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cos x 1 1 cos 2 x cos 2 x cos2 x 2 cos x 1 1 0 2 cos2 x 2 cos x 0 2 cos xcos x 1 0 Setting each factor equal to zero produces Write original equation. Square each side. Pythagorean identity Rewrite equation. Combine like terms. Factor. 2 cos x 0 cos x 0 x, 2 and cos x 1 0 cos x 1 x. 3 2 Because you squared the original equation, check for extraneous solutions. Check x /2 1? cos sin 2 2 0 1 1 x 3/2 3 2 0 1 1 1? sin Check cos Check x cos 1? 1 1 0 sin Substitute 2 for x. Solution checks. ✓ 3 2 Substitute 32 for x. Solution does not check. x. for Substitute Solution checks. ✓ x 32 x 2 Of the three possible solutions, 0, 2, the only two solutions are is extraneous. So, in the interval and x. Now try Exercise 33. You square each side of the equation in Example 6 because the squares of the sine and cosine functions are related by a Pythagorean identity. The same is true for the squares of the secant and tangent functions and the cosecant and cotangent functions. Exploration Use a graphing utility to confirm the solutions found in Example 6 in two different ways. Do both methods produce the same -values? Which method do you prefer? Why? x 1. Graph both sides of the equation and find the x -coordinates of the points at which the graphs intersect. y cos x 1 y sin x Right side: Left side: 2. Graph the equation y cos x 1 sin x x and find the -intercepts of the graph. 333202_0503.qxd 12/5/05 9:03 AM Page 394 394 Chapter 5 Analytic Trigonometry Functions Involving Multiple Angles The next two examples involve trigonometric functions of multiple angles of the ku forms sin To solve equations of these forms, first solve the equation for then divide your result by and cos ku, ku. k. Example 7 Functions of Multiple Angles Solve 2 cos 3t 1 0. Solution 2 cos 3t 1 0 2 cos 3t 1 cos 3t 1 2 0, 2, In the interval solutions, so, in general, you have you know that Write original equation. Add 1 to each side. Divide each side by 2. 3t 3 and 3t |
53 are the only 3t 3 2n and 3t 5 3 2n. Dividing these results by 3, you obtain the general solution t 9 2n 3 and t 5 9 2n 3 General solution where n is an integer. Now try Exercise 35. Example 8 Functions of Multiple Angles Solve 3 tan x 2 3 0. Solution 3 tan x 2 3 0 3 tan tan 3 x 2 x 2 0,, 1 In the interval general, you have 3 4 x 2 n. Write original equation. Subtract 3 from each side. Divide each side by 3. you know that x2 34 is the only solution, so, in Multiplying this result by 2, you obtain the general solution x 3 2 2n where n is an integer. Now try Exercise 39. General solution 333202_0503.qxd 12/5/05 9:03 AM Page 395 Section 5.3 Solving Trigonometric Equations 395 Using Inverse Functions In the next example, you will see how inverse trigonometric functions can be used to solve an equation. Example 9 Using Inverse Functions Solve sec2 x 2 tan x 4. Solution sec2 x 2 tan x 4 1 tan2 x 2 tan x 4 0 tan2 x 2 tan x 3 0 tan x 3tan x 1 0 Write original equation. Pythagorean identity Combine like terms. Factor. Setting each factor equal to zero, you obtain two solutions in the interval 2, 2. [Recall that the range of the inverse tangent function is 2, 2.] tan x 3 0 tan x 3 x arctan 3 and tan x 1 0 tan x 1 x 4 Finally, because adding multiples of tan x has a period of, you obtain the general solution by x arctan 3 n and x 4 n General solution n where arctan 3. is an integer. You can use a calculator to approximate the value of Now try Exercise 59. W RITING ABOUT MATHEMATICS Equations with No Solutions One of the following equations has solutions and the other two do not. Which two equations do not have solutions? a. b. c. sin2 x 5 sin x 6 0 sin2 x 4 sin x 6 0 sin2 x 5 sin x 6 0 Find conditions involving the constants b and c that will guarantee that the equation sin2 x b sin x c 0 has at least one solution on some interval of length 2. 333202_0503.qxd 12/ |
5/05 9:03 AM Page 396 396 Chapter 5 Analytic Trigonometry 5.3 Exercises VOCABULARY CHECK: Fill in the blanks. 1. The equation 2 sin 1 0 has the solutions 7 6 2n and 11 6 2n, which are called ________ solutions. 2. The equation 2 tan2 x 3 tan x 1 0 is a trigonometric equation that is of ________ type. 3. A solution to an equation that does not satisfy the original equation is called an ________ solution. PREREQUISITE SKILLS REVIEW: Practice and review algebra skills needed for this section at www.Eduspace.com. In Exercises 1– 6, verify that the the equation. x -values are solutions of 1. 2 cos x 1 0 (a) x 2. sec x 2 0 3 3. 4. (a) x 3 3 tan2 2x 1 0 (a) x 12 2 cos2 4x 1 0 (a) x 16 5. 2 sin2 x sin x 1 0 (a) x 6. 2 csc4 x 4 csc2 x 0 (a) x 6 (b) x 5 3 (b) x 5 3 (b) x 5 12 (b) x 3 16 (b) x 7 6 (b) x 5 6 In Exercises 7–20, solve the equation. 7. 9. 11. 13. 14. 15. 17. 19. 8. 10. 12. 2 cos x 1 0 3 csc x 2 0 3 sec2 x 4 0 sin xsin x 1 0 3 tan2 x 1tan2 x 3 0 4 cos2 x 1 0 16. 2 sin2 2x 1 tan 3xtan x 1 0 20. 18. 2 sin x 1 0 tan x 3 0 3 cot2 x 1 0 sin2 x 3 cos2 x tan2 3x 3 cos 2x2 cos x 1 0 26. 28. sec x csc x 2 csc x sec x tan x 1 25. 27. 29. 30. 31. 32. 33. 34. sec2 x sec x 2 2 sin x csc x 0 2 cos2 x cos x 1 0 2 sin2 x 3 sin x 1 0 2 sec2 x tan2 x 3 0 cos x sin x tan x 2 csc x cot x 1 sin x 2 cos x 2 In Exercises 35– 40, solve |
the multiple-angle equation. 35. 37. 39. cos 2x 1 2 tan 3x 1 x 2 cos 2 2 36. sin 2x 3 2 38. 40. sec 4x 2 x 2 sin 3 2 In Exercises 41– 44, find the -intercepts of the graph. x 41. y sin x 2 1 42. y sin x cos x y 3 2 1 −2 −1 1 2 3 4 −2 43. 3 y tan2x 44. 4 y sec4x 8 y 2 1 −3 −1 1 3 −2 x x x x In Exercises 21–34, find all solutions of the equation in the interval [0, 2. −3 −1 1 3 −2 21. 23. cos3 x cos x 3 tan3 x tan x 22. 24. sec2 x 1 0 2 sin2 x 2 cos x 333202_0503.qxd 12/5/05 9:03 AM Page 397 In Exercises 45– 54, use a graphing utility to approximate the solutions (to three decimal places) of the equation in the interval [0, 2. 50. x cos x 1 0 − π 45. 46. 47. 48. 49. 51. 52. 53. 54. 2 sin x cos x 0 4 sin3 x 2 sin2 x 2 sin x 1 0 1 sin x cos x cos x 1 sin x 4 3 cos x cot x 1 sin x x tan x 1 0 sec2 x 0.5 tan x 1 0 csc2 x 0.5 cot x 5 0 2 tan2 x 7 tan x 15 0 6 sin2 x 7 sin x 2 0 In Exercises 55–58, use the Quadratic Formula to solve the equation in the interval Then use a graphing utility to approximate the angle [0, 2. x. 55. 56. 57. 58. 12 sin2 x 13 sin x 3 0 3 tan2 x 4 tan x 4 0 tan2 x 3 tan x 1 0 4 cos2 x 4 cos x 1 0 In Exercises 59–62, use inverse functions where needed to find all solutions of the equation in the interval [0, 2. 59. 60. 61. 62. tan2 x 6 tan x 5 0 sec2 x tan x 3 0 2 cos2 x 5 cos x 2 0 2 sin2 x 7 sin x 3 0 In Exercises 63 and 64, (a) |
use a graphing utility to graph the function and approximate the maximum and minimum points on the graph in the interval and (b) solve the trigonometric equation and demonstrate that its x -coordinates of the maximum and solutions are the f. minimum points of (Calculus is required to find the trigonometric equation.) [0, 2, Function f x sin x cos x f x 2 sin x cos 2x 63. 64. Trigonometric Equation cos x sin x 0 2 cos x 4 sin x cos x 0 Fixed Point positive fixed point of the function function [ f. f c c.] is a real number such that In Exercises 65 and 66, find the smallest A fixed point of a c f 65. f x tan x 4 66. f x cos x Section 5.3 Solving Trigonometric Equations 397 67. Graphical Reasoning Consider the function given by f x cos 1 x and its graph shown in the figure. y 2 1 −2 π x (a) What is the domain of the function? (b) Identify any symmetry and any asymptotes of the graph. (c) Describe the behavior of the function as x → 0. (d) How many solutions does the equation cos 1 x 0 have in the interval 1, 1? Find the solutions. (e) Does the equation have a greatest solution? If so, approximate the solution. If not, explain why. cos1x 0 68. Graphical Reasoning Consider the function given by f x sin x x and its graph shown in the figure. y 3 2 −1 −2 −3 − π π x (a) What is the domain of the function? (b) Identify any symmetry and any asymptotes of the graph. (c) Describe the behavior of the function as x → 0. (d) How many solutions does the equation sin x x 0 have in the interval 8, 8? Find the solutions. 333202_0503.qxd 12/5/05 9:03 AM Page 398 398 Chapter 5 Analytic Trigonometry 69. Harmonic Motion A weight is oscillating on the end of a spring (see figure). The position of the weight relative to the cos 8t 3 sin 8t, point of equilibrium is given by where is the displacement (in meters) and is the time (in seconds). Find the times when the weight is at the point of 0 ≤ t ≤ 1. |
