, and D on the graph of y = A sin B(x À C) + D. After having done the launch questions, your memory has probably been refreshed on what you learned in secondary school about the graphs of trigonometric functions. By reading this section, you will learn more details about these graphs. 12.5.1 The Graphs of Sin θ and Cos θ Thus far, we have examined the graphs of exponential and polynomial functions. It is only natural to wonder what the graph of a trigonometric function looks like. Let us proceed to find out. We already mentioned that, when finding the sine of an angle, θ, we are allowed to pick any point on the terminal side of θ. We will always get the same answer for sin θ. That follows by similar triangles, as Figure 12.39 shows: y-axis P’ (x,y) P (x, y) y1 ry x-axis qx x1 Figure 12.39

596 Chapter 12 Trigonometry So, if we can take any point on the terminal side of an angle, why not take a point whose dis- tance from the origin is 1? (See Figure 12.40.) y -axis P (x, y) r =1 y qx x-axis Figure 12.40 Since sin y ¼ y ¼ y 1 and cos y ¼ x ¼ x; 1 we see that, when we choose a point on the terminal side of the angle drawn in a circle whose radius is 1, the y coordinate represents sin θ and the x coordinate is cos θ. That leads us to consider taking points on the circle whose center is at the origin and whose radius is 1. (See Figure 12.41.) y-axis P 1 q y = sin q x-axis x = cos q Figure 12.41 Since the y-coordinate of P is sin θ, by looking at the y coordinate of P as θ varies, we get a picture of how sin θ varies with θ and if we look at the x coordinate of P, we see how cos θ varies. Let us first concentrate on sin θ. We notice that, as θ goes from 0 to 90, the y coordinate, which is sin θ, varies from 0 to 1. As θ varies from 90 to 180, y = sin θ decreases from 1 to 0. As θ varies from 180 to 270, y = sin θ decreases from 0 to À1 and as θ varies from 270 to 360, y = sin θ increases from À1 to 0. In Figure 12.42 you see a graph of how sin θ varies as θ varies from 0 to

12.5 Graphing Trigonometric Curves 597 360.We have traced out what is called one complete cycle of the sine curve. The period is the number of degrees it takes to complete one cycle. Thus, the period of the function y = sin θ is 360 degrees or 2π radians. As you already know, the sine curve looks like a wave and repeats once the angles go beyond 360. 1 50 75 100 125 150 175 200 225 250 275 300 325 350 0.8 theta degrees 0.6 0.4 0.2 0 0 25 −0.2 −0.4 −0.6 −0.8 −1 Figure 12.42 The graph of y = sin θ, 0° θ 360°. In the Student Learning Opportunities you will use a similar process to find the graph of the cosine curve, which we show in Figure 12.43. 1 0.8 0.6 0.4 0.2 0 0 25 50 75 100 125 150 175 200 225 250 275 300 325 350 −0.2 theta degrees −0.4 −0.6 −0.8 −1 Figure 12.43 The graph of y = cos θ, 0° θ 360°.

598 Chapter 12 Trigonometry In Figures 12.44 and 12.45, we see the graphs of sin θ and cos θ with the θ-axis marked off in radians. y 0 p/2 p 3p /2 2p radians Figure 12.44 The graph of y = sin θ, 0 θ 2π. y 0 p/2 p 3p /2 2p q Figure 12.45 The graph of y = cos θ, 0 θ 2π. Whether you choose to graph the function in degree or radians makes no difference. Any function that repeats over and over is called periodic. Thus, f(θ) = sin θ and g(θ) = cos θ are periodic, but so is the function shown in Figure 12.46, where the graph continues forever in both directions. y x Figure 12.46 It is interesting to note that any periodic behavior, even the one just illustrated, can be studied using sines and cosines. If you are familiar with Fourier Series, you will understand what we mean. However, this is beyond the scope of this book. Sticking to secondary school material, we will focus only on the periodic functions of sine, cosine, tangent, and their reciprocals. When we graph y = 5 sin θ, it looks just like the graph of y = sin θ, only instead of reaching a maximum of 1 and a minimum of À1, it reaches a maximum of 5 and a minimum of À5. We see that its period remains 360. The graph of y = À5 sin θ is just a reflection of the graph of y = 5 sin θ about the θ-axis. Its graph is shown in Figure 12.47.

12.5 Graphing Trigonometric Curves 599 y y = −5 sin q 5 4 90 180 270 360 q 3 2 1 0 −1 −2 −3 −4 −5 Figure 12.47 When we graph the function y = sin 2θ, it is essentially traced out twice as fast as y = sin θ. So, in 360 degrees, you would have two cycles and the period of the graph would be 180. The frequency is the number of cycles traced out in 360 degrees. Thus, y = sin 2θ has a frequency of 2. In Fig- ure 12.48 we see the graph of y = sin 2θ for 0 θ 360. Observe that there are two full sine cycles in 360 degrees. y y = sin 2q 1 0.8 90 270 360 0.6 q 0.4 180 0.2 00 −0.2 −0.4 −0.6 −0.8 −1 Figure 12.48 Note: In general, when we graph functions, we use x to label the horizontal axis, and y to label the vertical axis, though we have been labeling the horizontal axis as the θ-axis. From here on in, we will suppress θ and use x when it doesn’t cause confusion. In general, the graph of y = A sin Bx, where B is positive, has a period of 360 degrees (or 2p radians B B if we are measuring in radians). It reaches a maximum height of |A| and a minimum height of À|A|. The graph of y = D + A sin Bx raises or lowers the graph of y = A sin Bx by D, that is, translates it up or down. If D is positive, the graph is translated up, while if it is negative, it is translated down. Thus, the graph of y = 5 + 3 sin x raises the graph of y = 3 sin x by 5, while y = À3 + sin x lowers the graph of y = sin x by 3. A vertical translation of the graph does not affect its period. The graph of y = sin 4(x À 16) is the graph of y = sin 4x translated to the right 16 degrees as one can see by making a table of values for x and y. So one of its cycles begins at the point (16, 0). The graph of y = sin 4(x + 16) is the graph of y = sin 4x translated to the left by 16 degrees. So one of its cycles begins at (À16, 0). You can check these statements by graphing all three functions on your calculator.

600 Chapter 12 Trigonometry We summarize our observations in the theorem. Theorem 12.14 Suppose that C is positive. The graph of y = D + A sin B(x À C) is the graph of y = A sin Bx translated vertically a distance D and horizontally a distance C to the right. The graph of y = D + A sin B(x + C) is the graph of y = A sin Bx translated vertically a distance D and horizontally a distance C to the left. The period of all of these graphs is 360 degrees. If B(x À C) is measured in radians, then the B period of all of these graphs is 2Bp. The number C is called the phase shift. Here is an illustration. Example 12.15 Graph the function y = 4 + 5 sin 3x for x between 0 and 360 degrees. Do it without a calculator. Solution: If we can graph y = 5 sin 3x, then we can graph y = 4 + 5 sin 3x. The latter graph will be the former graph raised by 4 units. So we turn to graphing y = 5 sin 3x. Its period is 360 ¼ 360 or 120 B 3 degrees. Thus, one complete cycle will cover 120 degrees. Since we want the graph between 0 and 360, there will be three full cycles. The maximum height of y = 5 sin 3x is 5 and the minimum height is À5. Translating this graph vertically by 4 units, our graph looks like that in Figure 12.49. y 9 4.0 −1.0 0 120 240 x Figure 12.49 360 The maximum height is 9 and the minimum height is À1. With the power of today’s graphing calculators, you may be wondering about the value of knowing how to graph these equations by hand. Our students always question this. But experience shows that dependence on the calculator is dangerous. For example, the future teacher can see this by asking a class to graph y = x + 100 on the calculator. Since the graph is not visible in the standard window, you will hear some students who do not know that the graph is a line and who graphed it in the standard window claim that “There is no graph.” Of course, we know there is one. With trig- onometric functions, graphing equations on the calculator can lead to even stranger results. We show what can happen when you graph the same trigonometric function in different windows. The results are striking.

12.5 Graphing Trigonometric Curves 601 Example 12.16 (a) Graph y = sin 30x in each of the following windows: (a) [À5, 5] × [À1, 1] (b) [À0.01, 0.01] × [À1, 1] (c) [0, 0.107] × [À1, 1] (d) [À16, 16] × [À1, 1] (e) [À7, 7] × [À1, 1] (f) [À20, 20] × [À1, 1]. (x is measured in radians.) Solution: Here are the 6 graphs. (a) y1 0.5 0 −5 −2.5 0 2.5 5 x −0.5 −1 Figure 12.50 (b) y 0.25 0.125 −0.01 −0.005 0 0.005 0.01 0 x −0.125 −0.25 Figure 12.51 (c) y1 0.75 0.5 0.25 0.025 0.05 0.075 0.1 x 0 Figure 12.52 0

602 Chapter 12 Trigonometry (d) y1 0.5 0 −15 −10 −5 0 5 10 15 x −0.5 −1 Figure 12.53 (e) y1 0.5 0 −5 −2.5 0 2.5 5 x −0.5 −1 Figure 12.54 (f) y1 0.5 0 −20 −10 0 10 20 x −0.5 −1 Figure 12.55 We hope we made our point. Technology is a wonderful tool but it cannot “see” like we can or “think” like we can. We need to have knowledge and understanding to know when something is wrong. That is why knowledge of the basic sine, cosine, and tangent graphs is important!

12.5 Graphing Trigonometric Curves 603 12.5.2 The Graph of y = Tan θ In Section 11.3 we defined sin θ to be y and cos θ to be xr , where (x, y) is any point on the terminal r side of θ, and r > 0 is the distance of that point to the origin. If we divide the expression for sin θ by cos θ, assuming that cos θ is not zero, we get that y sin y ¼ r ¼ y : cos y x x r Since y is tan θ, we see that x tan y ¼ sin y when cos y 6¼ 0: cos y Thus, from the values of sin θ and cos θ, we can get the values of tan θ and we can easily draw the graph of tan θ. Since sin 0 = 0, and cos 0 = 1, we immediately see that tan 0 ¼ 0 ¼ 0. As θ goes 1 from 0 to 90, sin θ increases to 1, and cos θ decreases to 0. When the numerator of a fraction increases while the denominator decreases, the fraction increases. Thus, the quotient of them, tan θ increases. Since sin θ is approaching 1 and cos θ is approaching 0, as θ approaches 90, tan θ, their quotient, gets larger and larger and goes to 1. Tan 90 does not exist, as you can easily show. Thus, as x approaches 90, the graph of tan θ approaches 1, and x = 90 is a vertical asymptote. Similarly, as x approaches À90, y = tan θ approaches À1. Thus, the graph of tan θ for À90 < θ < 90 appears as shown in Figure 12.56. y 50 25 0q −25 −50 −90° 90° Figure 12.56 Once we pass 90 and move to 270, we find by calculation that the graph repeats. In fact, the graph repeats every 180 as shown. Thus, the graph has a period θ of 180 or π radians. The graph of y = tan θ is shown in Figure 12.57.