equilibrium y 0 y 1 12 for y t 74. Projectile Motion A sharpshooter intends to hit a target at a distance of 1000 yards with a gun that has a muzzle velocity of 1200 feet per second (see figure). Neglecting air resistance, determine the gun’s minimum angle of elevation if the range is given by r Equilibrium y r 1 32 2 sin 2. v0 θ r = 1000 yd Not drawn to scale 70. Damped Harmonic Motion The displacement from equilibrium of a weight oscillating on the end of a spring is given by is the where displacement (in feet) and is the time (in seconds). Use a graphing utility to graph the displacement function for 0 ≤ t ≤ 10. Find the time beyond which the displacement does not exceed 1 foot from equilibrium. y 1.56e0.22t cos 4.9t, t y 71. Sales The monthly sales S (in thousands of units) of a seasonal product are approximated by S 74.50 43.75 sin t 6 t is the time (in months), with where corresponding to January. Determine the months when sales exceed 100,000 units. t 1 75. Ferris Wheel A Ferris wheel is built such that the height (in feet) above ground of a seat on the wheel at time (in t h minutes) can be modeled by ht 53 50 sin 16 t. 2 The wheel makes one revolution every 32 seconds. The ride begins when t 0. (a) During the first 32 seconds of the ride, when will a person on the Ferris wheel be 53 feet above ground? (b) When will a person be at the top of the Ferris wheel for the first time during the ride? If the ride lasts 160 seconds, how many times will a person be at the top of the ride, and at what times? 72. Sales The monthly sales S (in hundreds of units) of skiing equipment at a sports store are approximated by Model It S 58.3 32.5 cos t 6 t is the time (in months), with where corresponding to January. Determine the months when sales exceed 7500 units. t 1 73. Projectile Motion A batted baseball leaves the bat at an angle of with the horizontal and an initial velocity of v0 feet per second. The ball is caught by an outfielder 300 feet from home plate (see figure). Find if 2 sin 2. the range of a projectile is given by 100 r 1 r 32 v |
0 θ r = 300 ft Not drawn to scale 76. Data Analysis: Unemployment Rate The table in the United States shows the unemployment rates t for selected years from 1990 through 2004. The time corresponding to is measured in years, with 1990. (Source: U.S. Bureau of Labor Statistics) t 0 r Time, t Rate, r Time, t Rate, r 0 2 4 6 5.6 7.5 6.1 5.4 8 10 12 14 4.5 4.0 5.8 5.5 (a) Create a scatter plot of the data. 333202_0503.qxd 12/5/05 9:03 AM Page 399 Model It (co n t i n u e d ) (b) Which of the following models best represents the data? Explain your reasoning. (1) (2) (3) (4) r 1.24 sin0.47t 0.40 5.45 r 1.24 sin0.47t 0.01 5.45 r sin0.10t 5.61 4.80 r 896 sin0.57t 2.05 6.48 (c) What term in the model gives the average unemployment rate? What is the rate? (d) Economists study the lengths of business cycles such as unemployment rates. Based on this short span of time, use the model to find the length of this cycle. (e) Use the model to estimate the next time the unemployment rate will be 5% or less. Section 5.3 Solving Trigonometric Equations 399 80. If you correctly solve a trigonometric equation to the statethen you can finish solving the equation sin x 3.4, ment by using an inverse function. In Exercises 81 and 82, use the graph to approximate the number of points of intersection of the graphs of and y2. y1 81. 2 sin x 3x 1 y1 y2 y 4 3 2 1 y1 y2 π 2 x 82. y1 y2 2 sin x 1 2 x 1 y y2 y1 π 2 4 3 2 1 −3 −4. x 77. Geometry The area of a rectangle (see figure) inscribed Skills Review is given by In Exercises 83 and 84, solve triangle missing angle measures and side lengths. ABC by finding all in one arc of the graph of A 2x cos x, 0 < x < y cos 1 83. B 66° 22.3 x |
page 405, you can use an identity to rewrite a trigonometric expression in a form that helps you analyze a harmonic motion equation. Using Sum and Difference Formulas In this and the following section, you will study the uses of several trigonometric identities and formulas. Sum and Difference Formulas sinu v sin u cos v cos u sin v sinu v sin u cos v cos u sin v cosu v cos u cos v sin u sin v cosu v cos u cos v sin u sin v tanu v tan u tan v 1 tan u tan v tanu v tan u tan v 1 tan u tan v For a proof of the sum and difference formulas, see Proofs in Mathematics on page 424. Exploration cos x cos 2 Use a graphing utility to graph in the same viewing window. What can you conclude about the graphs? Is it true that cos 2? sin x sin 4 y1 in the same viewing window. What can you conclude about the graphs? Is it true that cosx 2 cos x Use a graphing utility to graph cosx 2 sin x sin 4? sinx 4 sinx 4 and and y1 y2 y2 Richard Megna/Fundamental Photographs Examples 1 and 2 show how sum and difference formulas can be used to find exact values of trigonometric functions involving sums or differences of special angles. Example 1 Evaluating a Trigonometric Function Find the exact value of cos 75. Solution To find the exact value of cos Consequently, the formula for cos 75 cos30 45 75, cosu v use the fact that yields 75 30 45. cos 30 cos 45 sin 30 sin 45. Try checking this result on your calculator. You will find that cos 75 0.259. Now try Exercise 1. 333202_0504.qxd 12/5/05 9:04 AM Page 401 Historical Note Hipparchus, considered the most eminent of Greek astronomers, was born about 160 B.C. in Nicaea. He was credited with the invention of trigonometry. He also derived the sum and difference sinA ± B formulas for cosA ± B. and FIGURE 5.7 Section 5.4 Sum and Difference Formulas 401 Example 2 Evaluating a Trigonometric Expression Find the exact value of sin 12. Solution Using the fact that 12 3 4 sinu v, you obtain together with the formula for sin 3 4 sin 12 sin cos cos sin. Now try Exercise 3. Example 3 Evaluating a Trigonometric Expression Find the exact value of |
sin 42 cos 12 cos 42 sin 12. Solution Recognizing that this expression fits the formula for sinu v, you can write sin 42 cos 12 cos 42 sin 12 sin42 12 sin 30 1 2. Now try Exercise 31. Example 4 An Application of a Sum Formula Write cosarctan 1 arccos x as an algebraic expression. Solution This expression fits the formula for v arccos x are shown in Figure 5.7. So cosu v. Angles u arctan 1 and cosu v cosarctan 1 cosarccos x sinarctan 1 sinarccos. Now try Exercise 51. 333202_0504.qxd 12/5/05 9:04 AM Page 402 402 Chapter 5 Analytic Trigonometry Example 5 shows how to use a difference formula to prove the cofunction identity cos 2 x sin x. Example 5 Proving a Cofunction Identity Prove the cofunction identity cos 2 x sin x. Solution Using the formula for x cos cos 2 cosu v, you have 2 cos x sin 2 sin x 0cos x 1sin x sin x. Now try Exercise 55. Sum and difference formulas can be used to rewrite expressions such as sin n cos n, 2 2 is an integer where and n as expressions involving only reduction formulas. sin or cos. The resulting formulas are called Example 6 Deriving Reduction Formulas Simplify each expression. a. cos 3 2 b. tan 3 Solution a. Using the formula for cos 3 2 you have cos cos cosu v, 3 2 cos 0 sin 1 sin. sin sin 3 2 b. Using the formula for tanu v, you have tan 3 tan tan 3 1 tan tan 3 tan 0 1 tan 0 tan. Now try Exercise 65. 333202_0504.qxd 12/5/05 9:04 AM Page 403 Section 5.4 Sum and Difference Formulas 403 Example 7 Solving a Trigonometric Equation Find all solutions of sinx sinx 4 1 4 in the interval 0, 2. Solution Using sum and difference formulas, rewrite the equation as sin x cos cos x sin sin x cos cos x sin 1 4 4 4 4 1 2 sin x cos 4 1 2sin x2 2 sin x 1 2 2 2 sin x. π π 2 x 2 π So, the only solutions in the interval 0, 2 are x 5 4 and x 7. 4 y 3 2 1 −1 −2 −3 ( y = sin |