604 Chapter 12 Trigonometry y q 50 25 0 −25 −50 −270° −90° 90° 270° Figure 12.57 Student Learning Opportunities 1 Write the equations for each of the following graphs (Figures 12.58–12.60). (a)* y 4 3 2 1 p x −1 2p −2 Figure 12.58 (b) y 4 3 2 1 2π x π –1 –4 Figure 12.59

12.5 Graphing Trigonometric Curves 605 (c) y 4 2 p /4 p 2p 3p x −2 Figure 12.60 2 For each of the following graphs, find a trig function of the form y = D + A sin B(x À C) or y = D + A cos B(x À C) that fits the graph. Then check with your calculator that you are correct (Figures 12.61–12.64). (a)* y 6 4 2 x −5 5 Figure 12.61 (b) y 4 2 x −5 5 −2 Figure 12.62 (c) y 4 2 x −5 5 −2 Figure 12.63

606 Chapter 12 Trigonometry (d) 4 2 −5 5 −2 Figure 12.64 ÀÁ 3 (C) When graphing y ¼ sin x À p , many of your students make the mistake of shifting the 2 curve y ¼ sin x; p radians to the left. They claim that since À p occurs, the curve should be 2 2 shifted in the negative direction. How do you help your students understand why this is incorrect? 4* Show that the graph of y = sin(2π/x) for x in (0, 1) has infinitely many zeroes in that interval. Now graph this on your calculator. Are you seeing this? If not, try zooming in a few times. What are you seeing? What lesson can you learn from this? 12.6 Modeling With Trigonometric Functions LAUNCH Examine the following scenarios and decide what they have in common: (a) Imagine a bicycle wheel whose radius is one unit, with a marker attached to the rim of the rear wheel. As the wheel rotates, the height h(t) of the marker above the center of the wheel is measured. (b) Due to the tidal changes, the depth of water at my favorite surfing spot varies from 5 ft to 15 ft daily depending on the time of day, and this variation of depth doesn’t change from day to day. A description is given of the depth of the water as a function of time. (c) A spring is hanging from the ceiling. Young Jack comes around and pulls the spring and lets it go and it starts to oscillate. We hope that in examining the scenarios in the launch, you realized that each of the situations represented cyclical motion and could be modeled by a trigonometric function. As we have pointed out in the introduction to this chapter, trigonometric functions are widely applied. In this section we will describe how they can be used to model some very interesting

12.6 Modeling With Trigonometric Functions 607 situations that occur in reality, some of which affect us on a daily basis. We will now give several examples. Example 12.17 As shown in Figure 12.65, a spring is hanging from the ceiling and attached to it is a block. It is in equilibrium, meaning it is not moving. The block is pulled down a distance of 5 inches and then is released. Equilibrium 5 inches Spring stretched 5 inches beyond its equilibrium length. Figure 12.65 The spring starts to oscillate, and the position of the bottom of the block relative to its equilibrium position is graphed over time. Figure 12.66 shows the resulting graph for the first 3 seconds. How can we model the motion with a trigonometric function? 5 time (seconds) 0 0.5 1 1.5 2 2.5 3 0 −5 −10 −15 Figure 12.66 Solution: We note that the graph of the motion begins at a height of À5. This simply represents the fact that initially the bottom of the block is 5 inches below its natural, or equilibrium, position, which we are taking to be at a height of 0. A distance of + 5 would mean it is 5 inches above its natural position. For the first three seconds, the graph does appear to be a sine curve that has been translated, so its equation will be of the form y = D + A sin B(x À C). Only in this case, the x-axis no longer represents angles, rather time, which we denote by t. It seems that, during the 3 second period, the graph completes three cycles. Thus, the period of the graph, the time needed to go through one cycle, is 1 second. Since the period is 2p and this is 1 B second, 2p ¼ 1 and hence B = 2π. In general, to find |A|, take half the difference of the maximum and B

608 Chapter 12 Trigonometry minimum value of the curve. This yields jAj ¼ 1 ð5 À À15Þ ¼ 10: Since the curve looks like a trans- 2 lated sine curve and not a reflected sine curve, A should be positive. Since the graph starts 5 units below the x-axis, we have translated the sine curve down 5 inches. Thus, D = À5. Finally, since the cycle begins at the origin, C = 0. So our graph is y = À5 + 10 sin 2π(tÀ0), where t represents time, or just À5 + 10 sin 2πt. Our model seems to give us the faulty impression that the spring continues to oscillate forever, which we know does not happen. Since friction dampens the oscillations and eventually stops the spring from oscillating, a more accurate model is needed. In fact, such models exist and they typ- ically consist of a product of two functions. One function is the sine function and the other is the function eÀkt where k is positive. The correct value of k depends on something called the spring constant which measures the stiffness of the spring. The details of all this involve differential equa- tions which we will not describe. Rather, we just show you the graph of the damped oscillation y = eÀ2t(À5 + 10 sin 2πt) which is typical of models of these situations. (See Figure 12.67.) y 2 0 5 0 1.25 2.5 3.75 x −2 −4 Figure 12.67 This certainly is a better model of what is happening. Example 12.18 The number of hours of daylight in a city located at a latitude of 40 on the 15th of the month is shown in the following table: Month Hours of Daylight January 9.6167 February 10.7 March 11.883 April 13.223 May 15.367 June 15 July 14.817 August 13.8 September 12.517 October 11.167 November 10.017 December 9.33 Model this with a trigonometric function.