x + + sin FIGURE 5.8 You can confirm this graphically by sketching the graph of y sinx sinx 4 1 4 for 0 ≤ x < 2, as shown in Figure 5.8. From the graph you can see that the and 74. x- intercepts are 54 Now try Exercise 69. The next example was taken from calculus. It is used to derive the derivative of the sine function. Example 8 An Application from Calculus Verify that sinx h sin x h h 0. where cos x sin h h sin x1 cos h h Solution Using the formula for sinx h sin x h you have sinu v, sin x cos h cos x sin h sin x h cos x sin h sin x1 cos h h sin x1 cos h cos xsin h h Now try Exercise 91. h. 333202_0504.qxd 12/5/05 9:04 AM Page 404 404 Chapter 5 Analytic Trigonometry 5.4 Exercises VOCABULARY CHECK: Fill in the blank to complete the trigonometric identity. 2. 1. sinu v ________ tanu v ________ cosu v ________ 3. 5. cosu v ________ sinu v ________ tanu v ________ 4. 6. PREREQUISITE SKILLS REVIEW: Practice and review algebra skills needed for this section at www.Eduspace.com. 27. 28. 29. 30. sin 3 cos 1.2 cos 3 sin 1.2 cos cos 7 sin sin 7 5 5 tan 2x tan x 1 tan 2x tan x cos 3x cos 2y sin 3x sin 2y In Exercises 31–36, find the exact value of the expression. 31. 32. sin 330 cos 30 cos 330 sin 30 cos 15 cos 60 sin 15 sin 60 33. sin cos cos 12 16 4 3 16 12 sin 4 3 16 16 cos cos sin sin 34. 35. 36. tan 25 tan 110 1 tan 25 tan 110 tan54 tan12 1 tan54 tan12 In Exercises 1– 6, find the exact value of each expression. 1. (a) 2. (a) 3. (a) 4. (a) 5. (a) 6. (a) cos120 45 sin135 30 cos 3 4 5 sin3 6 4 sin7 6 sin315 60 3 (b) (b) (b) cos 120 cos 45 sin 135 cos 30 |
cos cos (b) sin (b) sin sin sin (b) sin 315 sin 60 3 5 6 3 4 3 4 7 6 In Exercises 7–22, find the exact values of the sine, cosine, and tangent of the angle by using a sum or difference formula. 165 135 30 255 300 45 7 12 4 3 7. 9. 11. 13. 15. 17. 19. 21. 6 5 6 105 60 45 195 225 30 11 3 4 12 9 17 4 12 285 165 13 12 13 12 8. 10. 12. 14. 16. 18. 20. 22. 12 105 15 7 12 5 12 6 4 In Exercises 37–44, find the exact value of the trigonometric u (Both and function given that v are in Quadrant II.) cos v 3 5. sin u 5 13 and 37. 39. 41. 43. sinu v cosu v tanu v secv u 38. 40. 42. 44. cosu v sinv u cscu v cotu v In Exercises 23–30, write the expression as the sine, cosine, or tangent of an angle. 23. 24. 25. 26. cos 25 cos 15 sin 25 sin 15 sin 140 cos 50 cos 140 sin 50 tan 325 tan 86 1 tan 325 tan 86 tan 140 tan 60 1 tan 140 tan 60 In Exercises 45–50, find the exact value of the trigonometric u function given that and are in Quadrant III.) cos v 4 5. sin u 7 25 (Both and v cosu v tanu v secu v 45. 47. 49. 46. sinu v cotv u 48. 50. cosu v 333202_0504.qxd 12/5/05 9:04 AM Page 405 Section 5.4 Sum and Difference Formulas 405 In Exercises 51–54, write the trigonometric expression as an algebraic expression. Model It 51. 53. 54. sinarcsin x arccos x cosarccos x arcsin x cosarccos x arctan x 52. sinarctan 2x arccos x In Exercises 55– 64, verify the identity. 55. sin3 x sin x 56. sin 2 x cos x 75. Harmonic Motion A weight is attached to a spring suspended vertically from a ceiling. When a driving the weight moves force is applied to the system, vertically |
from its equilibrium position, and this motion is modeled by y 1 3 sin 2t 1 4 cos 2t cos x sin x cos x 3 sin x sin x 1 2 6 cos5 x 2 2 4 cos sin 0 2 1 tan tan 1 tan 4 cosx y cosx y cos2 x sin2 y sinx y sinx y) sin2 x sin2 y sinx y sinx y 2 sin x cos y cosx y cosx y 2 cos x cos y 57. 58. 59. 60. 61. 62. 63. 64. In Exercises 65 –68, simplify the expression algebraically and use a graphing utility to confirm your answer graphically. cos3 2 sin3 2 cos x tan x 67. 66. 65. 68. In Exercises 69 –72, find all solutions of the equation in the interval [0, 2. sinx sinx 1 sinx 3 3 1 sinx 2 6 6 1 cosx cosx tanx 2 sinx 0 4 4 69. 70. 71. 72. In Exercises 73 and 74, use a graphing utility to approximate the solutions in the interval 1 cosx [0, 2. 73. cosx 4 0 4 tanx cosx 2 74. is the distance from equilibrium (in feet) and y where is the time (in seconds). t (a) Use the identity a sin B b cos B a2 b2 sinB C C arctanba, where in the form y a2 b2 sinBt C. a > 0, to write the model (b) Find the amplitude of the oscillations of the weight. (c) Find the frequency of the oscillations of the weight. 76. Standing Waves The equation of a standing wave is obtained by adding the displacements of two waves traveling in opposite directions (see figure). Assume that each of the waves has amplitude. If the models for these waves are and wavelength period A, T, y1 A cos 2 t T x and y2 A cos 2 t T x show that y1 y2 2A cos 2t T cos 2x. = T1 8 t = T2 8 y + y y1 2 2 1 y 2 y + y y1 2 2 1 y + y y1 2 2 1 y 2 y 2 333202_0504.qxd 12/5/05 9:04 AM Page 406 406 Chapter 5 Analytic Trigonometry Synthesis True or False? |
statement is true or false. Justify your answer. In Exercises 77–80, determine whether the 77. 78. 79. sinu ± v sin u ± sin v cosu ± v cos u ± cos v sin x cosx 2 sinx 80. cos x 2 n 81. 82. is an integer In Exercises 81–84, verify the identity. cosn 1n cos, sinn 1n sin, a sin B b cos Ba2 b2 sinB C, where a sin B b cos Ba2 b2 cosB C, where C arctanab C arctanba is an integer a > 0 b > 0 and and 83. 84. n In Exercises 85–88, use the formulas given in Exercises 83 and 84 to write the trigonometric expression in the following forms. (a) 85. 87. a2 b2 sinB C (b) a2 b2 cosB C sin cos 12 sin 3 5 cos 3 86. 88. 3 sin 2 4 cos 2 sin 2 cos 2 In Exercises 89 and 90, use the formulas given in Exercises 83 and 84 to write the trigonometric expression in the form a sin B b cos B. 2 sin 89. 90. 5 cos 3 4 2 (c) Use a graphing utility to graph the functions and f g. (d) Use the table and the graphs to make a conjecture about the values of the functions and as g f h → 0. In Exercises 93 and 94, use the figure, which shows two lines whose equations are y1 m1x b1 and y2 m2 x b2. Assume that both lines have positive slopes. Derive a formula for the angle between the two lines.Then use your formula to find the angle between the given pair of lines. y 6 4 θ y1 = m1x + b1 −2 2 4 x y2 = m2x + b2 93. y x and 94. y x and y 3 x y 1 3 x 95. Conjecture Consider the function given by f sin2 sin2 4. 4 Use a graphing utility to graph the function and use the graph to create an identity. Prove your conjecture. 91. Verify the following identity used in calculus. 96. Proof cosx h cos x h cos xcos h 1 h sin x sin h h 92. Exploration Let in the identity in Exercise 91 f x |
6 g and define the functions and as follows. f h cos6 h cos6 h sin cos h 1 sin h h h gh cos 6 6 (a) What are the domains of the functions and f g? (b) Use a graphing utility to complete the table. 0.01 0.02 0.05 0.1 0.2 0.5 h f h gh (a) Write a proof of the formula for (b) Write a proof of the formula for sinu v. sinu v. Skills Review In Exercises 97–100, find the inverse function of Verify f 1f x x. that f f 1x x and f. 97. f x 5x 3 98. f x 7 x 8 99. f x x2 8 100. f x x 16 In Exercises 101–104, apply the inverse properties of and to simplify the expression. e x ln x 101. 103. log3 34x3 eln6x3 log8 83x 2 102. 104. 12x eln xx2 333202_0505.qxd 12/5/05 9:06 AM Page 407 Section 5.5 Multiple-Angle and Product-to-Sum Formulas 407 5.5 Multiple Angle and Product-to-Sum Formulas What you should learn • Use multiple-angle formulas to rewrite and evaluate trigonometric functions. • Use power-reducing formulas to rewrite and evaluate trigonometric functions. • Use half-angle formulas to rewrite and evaluate trigonometric functions. • Use product-to-sum and sum-to-product formulas to rewrite and evaluate trigonometric functions. • Use trigonometric formulas to rewrite real-life models. Why you should learn it You can use a variety of trigonometric formulas to rewrite trigonometric functions in more convenient forms. For instance, in Exercise 119 on page 417, you can use a double-angle formula to determine at what angle an athlete must throw a javelin. Multiple-Angle Formulas In this section, you will study four other categories of trigonometric identities. 1. The first category involves functions of multiple angles such as sin ku and cos ku. 2. The second category involves squares of trigonometric functions such as sin2 u. 3. The third category involves functions of half-angles such as 4. The fourth category involves products of trigonometric functions such as sinu2. sin u cos v. You should learn the |