12.6 Modeling With Trigonometric Functions 609 Solution: To get an idea of the model that will work best we start by plotting the points. A scatterHours of daylight plot of the data is shown in Figure 12.68, with month 1 being January. 18 16 14 12 10 8 6 4 2 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 Month Figure 12.68 The points seem to outline half a sine curve translated in the vertical direction. Therefore, we know we are looking to find the values of A, B, C, and D in the equation y = D + A sin B(x À C). Since the graph has been translated 9.6167, D = 9.6167. The 11-month period from January 15th to December 15th graphed is half of the sine curve’s period. So, the sine curve has a period of 22 months. Since 2p is B the period, 2p ¼ 22, so B ¼ p : Since there is a phase shift here of 1, C = 1, and the value of A which B 11 measures the rise of the curve is 15.367 À 9.33 = 6.037. So our model sine curve is y ¼ 9:6167 þ 6:037 sin p ðx À 1Þ: You will note that, since the data points are not symmetric, our model, like 11 most models, is only an approximation to what is happening. Models rarely are perfect fits. Example 12.19 In Figure 12.69 we have a schematic of a Ferris wheel. It takes one minute to make a complete revolution. When the wheel begins turning, the point P is at the bottom, which is 6 feet off the ground. How high will P be t seconds after the wheel begins turning if the Ferris wheel has an outer radius of 50 feet and turns counterclockwise? 50 ft P 6 ft Figure 12.69 Solution: Since P starts at a height of 6 feet, reaches a maximum height of 106 feet at 30 seconds, and then returns, its motion is cyclical and we can picture it as in Figure 12.70.

610 Chapter 12 Trigonometry H 106 6 t 30 seconds Figure 12.70 It looks like a cosine curve reflected and translated. So, we will model the height by a curve of the form D + A cos B(xÀC). Since it takes 60 seconds to complete a cycle, the period is 60 seconds. Because the period is calculated by computing 2Bp, we have that 2p ¼ 60 and B ¼ p : The ampli- B 30 tude, jAj, which is half the difference between the maximum and minimum points is 1 ð106 À 6Þ ¼ 2 50 feet. coFus rt3hp0ermt. oIrfew, esignrcaephitedisthais,reitflwecoteudld cosine curve, A is negative. So, our model is y¼ À50 begin at a height of À50. Since the Ferris wheel begins at a height of 6 and our model begins at a height of À50, we must translate our graph up 56 units. So D = 56. There is no phase shift here, so C = 0. Our model is then H ¼ 56 À 50 cos p t , where H is the height off the ground and t is the number of seconds. 30 Sound is caused by vibration and is believed to be wavelike in nature. Thus, when we strike the middle “A” on the piano, we are causing a string inside the piano to vibrate with a frequency of 440 hertz. (A hertz is one complete cycle per second, so 440 hertz means 440 cycles per second.) Every note has a different frequency for which there is a formula based on the note’s proximity to middle A. That formula is F ¼ 2n Á 440 hz where n is the number of half steps from middle A, 12 with n being positive if the note is above middle A and negative if n is below middle A. Thus, the C above middle A (for those familiar with the piano keyboard) is 3 half steps above A and has fre- quency of approximately F = 23/12Á440 % 523 hertz. Similarly the F below middle A is 4 half-steps below A, so its frequency is approximately 2À4/12Á440 % 349 hertz. Interestingly, history indicates that one of the first persons to explore the rich mathematical theory behind music was Pythagoras. When we turn on the radio and tune into a station, we are really experiencing trigonometry first hand. Radio signals are sent out in the form of sine and cosine waves. So, when we turn our dial to say 1010, we are trying to access a signal, which is being broadcast at a frequency of 1010 kilohertz. The prefix “kilo” in “kilohertz” means thousand. These are very high frequency waves. The words and music from the broadcasting station are low frequency waves (compared with the broadcasting waves). These low frequency waves are “carried” on the high frequency signals, which change or “modulate” the signal. One kind of change that occurs is that the amplitude of the carrier wave changes as a result of putting this other wave coming from the broadcast studio on top of it. The resulting signal is an amplitude modulated signal, which we know as AM radio.

12.6 Modeling With Trigonometric Functions 611 This modulation is achieved by combining the carrier wave and the signal wave. The net result is an amplitude modified curve which is sent over the airwaves. Inside your radio is a decoder that strips off the carrier wave and leaves the signal wave. Another type of modulation is the FM signal, where we modulate the frequency of the carrier wave. FM is less prone to distortion from noise. Each method of transmission, however, has its advantages which we won’t discuss here. Another application we will briefly mention affects us on a daily basis: electricity! Did you know that the electricity in your home consists of something called alternating current whose behavior is represented by sine curves? Through the use of a transformer, electrical energy can be transmitted over long distances. Thus, as a result of this engineering feat with trigonometric functions, we have all the electrical conveniences in our homes. Isn’t this surprising? After all, who would think that trigonometry had anything to do with such things as the lights in our house? Finally, department stores place devices by their doors to check if a person is shoplifting. Believe it or not, the device you walk through as you enter or leave a store that sounds an alarm if you are stealing an item, works on trigonometric principles. Trigonometry has also permeated the medical field in such instruments as MRI and CAT scan machines. In fact, new applications of trigonometry in technology are appearing every day in your favorite devices such as computers, cell phones, mp3 players, calculators, cable TVs, and cars. While students may be impressed to learn about all of these uses of trigonometry, nothing convinces them more than to see exactly how the trigonometric principles are applied. Therefore, in the next section we give a detailed example of how trigonomet- ric identities studied in high school are applied to robotics. 12.6.1 Robotics While a child may envision a robot as a toy similar to the one pictured in Figure 12.71, Figure 12.71