double-angle formulas because they are used often in trigonometry and calculus. For proofs of the formulas, see Proofs in Mathematics on page 425. Double-Angle Formulas sin 2u 2 sin u cos u tan 2u 2 tan u 1 tan2 u cos 2u cos2 u sin2 u 2 cos 2 u 1 1 2 sin2 u Example 1 Solving a Multiple-Angle Equation Solve 2 cos x sin 2x 0. Solution Begin by rewriting the equation so that it involves functions of Then factor and solve as usual. x rather than 2x. Mark Dadswell/Getty Images 2 cos x sin 2x 0 2 cos x 2 sin x cos x 0 2 cos x1 sin x 0 2 cos x 0 and 1 sin Write original equation. Double-angle formula Factor. Set factors equal to zero. Solutions in 0, 2 So, the general solution is x 2 2n and x 3 2 2n where n is an integer. Try verifying these solutions graphically. Now try Exercise 9. 333202_0505.qxd 12/5/05 9:06 AM Page 408 408 Chapter 5 Analytic Trigonometry Example 2 Using Double-Angle Formulas to Analyze Graphs Use a double-angle formula to rewrite the equation y 4 cos2 x 2. Then sketch the graph of the equation over the interval 0, 2. Solution Using the double-angle formula for as cos 2u, you can rewrite the original equation y 4 cos2 x 2 22 cos2 x 1 2 cos 2x. Write original equation. Factor. Use double-angle formula. Using the techniques discussed in Section 4.5, you can recognize that the graph of this function has an amplitude of 2 and a period of The key points in the interval are as follows. 0,. Maximum 0, 2 Intercept, 0 4 Minimum, 2 2 Intercept, 0 3 4 Maximum, 2 y = 4 cos2x − 2 π x π2 y 2 1 −1 −2 FIGURE 5.9 Two cycles of the graph are shown in Figure 5.9. Now try Exercise 21. Example 3 Evaluating Functions Involving Double Angles y θ −4 −2 2 4 6 x −2 −4 −6 −8 −10 −12 13 (5, −12) Use the following to find 3 2 cos 5 13, < < 2 sin 2, cos 2, and tan 2. Solution From Figure 5.10, you can see that |
each of the double-angle formulas, you can write sin yr 1213. Consequently, using sin 2 2 sin cos 212 13 cos 2 2 cos2 1 2 25 169 tan 2 sin 2 cos 2 120 119. 120 5 169 13 1 119 169 FIGURE 5.10 Now try Exercise 23. The double-angle formulas are not restricted to angles and. and Other are also valid. Here are two and 3, 6 2 4 2 or double combinations, such as examples. sin 4 2 sin 2 cos 2 and cos 6 cos2 3 sin2 3 By using double-angle formulas together with the sum formulas given in the preceding section, you can form other multiple-angle formulas. 333202_0505.qxd 12/5/05 9:06 AM Page 409 Section 5.5 Multiple-Angle and Product-to-Sum Formulas 409 Example 4 Deriving a Triple-Angle Formula sin 3x sin2x x sin 2x cos x cos 2x sin x 2 sin x cos x cos x 1 2 sin2 xsin x 2 sin x cos2 x sin x 2 sin3 x 2 sin x1 sin2 x sin x 2 sin3 x 2 sin x 2 sin3 x sin x 2 sin3 x 3 sin x 4 sin3 x Now try Exercise 97. Power-Reducing Formulas The double-angle formulas can be used to obtain the following power-reducing formulas. Example 5 shows a typical power reduction that is used in calculus. Power-Reducing Formulas sin2 u 1 cos 2u 2 cos2 u 1 cos 2u 2 tan2 u 1 cos 2u 1 cos 2u For a proof of the power-reducing formulas, see Proofs in Mathematics on page 425. Example 5 Reducing a Power Rewrite sin4 x as a sum of first powers of the cosines of multiple angles. Solution Note the repeated use of power-reducing formulas. sin4 x sin2 x2 1 cos 2x 2 2 Property of exponents Power-reducing formula cos 2x cos2 2x Expand. 1 2 cos 2x 1 cos 4x 2 1 2 cos 2x 1 8 1 8 cos 4x Power-reducing formula Distributive Property 3 4 cos 2x cos 4x Factor out common factor. Now try Exercise 29. 333202_0505.qxd 12/5/05 9:06 AM Page 410 410 Chapter 5 Analytic Trigonometry Half-Angle Formulas You can derive some useful alternative |
forms of the power-reducing formulas by replacing with The results are called half-angle formulas. u2. u Half-Angle Formulas ±1 cos u ±1 cos u cos sin u 2 2 u 2 2 tan u 2 1 cos u sin u sin u 1 cos u The signs of sin u 2 and cos u 2 depend on the quadrant in which u 2 lies. Example 6 Using a Half-Angle Formula Find the exact value of sin 105. DMS To find the exact value of a trigonometric function with an angle measure in form using a half-angle formula, first convert the angle measure to decimal degree form. Then multiply the resulting angle measure by 2. Solution Begin by noting that sinu2 105 and the fact that is half of 105 210. Then, using the half-angle formula for lies in Quadrant II, you have 2 sin 105 1 cos 210 1 cos 30 1 32 2 2 3 2 2. The positive square root is chosen because sin is positive in Quadrant II. Now try Exercise 41. Use your calculator to verify the result obtained in Example 6. That is, evaluate 2 3 2. sin 105 and sin 105 0.9659258 2 3 2 0.9659258 You can see that both values are approximately 0.9659258. 333202_0505.qxd 12/5/05 9:06 AM Page 411 Section 5.5 Multiple-Angle and Product-to-Sum Formulas 411 Example 7 Solving a Trigonometric Equation Find all solutions of 2 sin2 x 2 cos 2 x 2 in the interval 0, 2. Algebraic Solution Write original equation. 2 Half-angle formula Simplify. Simplify. Pythagorean identity Simplify. Factor. equal to zero, you find 2 sin2 x 2 cos 2 2 sin2 x 2±1 cos x 2 sin2 x 21 cos x 2 x 2 2 2 sin2 x 1 cos x 2 1 cos2 x 1 cos x cos2 x cos x 0 cos xcos x 1 0 By setting the factors and that the solutions in the interval cos x cos x 1 0, 2 are x, 2 x 3, 2 and x 0. Now try Exercise 59. Graphical Solution Use a graphing utility set in radian mode to graph y 2 sin2 x 2 cos2x2, as shown in Figure 5.11. Use the zero or root feature or the zoom and trace features to approximate the intercepts in the interval |
0, 2 to be x- x 0, x 1.571, 2 and x 4.712 3. 2 These values are the approximate solutions of 2 sin2 x 2 cos2x2 0 interval 0, 2. the in 3 2( ) y = 2 − sin2x − 2 cos 2 x − 2 −1 FIGURE 5.11 2 Product-to-Sum Formulas Each of the following product-to-sum formulas is easily verified using the sum and difference formulas discussed in the preceding section. Product-to-Sum Formulas sin u sin v 1 2 cos u cos v 1 2 sin u cos v 1 2 cos u sin v 1 2 cosu v cosu v cosu v cosu v sinu v sinu v sinu v sinu v Product-to-sum formulas are used in calculus to evaluate integrals involving the products of sines and cosines of two different angles. 333202_0505.qxd 12/5/05 9:06 AM Page 412 412 Chapter 5 Analytic Trigonometry Example 8 Writing Products as Sums Rewrite the product cos 5x sin 4x as a sum or difference. Solution Using the appropriate product-to-sum formula, you obtain cos 5x sin 4x 1 2 1 sin5x 4x sin5x 4x 2 sin 9x 1 2 sin x. Now try Exercise 67. Occasionally, it is useful to reverse the procedure and write a sum of trigonometric functions as a product. This can be accomplished with the following sum-to-product formulas. Sum-to-Product Formulas sin u sin v 2 sinu v 2 sin u sin v 2 cosu v 2 cos u cos v 2 cosu v 2 cos u cos v 2 sinu v 2 cosu v 2 sinu v 2 cosu v 2 sinu v 2 For a proof of the sum-to-product formulas, see Proofs in Mathematics on page 426. Example 9 Using a Sum-to-Product Formula Find the exact value of cos 195 cos 105. Solution Using the appropriate sum-to-product formula, you obtain cos 195 cos 105 2 cos195 105 2 cos195 105 2 2 cos 150 cos 45 2 2 2 3 2 6 2. Now try Exercise 83. 333202_0505.qxd 12/7/05 4:13 PM Page 413 y = sin 5x + sin 3x x π3 2 y 2 1 FIGURE 5.12 Section 5.5 Multiple |
-Angle and Product-to-Sum Formulas 413 Example 10 Solving a Trigonometric Equation Solve sin 5x sin 3x 0. Solution 2 sin5x 3x 2 sin 5x sin 3x 0 cos5x 3x 0 2 2 sin 4x cos x 0 Write original equation. Sum-to-product formula Simplify. 2 sin 4x equal to zero, you can find that the solutions in the By setting the factor 0, 2 interval, x 0, are cos x 0 The equation the solutions are of the form yields no additional solutions, and you can conclude that x n 4 is an integer. You can confirm this graphically by sketching the graph of as shown in Figure 5.12. From the graph you can see that n where y sin 5x sin 3x, the x- intercepts occur at multiples of 4. Now try Exercise 87. Example 11 Verifying a Trigonometric Identity Verify the identity sin t sin 3t cos t cos 3t tan 2t. Solution Using appropriate sum-to-product formulas, you have sin t sin 3t cos t cos 3t 2 2 sint 3t 2 cost 3t 2 cost 3t cost 3t 2 2 2 sin2t cost 2 cos2t cost sin 2t cos 2t tan 2t. Now try Exercise 105. 333202_0505.qxd 12/5/05 9:06 AM Page 414 414 Chapter 5 Analytic Trigonometry Application Example 12 Projectile Motion Ignoring air resistance, the range of a projectile fired at an angle with the horizontal and with an initial velocity of feet per second is given by v0 θ FIGURE 5.13 r 1 16 2 sin cos v0 r where is the horizontal distance (in feet) that the projectile will travel. A place kicker for a football team can kick a football from ground level with an initial velocity of 80 feet per second (see Figure 5.13). Not drawn to scale a. Write the projectile motion model in a simpler form. b. At what angle must the player kick the football so that the football travels 200 feet? c. For what angle is the horizontal distance the football travels a maximum? Solution a. You can use a double-angle formula to rewrite the projectile motion model as r 1 32 1 32 22 sin cos v0 Rewrite original projectile motion model. 2 sin 2. v0 Rewrite model using a double-angle formula. b. 2 sin 2 v |
0 r 1 32 200 1 32 200 200 sin 2 1 sin 2 802 sin 2 Write projectile motion model. Substitute 200 for and 80 for r v0. Simplify. Divide each side by 200. You know that 4 45, Because 45 at an angle of c. From the model 2 2, so dividing this result by 2 produces 4. you can conclude that the player must kick the football so that the football will travel 200 feet. r 200 sin 2 r 200 maximum range is 45. to an angle of produce a maximum horizontal distance of 200 feet. you can see that the amplitude is 200. So the feet. From part (b), you know that this corresponds will Therefore, kicking the football at an angle of 45 Now try Exercise 119. W RITING ABOUT MATHEMATICS Deriving an Area Formula Describe how you can use a double-angle formula or a half-angle formula to derive a formula for the area of an isosceles triangle. Use a labeled sketch to illustrate your derivation. Then write two examples that show how your formula can be used. 333202_0505.qxd 12/5/05 9:06 AM Page 415 Section 5.5 Multiple-Angle and Product-to-Sum Formulas 415 5.5 Exercises VOCABULARY CHECK: Fill in the blank to complete the trigonometric formula. 1. sin 2u ________ 3. cos 2u ________ 5. sin u 2 ________ 2. 4. 6. 1 cos 2u 2 1 cos 2u 1 cos 2u u 2 tan ________ ________ ________ 7. cos u cos v ________ sin u sin v ________ sin u cos v ________ cos u cos v ________ PREREQUISITE SKILLS REVIEW: Practice and review algebra skills needed for this section at www.Eduspace.com. 10. 8. 9. In Exercises 1– 8, use the figure to find the exact value of the trigonometric function. 25. tan u 3 4, 0 < u < 2 26. cot u 4, 27. sec u 5 2, 28. csc u 3 < In Exercises 29–34, use the power-reducing formulas to rewrite the expression in terms of the first power of the cosine. 29. 31. 33. cos4 x sin2 x cos2 x sin2 x cos4 x 30. 32. 34. sin8 x |
sin4 x cos4 x sin4 x cos2 x In Exercises 35–40, use the figure to find the exact value of the trigonometric function. 1 θ 4 2. 4. 6. 8. tan sin 2 sec 2 cot 2 1. 3. 5. 7. sin cos 2 tan 2 csc 2 In Exercises 9–18, find the exact solutions of the equation in the interval [0, 2. 9. 11. 13. 15. 17. sin 2x sin x 0 4 sin x cos x 1 cos 2x cos x 0 tan 2x cot x 0 sin 4x 2 sin 2x 10. 12. 14. 16. 18. sin 2x cos x 0 sin 2x sin x cos x cos 2x sin x 0 tan 2x 2 cos x 0 sin 2x cos 2x2 1 In Exercises 19–22, use a double-angle formula to rewrite the expression. 19. 21. 22. 6 sin x cos x 4 8 sin2 x cos x sin xcos x sin x 20. 6 cos2 x 3 In Exercises 23–28, find the exact values of using the double-angle formulas. and tan 2u sin 2u, cos 2u, 35. cos 23. 24., sin u 4 5 cos 15 2 2 2 37. tan 39. csc 8 36. sin 38. sec 40. cot 2 2 2 333202_0505.qxd 12/5/05 9:06 AM Page 416 416 Chapter 5 Analytic Trigonometry In Exercises 41– 48, use the half-angle formulas to determine the exact values of the sine, cosine, and tangent of the angle. 71. 73. sinx y sinx y cos sin 72. 74. sinx y cosx y sin sin 41. 43. 45. 47. 75 112 30 8 3 8 42. 44. 46. 48. 165 67 30 12 7 12 In Exercises 49–54, find the exact values of cosu/2, using the half-angle formulas. and sinu/2, tanu/2 49. 50. 51., sin u 5 13 cos u 3 5 tan 52. cot u 3, < u < 3 2 53. 54. csc u 5, 3 sec < In Exercises 55–58, use the half-angle formulas to simplify the expression. 55. 57. |
2 1 cos 6x 1 cos 8x 1 cos 8x 56. 58. 1 cos 4x 1 cosx 1 2 2 In Exercises 59–62, find all solutions of the equation in [0, 2. the interval Use a graphing utility to graph the equation and verify the solutions. 59. sin x 2 61. cos x 2 cos x 0 sin x 0 60. sin x 2 62. tan x 2 cos x 1 0 sin x 0 In Exercises 63–74, use the product-to-sum formulas to write the product as a sum or difference. 63. 6 sin cos 4 4 65. 67. 69. 10 cos 75 cos 15 cos 4 sin 6 5 cos5 cos 3 64. 66. 68. 70. 5 6 4 cos sin 3 6 sin 45 cos 15 3 sin 2 sin 3 cos 2 cos 4 In Exercises 75–82, use the sum-to-product formulas to write the sum or difference as a product. 76. 78. sin 3 sin sin x sin 5x 75. 77. 79. 80. 81. 82. sin 5 sin 3 cos 6x cos 2x sin sin cos 2 cos cos cos 2 2 sinx sinx 2 2 In Exercises 83–86, use the sum-to-product formulas to find the exact value of the expression. 83. 85. sin 60 sin 30 cos cos 3 4 4 84. cos 120 cos 30 86. sin 5 4 sin 3 4 In Exercises 87–90, find all solutions of the equation in the interval Use a graphing utility to graph the equation and verify the solutions. [0, 2. 87. sin 6x sin 2x 0 88. cos 2x cos 6x 0 89. cos 2x sin 3x sin x 1 0 90. sin2 3x sin2 x 0 In Exercises 91–94, use the figure and trigonometric identities to find the exact value of the trigonometric function in two ways. 3 β 4 α 12 5 91. 93. sin2 sin cos 92. 94. cos2 cos sin In Exercises 95–110, verify the identity. 95. 96. csc 2 csc 2 cos sec 2 sec2 2 sec2 cos2 2 sin2 2 cos 4 cos4 x sin4 x cos 2x 98. 99. sin x cos x2 1 sin 2x 97. 333202_0505.qxd 12/5/05 9:06 |
AM Page 417 100. sin cos 3 3 1 sin 2 2 3 101. 102. 103. 1 cos 10y 2 cos 2 5y cos 3 cos u 2 1 4 sin2 ± 2 tan u sec tan u sin u 104. tan u 2 csc u cot u Section 5.5 Multiple-Angle and Product-to-Sum Formulas 417 120. Geometry The length of each of the two equal sides of an isosceles triangle is 10 meters (see figure). The angle between the two sides is. θ 10 m 10 m tan cot x ± y 2 x y 2 cot 3x cot t 105. 106. 107. 108. 109. sin x ± sin y cos x cos y sin x sin y cos x cos y cos 4x cos 2x sin 4x sin 2x cos t cos 3t sin 3t sin t sin 6 x sin 6 x cos x 110. cos 3 x cos 3 x cos x (a) Write the area of the triangle as a function of 2. (b) Write the area of the triangle as a function of. such that the area is a Determine the value of maximum. Model It M 121. Mach Number The mach number of an airplane is the ratio of its speed to the speed of sound. When an airplane travels faster than the speed of sound, the sound waves form a cone behind the airplane (see figure). The mach number is related to the apex angle of the cone by In Exercises 111–114, use a graphing utility to verify the identity. Confirm that it is an identity algebraically. sin 2 1 M. 111. 112. 113. 114. cos 3 cos3 3 sin2 cos sin 4 4 sin cos 1 2 sin2 cos 4x cos 2x2 sin 3x sin x cos 3x cos xsin 3x sin x tan 2x θ In Exercises 115 and 116, graph the function by hand in the interval by using the power-reducing formulas. [0, 2] f x sin2 x 115. 116. f x) cos2 x (a) Find the angle that corresponds to a mach number In Exercises 117 and 118, write the trigonometric expression as an algebraic expression. 117. sin2 arcsin x 118. cos2 arccos x 119. Projectile Motion The range of a projectile fired at an angle with the horizontal and with an initial velocity of v0 r 1 32 feet per |