612 Chapter 12 Trigonometry the truth is that robotics have revolutionized such areas as industry, medicine, public safety, and everyday life. For example, in industry and medicine robots and robotic arms are used to rivet bolts in cars, to spray paint automobiles, to locate parts in warehouses, to perform laparoscopic surgery, to disarm and dispose of bombs, and even to vacuum people’s homes. So, let us now examine the components of a typical robot. There is an upper arm and a lower arm connected by a joint. At the tip of the lower arm, one might find a hand or “riveter.” Of course, a robotic arm is three dimensional, but since studying that would entail rather advanced mathematics, we will study a two-dimensional robotic arm that is the basis for the operations of a three-dimensional robotic arm. That is, we will imagine a simplified version of the arm in which the robot is facing sideways. (See Figure 12.72.) The upper arm has length L1, while the lower arm has length L2. Throughout this discussion, think of the lower arm as that part of the arm on which the hand is. We will assume that our goal is to program the electronics in the arm to move it in such a way that Shoulder Upper arm Lower arm Figure 12.72 the tip of the lower arm will end at a certain point (x2, y2) where, say for example, a rivet must be placed. (See Figure 12.73.) To accomplish our goal, we assume that the robot’s shoulder is at a place we call the origin. When the tip of the lower arm, L2, reaches its destination (x2, y2) (where the riveter is), the tip of the upper arm, L1 will lie at some point (x1, y1). y (x2,y2) L2 L1 (x1,y1) x Figure 12.73 We will follow the usual convention that positive angles are those measured counterclockwise and negative angles are those measured clockwise. To make sure the tip of the arm is where we want it to be, the angles that will be relevant to us are angle α, that the upper arm makes with the hor- izontal axis and angle β, that the lower arm makes relative to the extension of the upper arm. Both angles are shown in Figure 12.74. Ultimately, we need to find general formulas which will allow us

12.6 Modeling With Trigonometric Functions 613 to find α and β which will move the robotic arm so that the tip of the lower arm is at (x2, y2). Let us proceed. y (x2,y2) L2 x β L1 (x1,y1) α Figure 12.74 The first thing we have to do is express x2 and y2 in terms of α, β, L1, and L2. We label the joint where the upper arm meets the lower arm as R, and the place where the hand might be placed as S. Looking at Figure 12.75, and remembering that R = (x1, y1), we see, that in triangle ORP; cos a ¼ OP ¼ x1 and so x1 = OP = L1 cos α. Similarly, in triangle RST, since the measure of angle SRT is α + β, OR L1 we have cosða þ bÞ ¼ RT RT so RT = L2 cos(α + β). Since, in triangle ORP; sin a ¼ PR ¼ Ly11, we RS ¼ OR L2 have y1 = PR = L1 sin α and similarly, in triangle RST, TS = L2 sin(α + β). y S(x2,y2) L2 L1 R β horizontal α x T α O L1cosα P Figure 12.75 ð12:23Þ Putting this all together, we get that x2 ¼ OP þ RT ¼ L1 cos a þ L2 cosða þ bÞ and y2 ¼ PR þ TS ¼ L1 sin a þ L2 sinða þ bÞ: ð12:24Þ Our goal is to use these equations to find α and β. To make things clearer, rather than do it in the abstract, we will take the specific case where (x2, y2) = (6, 2), L1 = 5 and L2 = 3. Thus, our equa- tions in the above display become 6 ¼ 5 cos a þ 3 cosða þ bÞ and ð12:25Þ 2 ¼ 5 sin a þ 3 sinða þ bÞ: ð12:26Þ

614 Chapter 12 Trigonometry When we square (12.25) and (12.26) we get the equations 36 ¼ 25 cos2 a þ 30 cos a cosða þ bÞ þ 9 cos2ða þ bÞ; 4 ¼ 25 sin2a þ 30 sin a sinða þ bÞ þ 9 sin2ða þ bÞ; and when we add the squares, using the relationship that sin2 x + cos2 x = 1 we get: 36 þ 4 ¼ 25ðcos2 a þ sin2 aÞ þ 30½cos a cosða þ bÞ þ sin a sinða þ bފ þ 9½ cos2ða þ bÞ þ sin2ða þ bފ or more simply, 40 ¼ 34 þ 30½cos a cosða þ bÞ þ sin a sinða þ bފ: Subtracting 34 from both sides and dividing by 30 yields the simpler equation 1 ¼ ½cos a cosða þ bÞ þ sin a sinða þ bފ: 5 Finally, recalling that cos x cos y + sin x sin y = cos(x À y), and taking x to be α and y to be α + β, this reduces to 1 ¼ cosðÀbÞ: 5 Since cos(Àβ) = cos(β), we finally have the very simple equation that cos b ¼ 1 ð12:27Þ 5 cosÀ1 ÀÁ 78:46: and we can solve for β to get b ¼ 1 % It is really quite amazing how the trigono- 5 metric identities made this all so easy! Now to find α we return to (12.25) and (12.26) Using the relationships that cos(α + β) = cos α cos β À sin α sin β, and that sin(α + β) = sin α cos β + cos α sin β, and after rearranging terms, (12.25) and (12.26) become 6 ¼ ð5 þ 3 cos bÞcos a À 3 sin b sin a and ð12:28Þ 2 ¼ 3 sin b cos a þ ð5 þ 3 cos bÞ sin a: ð12:29Þ 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi Now, we use the values, cos b ¼ 5 ; sin b ¼ þ 1 À cos2 b ¼ 24 (since β is acute), and the display 5 simplifies to pffiffiffi 6 ¼ 258pfficffiffios a À 6 5 6 sin a and 66 28 2 ¼ 5 cos a þ 5 sin a: We solve this syspteffiffimffi of equations simultaneously, say by multiplying the first equation by 28 and the second by 6 6 and adding, which allows us to solve for cos α. We get cos α % 0.9869. In a similar manner we get sin α % À0.1609. Since cos α is positive and sin α is negative, α is in the fourth quadrant, and turns out to be % À9.26. Notice how we used the signs of the functions in the quadrants also to find the quadrant α is in, once again emphasizing something that students in high school are often asked, “What quadrant must the angle be in if cosine is positive and sine is negative.” We have worked this problem for specific values of L1 and L2 and derived the values of α and β. If we did the same process with letters instead of numbers, we would find the following formulas