second is 2 sin 2 v0 r where is measured in feet. An athlete throws a javelin at 75 feet per second. At what angle must the athlete throw the javelin so that the javelin travels 130 feet? of 1. (b) Find the angle that corresponds to a mach number of 4.5. (c) The speed of sound is about 760 miles per hour. Determine the speed of an object with the mach numbers from parts (a) and (b). (d) Rewrite the equation in terms of. 333202_0505.qxd 12/5/05 9:06 AM Page 418 418 Chapter 5 Analytic Trigonometry 122. Railroad Track When two railroad tracks merge, the overlapping portions of the tracks are in the shapes of circular arcs (see figure). The radius of each arc (in feet) and the angle are related by r x 2 2r sin2. 2 Write a formula for x in terms of cos. r θ r θ x Synthesis True or False? In Exercises 123 and 124, determine whether the statement is true or false. Justify your answer. 123. Because the sine function is an odd function, for a negative number u, sin 2u 2 sin u cos u. 124. sin 1 cos u 2 u 2 u when is in the second quadrant. In Exercises 125 and 126, (a) use a graphing utility to graph the function and approximate the maximum and minimum points on the graph in the interval and (b) solve the trigonometric equation and verify that its solutions are the -coordinates of the maximum and minimum points of x f. (Calculus is required to find the trigonometric equation.) [0, 2 Function Trigonometric Equation 125. x f x 4 sin 2 cos x x 2 cos 2 sin x 0 126. f xcos 2x 2 sin x 2 cos x2 sin x 1 0 127. Exploration Consider the function given by f x sin4 x cos4 x. (a) Use the power-reducing formulas to write the function in terms of cosine to the first power. (b) Determine another way of rewriting the function. Use a graphing utility to rule out incorrectly rewritten functions. (c) Add a trigonometric term to the function so that it becomes a perfect square trinomial. Rewrite the function as a perfect square trinomial minus the term that you added. Use a |
graphing utility to rule out incorrectly rewritten functions. (d) Rewrite the result of part (c) in terms of the sine of a double angle. Use a graphing utility to rule out incorrectly rewritten functions. (e) When you rewrite a trigonometric expression, the result may not be the same as a friend’s. Does this mean that one of you is wrong? Explain. 128. Conjecture Consider the function given by f x 2 sin x2 cos2 1. x 2 (a) Use a graphing utility to graph the function. (b) Make a conjecture about the function that is an identity with f. (c) Verify your conjecture analytically. Skills Review In Exercises 129–132, (a) plot the points, (b) find the distance between the points, and (c) find the midpoint of the line segment connecting the points. 129. 130. 131. 132. 5, 2, 1, 4 4, 3, 6, 10, 4 0, 1 3, 5, 1, 3 1 3, 2 2 3 2 2 In Exercises 133–136, find (if possible) the complement and supplement of each angle. 133. (a) 134. (a) 135. (a) 55 109 18 (b) (b) (b) 162 78 9 20 136. (a) 0.95 (b) 2.76 137. Profit The total profit for a car manufacturer in October was 16% higher than it was in September. The total profit for the 2 months was $507,600. Find the profit for each month. 138. Mixture Problem A 55-gallon barrel contains a mixture with a concentration of 30%. How much of this mixture must be withdrawn and replaced by 100% concentrate to bring the mixture up to 50% concentration? 139. Distance A baseball diamond has the shape of a square in which the distance between each of the consecutive bases is 90 feet. Approximate the straight-line distance from home plate to second base. 333202_050R.qxd 12/5/05 9:07 AM Page 419 5 Chapter Summary What did you learn? Section 5.1 Recognize and write the fundamental trigonometric identities (p. 374). Use the fundamental trigonometric identities to evaluate trigonometric functions, simplify trigonometric expressions, and rewrite trigonometric expressions (p. 375). Section 5.2 Verify trigonometric identities (p. |
382). Section 5.3 Use standard algebraic techniques to solve trigonometric equations (p. 389). Solve trigonometric equations of quadratic type (p. 391). Solve trigonometric equations involving multiple angles (p. 394). Use inverse trigonometric functions to solve trigonometric equations (p. 395). Section 5.4 Use sum and difference formulas to evaluate trigonometric functions, verify identities, and solve trigonometric equations (p. 400). Section 5.5 Use multiple-angle formulas to rewrite and evaluate trigonometric functions (p. 407). Use power-reducing formulas to rewrite and evaluate trigonometric functions (p. 409). Use half-angle formulas to rewrite and evaluate trigonometric functions (p. 410). Use product-to-sum and sum-to-product formulas to rewrite and evaluate trigonometric functions (p. 411). Use trigonometric formulas to rewrite real-life models (p. 414). Chapter Summary 419 Review Exercises 1–6 7–24 25–32 33–38 39–42 43–46 47–50 51–74 75–78 79–82 83–92 93–100 101–106 333202_050R.qxd 12/5/05 9:07 AM Page 420 420 Chapter 5 Analytic Trigonometry 5 Review Exercises In Exercises 1–6, name the trigonometric function 5.1 that is equivalent to the expression. 1. 3. 5. 1 cos x 1 sec x cos x sin x 2. 4. 1 sin x 1 tan x 6. 1 tan 2 x In Exercises 7–10, use the given values and trigonometric identities to evaluate (if possible) all six trigonometric functions. 7. 8. 9. 10. cos x 4 5 13 3 sec, sin x 3 5, tan 2 3 x 2 2 9, sin 2 csc 2, sin x 2 2 sin 45 9 In Exercises 11–22, use the fundamental trigonometric identities to simplify the expression. 11. 13. 15. 17. 18. 19. 21. 1 cot2 x 1 tan2 xcsc2 x 1 sin 2 sin cos2 x cos2 x cot2 x tan2 csc2 tan2 tan x 12 cos x 12. 14. 16. tan 1 cos2 cot2 xsin2 x u cot 2 cos u 20. sec x tan x |
2 1 csc 1 1 csc 1 22. cos2 x 1 sin x 23. Rate of Change The rate of change of the function the expression Show that this expression can also be is given by f x csc x cot x csc2 x csc x cot x. written as 1 cos x sin2 x. 5.2 In Exercises 25–32, verify the identity. 25. 26. 27. 28. 29. 30. 31. 32. cos xtan2 x 1 sec x sec2 x cot x cot x tan x cosx cot 2 sin x x tan x 2 1 tan csc cos cot x 1 tan x csc x sin x sin5 x cos2 x cos2 x 2 cos4 x cos6 x sin x cos3 x sin2 x sin2 x sin4 x cos x 5.3 In Exercises 33–38, solve the equation. 34. 4 cos 1 2 cos 33. 35. 36. 37. 38. sin x 3 sin x 33 tan u 3 1 2 sec x 1 0 3 csc2 x 4 4 tan2 u 1 tan2 u In Exercises 39– 46, find all solutions of the equation in the interval [0, 2. 39. 40. 41. 43. 45. 2 cos2 x cos x 1 2 sin2 x 3 sin x 1 cos2 x sin x 1 2 sin 2x 2 0 cos 4xcos x 1 0 42. 44. 46. sin2 x 2 cos x 2 3 tan 3x 0 3 csc2 5x 4 In Exercises 47–50, use inverse functions where needed to find all solutions of the equation in the interval [0, 2. 47. 49. 50. sin2 x 2 sin x 0 tan2 tan 12 0 sec2 x 6 tan x 4 0 48. 2 cos2 x 3 cos x 0 5.4 In Exercises 51–54, find the exact values of the sine, cosine, and tangent of the angle by using a sum or difference formula. 24. Rate of Change The rate of change of the function f x 2sin x sin12 x cos x. is given by the expression Show that this expression can also be written as cot xsin x. 51. 53. 285 315 30 25 11 6 12 4 52. 54. 345 300 45 19 11 6 12 4 333202_050R.qxd 12/5/ |
05 9:07 AM Page 421 In Exercises 55–58, write the expression as the sine, cosine, or tangent of an angle. 55. 56. 57. sin 60 cos 45 cos 60 sin 45 cos 45 cos 120 sin 45 sin 120 tan 25 tan 10 1 tan 25 tan 10 58. tan 68 tan 115 1 tan 68 tan 115 In Exercises 59–64, find the exact value of the trigonometsin u 3 u ric function given that 4 and are in Quadrant II.) cos v 5 13. (Both and v sinu v cosu v cosu v 59. 61. 63. 60. 62. 64. tanu v sinu v tanu v In Exercises 65–70, verify the identity. sin x 66. 65. 2 cosx cot x tan x 2 cos 3x 4 cos3 x 3 cos x sin cos cos tan tan 67. 69. 70. sinx 3 2 cos x 68. sin x sin x In Exercises 71–74, find all solutions of the equation in the interval [0, 2. 71. 72. 73. 74. sinx cosx sinx 2 cosx 3 4 1 sinx 4 4 1 cosx 6 6 3 sinx cosx 3 0 4 2 5.5 cos 2u, In Exercises 75 and 76, find the exact values of using the double-angle formulas. and tan 2u sin 2u, 75. 76., sin u 4 5 cos < In Exercises 77 and 78, use double-angle formulas to verify the identity algebraically and use a graphing utility to confirm your result graphically. 77. 78. sin 4x 8 cos3 x sin x 4 cos x sin x tan2 x 1 cos 2x 1 cos 2x Review Exercises 421 In Exercises 79– 82, use the power-reducing formulas to rewrite the expression in terms of the first power of the cosine. 79. 81. tan2 2x sin2 x tan2 x 80. 82. cos2 3x cos2 x tan2 x In Exercises 83– 86, use the half-angle formulas to determine the exact values of the sine, cosine, and tangent of the angle. 83. 85. 75 19 12 84. 86. 15 17 12 sinu/2, In Exercises 87–90, find the exact values of cosu/2, 87. using |