12.6 Modeling With Trigonometric Functions 615 will always work to find α and β and tell us how to move our robotic arm to where we want it so that the tip of the lower arm is at (x2, y2). cos b ¼ x22 þ y22 À L21 À L22 ; ð12:30Þ 2L1L2 cos a ¼ ðL1 þ L2 cos bÞðx2Þ þ ðL2 sin bÞy2 ; and ð12:31Þ ðL1 þ L2 cos bÞ2 þ ðL2 sin bÞ2 sin a ¼ ðÀL2 sin bÞðx2Þ þ ðL1 þ L2 sin bÞy2 : ð12:32Þ ðL1 þ L2 cos bÞ2 þ ðL2 sin bÞ2 Using equations (12.23) and (12.24) equation (12.30) is the analog of equation (12.27), while equations (12.31) and (12.32) come from solving simultaneously the analogs of (12.28) and (12.29) respectively. And so, programming the robot becomes an easy matter. For example, if L1 = 5 ft. and L2 = 3 ft. and we want the tip of the robotic arm to land at (x2, y2) = (6, 2), as we required earlier, these formulas give us cos b ¼ 62 þ 22 À 52 À 32 ¼ 1 and so 2ð5Þð3Þ 5 b ¼ cosÀ162 þ 22 À 52 À 32 % 78:46; 2ð5Þð3Þ pffiffiffi where we have taken the principal value of the cosine. Since sin b ¼ 24 we substitute the values of 5 cos β and sin β in (12.31) and (12.32) to get:   pffiffiffiffiffiffi! 1 24 5 þ 3 Á 5 6þ 3 5 2 cos a ¼  12 pffiffiffiffiffiffi!2 % 0:98697 and 5 24 þ 3 Á 5 þ 35 pffiffiffiffiffiffi!   24 1 À3 Á 5 6þ 5 þ 3 Á 5 2 sin a ¼  12 pffiffiffiffiffiffi!2 % À0:16091; 5 5 þ 3 Á þ 3 24 5 which places α in the fourth quadrant since the sine is negative and the cosine is positive. Thus, α = sinÀ1 (À0.16091) % À9.26 and now we know α and β. Student Learning Opportunities 1* Suppose that in Example 12.17 that the period of the spring is 30 seconds. Write the equation of the motion. 2 Suppose that in Example 12.19 P is at the top of the wheel when the Ferris wheel starts. What function will describe the height?

616 Chapter 12 Trigonometry 3 Suppose that in Example 12.19 P is at the 3 o’clock position when the Ferris wheel starts. What function will describe the height? 4* Suppose that in Example 12.19 P is at the 9 o’ clock position when the Ferris wheel starts. What function will describe the height? 5 What is the function that describes the position of P if the Ferris wheel has a height of 120 feet, and when it begins, P is at the lowest point 6 feet off the ground? 6 In Example 12.19, write a function that describes how far P is horizontally from the center of the Ferris wheel at time t. 7 Use graph paper and a protractor to do the following problem: Let the upper arm, L1 = 5 units on the graph paper and the lower arm, L2 = 3 units on the graph paper, α = 45 and β = 30. Draw the arm in this configuration and then use (12.23) and (12.24) to find the coordinates (x2, y2) where the tip of the lower arm lands. Then check what you drew and what (12.23) and (12.24) predicted are approximately the same. Use Figure 5 to help you draw the arm correctly. (a) Answer the same question with α = À45, β = 30 leaving L1 and L2 the same. (b) Answer the same question with α = 45, β = À30 leaving L1 and L2 the same. 8 Given that the upper arm has length 5 and the lower arm length 3, describe the region of points that are reachable by the tip of the lower arm (Ignore the thickness of the arms and assume that each part of the arm can rotate up to 360). 9 Take L1 = 4 and L2 = 2. Use the formulas (12.30)–(12.32) to find the angles α and β needed to reach the point (5, 1). 10 Prove, using (12.23) and (12.24), that in general cos b ¼ x22 þ y22 À L21 À L22 : ð12:33Þ 2L1L2 (This is equation (12.30).) 12.7 Inverse Trigonometric Functions LAUNCH One of the following, (a) or (b) is false. Which one, and why? (a) sinÀ1(sin x) = x; (b) sin(sinÀ1y) = y. If you are like most people, the launch question probably has you a bit confused. As you read this section on inverse trigonometric functions, you will have a clearer picture of what is really meant by the inverse of a trigonometric function.