the half-angle formulas. tanu/2 5, 0 < u < 2 8, < u < 32 and sin u 3 tan u 5 cos u 2 7, 2 < u < sec u 6, 2 < u < 88. 89. 90. In Exercises 91 and 92, use the half-angle formulas to simplify the expression. 91. 1 cos 10x 2 92. sin 6x 1 cos 6x In Exercises 93–96, use the product-to-sum formulas to write the product as a sum or difference. 93. cos sin 6 6 94. 6 sin 15 sin 45 95. cos 5 cos 3 96. 4 sin 3 cos 2 In Exercises 97–100, use the sum-to-product formulas to write the sum or difference as a product. 97. 98. 99. 100. sin 4 sin 2 cos 3 cos 2 cosx sinx cosx 6 6 sinx 4 4 101. Projectile Motion A baseball leaves the hand of the person at first base at an angle of with the horizontal and 80 feet per second. The ball at an initial velocity of is caught by the person at second base 100 feet away. Find if the range of a projectile is v0 r r 1 32 2 sin 2. v0 333202_050R.qxd 12/5/05 9:07 AM Page 422 422 Chapter 5 Analytic Trigonometry 102. Geometry A trough for feeding cattle is 4 meters long and its cross sections are isosceles triangles with the two equal sides being meter (see figure). The angle between the two sides is a) Write the trough’s volume as a function of. 2 (b) Write the volume of the trough as a function of and such that the volume is determine the value of maximum. Harmonic Motion In Exercises 103–106, use the following information. A weight is attached to a spring suspended vertically from a ceiling. When a driving force is applied to the system, the weight moves vertically from its equilibrium position, and this motion is described by the model y 1.5 sin 8t 0.5 cos 8t where y is the distance from equilibrium (in feet) and the time (in seconds). t is 103. Use a graphing utility to graph the model. 104. Write the model in the form y a 2 b2 sinBt C. 105. Find the amplitude of the oscillations of the weight. 106. Find the frequency of the oscillations |
of the weight. Synthesis True or False? the statement is true or false. Justify your answer. In Exercises 107–110, determine whether 107. If < <, then cos < 0. 2 2 sinx y sin x sin y 4 sinx cosx 2 sin 2x 4 sin 45 cos 15 1 3 108. 109. 110. 111. List the reciprocal identities, quotient identities, and Pythagorean identities from memory. 112. Think About It If a trigonometric equation has an infinite number of solutions, is it true that the equation is an identity? Explain. 113. Think About It Explain why you know from observahas no solution if a sin x b 0 tion that the equation a < b. 114. Surface Area The surface area of a honeycomb is given by the equation S 6hs 3 2 h 2.4 where shown in the figure. sin, 0 < ≤ 90 s23 cos s 0.75 inches, inch, and is the angle θ h = 2.4 in. s = 0.75 in. (a) For what value(s) of is the surface area 12 square inches? (b) What value of gives the minimum surface area? In Exercises 115 and 116, use the graphs of to determine how to change one function to form the identity y1 y2. and y2 y1 115. y1 y2 sec2 2 cot2 x y 4 x 116. y1 y2 cos 3x cos x 2 sin x2 y y1 y2 x π− π y2 π y1 x 4 −4 In Exercises 117 and 118, use the zero or root feature of a graphing utility to approximate the solutions of the equation. y x 3 4 cos x 117. 118. y 2 1 2 x2 3 sin x 2 333202_050R.qxd 12/5/05 9:07 AM Page 423 5 Chapter Test Chapter Test 423 Take this test as you would take a test in class. When you are finished, check your work against the answers given in the back of the book. 1. If tan 3 2 cos < 0,. trigonometric functions of and use the fundamental identities to evaluate the other five 2. Use the fundamental identities to simplify csc2 1 cos2. 3. Factor and simplify sec4 x tan4 x sec2 x tan2 x. 4. Add and simplify cos sin sin cos. 5. Determine |
the values of, 6. Use a graphing utility to graph the functions Make a conjecture about y1 0 ≤ < 2, for which y1 and Verify the result algebraically. tan sec2 1 y2 and cos x sin x tan x y2. is true. sec x. In Exercises 7–12, verify the identity. 7. 9. 11. 12. sin sec tan csc sec sin cos sinn 1n sin, n sin x cos x2 1 sin 2x cot tan is an integer. 8. 10. sec2 x tan2 x sec2 x sec4 x sin x cosx 2 13. Rewrite sin4 x tan2 x 14. Use a half-angle formula to simplify the expression in terms of the first power of the cosine. sin 4 1 cos 4. 15. Write 16. Write 4 cos 2 sin 4 sin 3 sin 4 as a sum or difference. as a product. y (1, 2) u x In Exercises 17–20, find all solutions of the equation in the interval [0, 2. 17. 19. tan2 x tan x 0 4 cos2 x 3 0 18. 20. sin 2 cos 0 csc2 x csc x 2 0 21. Use a graphing utility to approximate the solutions of the equation 3 cos x x 0 accurate to three decimal places. cos 105 22. Find the exact value of using the fact that 105 135 30. FIGURE FOR 23 23. Use the figure to find the exact values of sin 2u, cos 2u, and tan 2u. 24. Cheyenne, Wyoming has a latitude of 41N. At this latitude, the position of the sun at sunrise can be modeled by t 1.4 D 31 sin 2 365 t is the time (in days) and D where represents the number of degrees north or south of due east that the sun rises. Use a graphing utility to determine the days on which the sun is more than north of due east at sunrise. represents January 1. In this model, 20 t 1 25. The heights modeled by h (in feet) of two people in different seats on a Ferris wheel can be h1 28 cos 10t 38 and h2 28 cos10t 38, 0 ≤ t ≤ 2 6 where t is the time (in minutes). When are the two people at the same height? 333202_050R.qxd 12/5/05 9:07 AM Page 424 |
Proofs in Mathematics Sum and Difference Formulas (p. 400) sinu v sin u cos v cos u sin v sinu v sin u cos v cos u sin v cosu v cos u cos v sin u sin v cosu v cos u cos v sin u sin v tanu v tan u tan v 1 tan u tan v tanu v tan u tan v 1 tan u tan 1, 0 ) Proof You can use the figures at the left for the proofs of the formulas for v In the top figure, let be the point (1, 0) and then use and, B x1, y1 and i 1, 2, and 3. for bottom figure, note that arcs AC cosu ± v. to locate the points 2 1 2 yi xi In the have the same length. So, line segments D x3, y3 For convenience, assume that BD and are also equal in length, which implies that on the unit circle. So, 0 < v < u < 2., C x2, y2 BD AC A u and 12 y2 x2 2 2x2 x2 x2 2 y2 1 y2 2 1 2x2 1 1 2x2 x2 2 y3 02 x3 x1 x1 2 2x1x3 2 x3 2 x1 x3 2 y3 1 1 2x1x3 x3x1 Finally, by substituting the values y3 y1 and The for formula u v u v y3y1. x2 you obtain cosu v sin u, sin v, y1 2 2 y3 2 y1 2y1y3 2 2y1y3 2 2x1x3 2 y1 2y1y3 cosu v, cos v, cosu v cos u cos v sin u sin v. can be established by considering cos u, x1 x3 and using the formula just derived to obtain A = (1, 0) x cosu v cosu v cos u cos v sin u sinv cos u cos v sin u sin v. You can use the sum and difference formulas for sine and cosine to prove the formulas for tanu ± v. tanu ± v sinu ± v cosu ± v sin u cos v ± cos u sin v cos u cos v sin u sin v sin u cos v ± cos u sin v cos u cos v cos u cos v sin u sin v cos u cos v Quotient |