12.7 Inverse Trigonometric Functions 617 As we have seen from previous sections, one of the most amazing features of trigonometry is that it allows us to measure the unmeasurable. In secondary school there are two types of problems that students are exposed to when they study trigonometry that illustrate this and are shown in the following examples. Example 12.20 A person standing 50 feet from a tree measures the angle from the ground to the top of a tree with a surveying instrument. She finds that angle θ is 23. Estimate the height of the tree. Example 12.21 An entrance ramp on a highway is to span 60 feet and rise 3 feet. (See Figure 12.71.) What angle must the ramp make with the ground to achieve this? In the first example, we look at the diagram and notice that we can solve this problem by using the tangent ratio. (See Figure 12.76.) tree 23º 50 ft Figure 12.76 If we call the height of the tree, x, then tan 23 ¼ x 50 and multiplying both sides by 50 yields x = 50 tan 23 = 21.224 ft, which is easily obtained by the calculator. In the second problem we have the reverse situation. We draw a picture of the ramp in Figure 12.77: ramp 3 ft θ 60 ft Figure 12.77 Now we have to find the angle. If we call the angle that the ramp must make with the horizontal, θ, then tan y ¼ 3 60 and we have to solve for θ. In secondary school, we are taught that, if we use the tanÀ1 button on the calculator, we can solve this equation. Doing so, we get that y ¼ tanÀ1 3 % 2:8. So we have 60 solved our problem. In a similar manner, when we wish to find a solution of, say an equation like sin θ = 0.5, we compute sinÀ1(0.5) to get 30. What are these functions, tanÀ1 and sinÀ1, and what does the expo- nent À1 in each case mean?

618 Chapter 12 Trigonometry First, we tell you what they are not. These do not stand for reciprocals. While it is true that in algebra, xÀ1 means 1/x, or the reciprocal of x, this is not true here. Thus, don’t think that sinÀ1 is the reciprocal of sin. The reciprocal of the sine function is given a specific name. It is called the cose- cant. In a similar manner, the reciprocal of the tangent is called the cotangent. So once again, tanÀ1 is not the reciprocal of tangent. The À1 exponent refers to the concept of inverse function that we discussed in Chapter 10. Let us explain. If one looks at the graph of sin x, one sees that the equation sin x ¼ 1 has many 2 solutions. In fact, it has infinitely many solutions as we see in Figure 12.78. 4 2 g (x) = 1 2 30 60 90 120 150 f (x) = sin(x) –2 –4 Figure 12.78 One of them is 30, and another is 150. If we add or subtract multiples of 360, we get all our so- lutions. Thus, the function y = sin x is not a 1-1 function, since different x0s can give rise to the same y. However, if we restrict x to be between À90 and 90, then the equation sin x ¼ 1 has only one 2 solution, namely x = 30. See Figure 12.79 where sin x has been restricted. y 1 0.5 0x –90 –60 –30 0 30 60 90 –0.5 –1 Figure 12.79 What we are saying is that, if the function y = sin x is restricted to angles between À90 and 90, then we get a 1-1 function and this function has an inverse. The inverse function, obtained by solving for x in terms of y is denoted by x = sinÀ1y or also as x = arcsin y. These are read “x equals inverse sin of y” and “x equals arc sine y” and both mean the same thing, namely, that x

12.7 Inverse Trigonometric Functions 619 is the angle (between À90 and 90) whose sine is y. Thus, the equations y = sin x and x = sinÀ1y are interchangeable when the angle x is restricted to between À90 and 90. Similarly, when the func- tion y = tan x is restricted to between À90 and 90 (not including these values), it too becomes 1-1 and we can talk about its inverse, denoted by x = tanÀ1 y. Again, this is read as x is the angle (À90 < x < 90), whose tangent is y. This brings us to the common practice in secondary school of inter- changing variables when writing an inverse function, which we mentioned in Chapter 10. As we said, in practical problems this makes no sense, as we will demonstrate in this situation. When x is an angle say in a triangle, y = sin x represents the ratio of two sides. There is a big difference between the angle, x, and the ratio, y, of two sides of a triangle. So other than for the sake of graph- ing, say on a graphing calculator, we should not interchange variables. When using the calculator to solve equations, we must be careful about getting all the solutions. The general rule is that, if f(x) is periodic and has period p, to find all solutions of f(x) = a, we find all solutions within one period and we can then generate the rest by adding multiples of p. Let us illustrate. Example 12.22 Solve the equation 2 sin x + 3 = 4. Solution: We solve the equation for sin x first to get sin x ¼ 21. The tendency at this point is to just compute x ÀÁ x = 30. We might now reason, “The graph of y = sin ¼ sinÀ1 1 on our calculator and get 2 x is periodic with period 360 and so once we have a set of solutions in one period, we can generate all the other solutions by adding multiples of 360. So our solutions are x = 30 + 360k where k = 0, ±1, ±2, and so on.” But we are not completely correct since we have missed infinitely many solu- tions. We have not yet found all the solutions in one period. When we press sinÀ1(1/2), the calcu- lator only gives us an angle between À90 and 90, and the region from À90 to 90 is only half of one period. A period for the function sin x is 360 degrees. We need to find all solutions in one period before we can be sure we have all the solutions. One way to find all the solutions in one period is to graph the functions Y1 = 2 sin x + 3 and Y2 = 4 on the same set of axes and see where they intersect. One will find, assuming that the calculator is in degree mode, that between 0 and 360 (one period) we get two solutions, x = 30 and x = 150. Now we can generate all solutions and they are x = 30 + 360k where k = 0, ±1, ±2, . . . and x = 150 + 360k where k = 0, ±1, ±2, . . .. A second way to find all solutions in one period is to use what is called the reference angle. The reference angle is the smaller angle that the terminal side of an angle makes with the x-axis. For example in Figure 12.80, on the right, we see an angle of 150. The smaller angle that this side makes with the x-axis is 30, which is the reference angle. Similarly, with an angle of 210 or 330 there is an associated reference angle of 30  since that is the smaller angle the terminal side of either angle makes with the x-axis. When you press an inverse trigonometric function, say sinÀ1 À 1 on the calculator, you will get an angle. If you ignore the sign of the angle, you will get the refer- 2 ence angle. Thus, if you press sinÀ1 À 1 you get À30 and ignoring the sign we get that the reference 2 angle of any such angle is 30. Returning to the problem we just solved, we need to compute sinÀ1 1 2 and we get 30. Our answer, 30, is positive and so we don’t have to change the sign.

620 Chapter 12 Trigonometry Now we argue that, if sin x ¼ 12, then since sin x is positive, x is either in quadrant 1 or 2. Using the x-axis as one side of an angle, we draw an angle of 30 in quadrant 1 and another in quadrant 2 as shown in Figure 12.80 yy 30° x 150° x 30° Figure 12.80 We then measure our angles starting from the positive x-axis. They are 30 and 150 and we can generate our solutions as before, x = 30 + 360k where k = 0, ±1, ±2, . . .. This method always works. (So does the graphing method.) One must be careful with quadrantal angles when solving equations. Example 12.23 Solve the equation sin 2x cos 3x À sin 2x = 0 for all values of x where 0 x < 360. Solution: It is a good idea to factor out sin 2x, for then we get sin 2xðcos 3x À 1Þ ¼ 0: This tells us that sin 2x = 0 or cos 3x À 1 = 0. Calling θ = 2x and referring to the graph of sin θ, we see it crosses the x-axis, when θ = 0 or θ = 180, 360, and 540. Since θ = 2x, we have 2x = 0, 180, 360, and 540, and so x = 0, 90, 180, and 270. Notice that 0 x < 360, which makes 0 θ( or 2x)<720. That is why we looked at angles up to, but not including 720. To solve the second equation we get cos 3x = 1. This yields 3x = cosÀ11 = 0 and again, referring to the cosine graph, we see that the graph of cos θ is 1 only when θ = 3x is 0 or 360 or 720. (Once again, if 0 x < 360, 0 3x < 1080, so we need to go up to, but not include 1080.) So solutions are x = 0, 120, and 240 for this latter equation. Therefore, the given equation has as its solutions 0, 90, 120, 180, 240, and 270. Memorization of the graphs of sin θ and cos θ will help in solving trigonometric equations. Student Learning Opportunities 1* (a) (C) You ask your students to solve the equation 4 sin x + 3 = 5 for all values of x. One of your students, Jacqueline, solves the equation as follows: 4 sin x þ 3 ¼ 5 4 sin x ¼ 2 sin x ¼ 1=2 x ¼ sin À1ð1=2Þ x ¼ 30 ðfrom the calculatorÞ Is she correct? Why or why not?

12.7 Inverse Trigonometric Functions 621 (b) Another student, Sung, offers Jacqueline some help. He says, “You almost have the solu- tion. The solution is x = 30 + 360k where k ¼ 0; Æ1; Æ2::::” Is Sung correct? If not, what is the correct answer? 2 After seeing Example 12.21, Melissa now thinks she understands what the sinÀ1, cosÀ1, and tanÀ1 buttons are used for. She says, “They are used only to find angles in a right triangle. So if you compute cosÀ1(0.5) you are finding the angle in a right triangle whose cosine is 0.5 and this is 60.” When you persist and ask her to find cosÀ1(À0.5) using the calculator she gets 150 and now she is baffled. How could a right triangle have a 150-degree angle? What is wrong with how Melissa is thinking about inverse trigonometric functions? 3 (C) (a) One of your students, Jadin, is asked to solve the equation y = tanÀ1x for x in terms of y. He solves for x as follows: y = tanÀ1x y ¼ 1 x tan tan y = x. Jadin checks the answer in the back of the book and sees he is right. How can you help Jadin understand what is wrong with his work? (b) Another student, Marta, solves the same problem as follows: y = tanÀ1x y ¼ 1 x tan y tan x = 1 tan x ¼ 1 y x = tan-1 1  y When Marta checks her answer in the back of the book, it doesn’t check. How will you help her to see what is wrong with her work? 4 Solve each of the following equations for all values of x: (a)* cos 3x ¼ 1 2 (b) sin 2x ¼ À 1 2 (c) 2 sin x À1 = 3 (d) 3 cos 4x = 1 (e) sin x cos x À sin x = 0 (f) 2 sin2 x À3 sin x + 2 = 0 (g)* 3 tan x À 4 = 0 (Remember, the period of tan x is 180, not 360.) (h) csc x = 2 (i) 3 sec x À1 = 5 5 Using the relationship that sin2 x + cos2 x = 1, reduce each of the following equations to one involving only either sin x or cos x. Then solve them. Check your answers graphically. (a) 1 + sin x = 2 cos2 x (b) cos x = sin x À 1 [Hint: Square both sides and be sure to check your answers.] (c) 3 sin 3x À 1 = 0.5