identity Sum and difference formulas Divide numerator and denominator by cos u cos v. D x y = (, 3 3 ) 424 333202_050R.qxd 12/5/05 9:07 AM Page 425 sin u cos v cos u cos v cos u cos v cos u cos v ± cos u sin v cos u cos v sin u sin v cos u cos v Write as separate fractions. sin u cos u ± sin v cos v sin v cos v 1 sin u cos u tan u ± tan v 1 tan u tan v Product of fractions Quotient identity Double-Angle Formulas sin 2u 2 sin u cos u tan 2u 2 tan u 1 tan2 u (p. 407) cos 2u cos2 u sin2 u 2 cos2 u 1 1 2 sin2 u Proof To prove all three formulas, let sin 2u sinu u v u in the corresponding sum formulas. sin u cos u cos u sin u 2 sin u cos u cos 2u cosu u cos u cos u sin u sin u cos2 u sin2 u tan 2u tanu u tan u tan u 1 tan u tan u 2 tan u 1 tan2 u Trigonometry and Astronomy Trigonometry was used by early astronomers to calculate measurements in the universe. Trigonometry was used to calculate the circumference of Earth and the distance from Earth to the moon. Another major accomplishment in astronomy using trigonometry was computing distances to stars. Power-Reducing Formulas sin2 u 1 cos 2u 2 (p. 409) cos2 u 1 cos 2u 2 tan2 u 1 cos 2u 1 cos 2u Proof To prove the first formula, solve for cos 2u 1 2 sin2 u, as follows. sin2 u in the double-angle formula cos 2u 1 2 sin2 u 2 sin2 u 1 cos 2u sin2 u 1 cos 2u 2 Write double-angle formula. Subtract cos 2u from and add 2 sin2 u to each side. Divide each side by 2. 425 333202_050R.qxd 12/7/05 4:14 PM Page 426 In a similar way you can prove the second formula, by solving for double-angle formula cos2 u in the cos 2u 2 cos2 u 1. To prove the third formula, use a quotient identity, as follows. tan2 u sin2 u cos2 u 1 cos 2u 2 1 cos 2u 2 1 cos 2u |
1 cos 2u Sum-to-Product Formulas sin u sin v 2 sinu v 2 sin u sin v 2 cosu v 2 cos u cos v 2 cosu v 2 cos u cos v 2 sinu v 2 (p. 412) cosu v 2 sinu v 2 cosu v 2 sinu v 2 Proof To prove the first formula, u x y2 and x u v in the product-to-sum formula. y u v. and Then substitute let v x y2 sin u cos v 1 2 1 2 sinu v sinu v sin x sin y sinx y 2 2 sinx y 2 cosx y 2 cosx y 2 sin x sin y The other sum-to-product formulas can be proved in a similar manner. 426 333202_050R.qxd 12/5/05 9:07 AM Page 427 P.S. Problem Solving This collection of thought-provoking and challenging exercises further explores and expands upon concepts learned in this chapter. 1. (a) Write each of the other trigonometric functions of terms of sin. (b) Write each of the other trigonometric functions of in in terms of cos. 2. Verify that for all integers n, cos2n 1 2 0. 3. Verify that for all integers n, sin12n 1 6 1 2. 4. A particular sound wave is modeled by t 30p2 t p3 t p5 t 30p6 t pt 1 4 p1 t 1 n pn where sin524nt, and t is the time (in seconds). (a) Find the sine components and use a graphing utility to graph each component. Then verify the graph of that is shown. pn p t y 1.4 y = p(t) s u s FIGURE FOR 5 v s w s 6. The path traveled by an object (neglecting air resistance) that feet, an initial velocity is given by is projected at an initial height of of feet per second, and an initial angle h0 v0 y 16 v0 2 cos2 x2 tan x h0 x y where and are measured in feet. Find a formula for the maximum height of an object projected from ground level at velocity and angle To do this, find half of the horizontal distance v0. 1 32 2 sin 2 v0 and then substitute it for where h0 of a projectile 7. Use the figure to derive the formulas for x 0. in the general model |
for the path sin, cos, and tan 2 2 2 t 0.006 where is an acute angle. −1.4 (b) Find the period of each sine component of p. Is p periodic? If so, what is its period? (c) Use the zero or root feature or the zoom and trace features of a graphing utility to find the -intercepts of the graph of over one cycle. p t (d) Use the maximum and minimum features of a graphing utility to approximate the absolute maximum and absolute minimum values of over one cycle. p s 5. Three squares of side are placed side by side (see figure). Make a conjecture about the relationship between the sum u v Prove your conjecture by using the identity for the tangent of the sum of two angles. and w. θ 2 1 1 cos θ θ sin θ (in pounds) on a person’s back when he or she is modeled by 8. The force F bends over at an angle F 0.6W sin 90 sin 12 where W is the person’s weight (in pounds). (a) Simplify the model. (b) Use a graphing utility to graph the model, where W 185 and 0 < < 90. (c) At what angle is the force a maximum? At what angle is the force a minimum? 427 333202_050R.qxd 12/5/05 9:07 AM Page 428 9. The number of hours of daylight that occur at any location on Earth depends on the time of year and the latitude of the location. The following equations model the numbers of and New hours of daylight in Seward, Alaska Orleans, Louisiana 60 latitude n 12. The index of refraction of a transparent material is the ratio of the speed of light in a vacuum to the speed of light in the material. Some common materials and their indices are air (1.00), water (1.33), and glass (1.50). Triangular prisms are often used to measure the index of refraction based on the formula 30 latitude. D 12.2 6.4 cost 0.2 182.6 D 12.2 1.9 cost 0.2 182.6 Seward New Orleans n 2. sin 2 sin 2 For the prism shown in the figure, 60. α θ Air ht L ig Prism (a) Write the index of refraction as a function of cot2. (b) Find for a |
prism made of glass. 13. (a) Write a sum formula for (b) Write a sum formula for sinu v w. tanu v w. 14. (a) Derive a formula for (b) Derive a formula for cos 3. cos 4. h 15. The heights engine can be modeled by (in inches) of pistons 1 and 2 in an automobile h1 3.75 sin 733t 7.5 and h2 3.75 sin 733t 4 3 7.5 where t is measured in seconds. (a) Use a graphing utility to graph the heights of these two pistons in the same viewing window for 0 ≤ t ≤ 1. (b) How often are the pistons at the same height? D In these models, daylight and represents the day, with to January 1. represents the number of hours of t 0 corresponding t (a) Use a graphing utility to graph both models in the same viewing window. Use a viewing window of 0 ≤ t ≤ 365. (b) Find the days of the year on which both cities receive the same amount of daylight. What are these days called? (c) Which city has the greater variation in the number of daylight hours? Which constant in each model would you use to determine the difference between the greatest and least numbers of hours of daylight? (d) Determine the period of each model. 10. The tide, or depth of the ocean near the shore, changes (in feet) of a bay can d throughout the day. The water depth be modeled by d 35 28 cos 6.2 t is the time in hours, with t where 12:00 A.M. t 0 corresponding to (a) Algebraically find the times at which the high and low tides occur. (b) Algebraically find the time(s) at which the water depth is 3.5 feet. (c) Use a graphing utility to verify your results from parts (a) and (b). 11. Find the solution of each inequality in the interval 0, 2. (a) (c) sin x ≥ 0.5 tan x < sin x (b) (d) cos x ≤ 0.5 cos x ≥ sin x 428 333202_0600.qxd 12/5/05 10:38 AM Page 429 Additional Topics in Trigonometry 6.1 6.2 6.3 6.4 6.5 Law of Sines Law of Cosines |
Vectors in the Plane Vectors and Dot Products Trigonometric Form of a Complex Number 66 The work done by a force, such as pushing and pulling objects, can be calculated using vector operations AT I O N S Triangles and vectors have many real-life applications. The applications listed below represent a small sample of the applications in this chapter. • Bridge Design, Exercise 39, page 437 • Glide Path, Exercise 41, page 437 • Surveying, Exercise 31, page 444 • Paper Manufacturing, Exercise 45, page 445 • Revenue, Exercise 65, page 468 • Cable Tension, • Work, Exercises 79 and 80, page 458 Exercise 73, page 469 • Navigation, Exercise 84, page 459 429 333202_0601.qxd 12/5/05 10:40 AM Page 430 430 Chapter 6 Additional Topics in Trigonometry 6.1 Law of Sines What you should learn • Use the Law of Sines to solve oblique triangles (AAS or ASA). • Use the Law of Sines to solve oblique triangles (SSA). • Find the areas of oblique triangles. • Use the Law of Sines to model and solve real-life problems. Why you should learn it You can use the Law of Sines to solve real-life problems involving oblique triangles. For instance, in Exercise 44 on page 438, you can use the Law of Sines to determine the length of the shadow of the Leaning Tower of Pisa. Introduction In Chapter 4, you studied techniques for solving right triangles. In this section and the next, you will solve oblique triangles—triangles that have no right and angles. As standard notation, the angles of a triangle are labeled a, their opposite sides are labeled as shown in Figure 6.1. B,A, and and C, b, c, C b a A c FIGURE 6.1 B To solve an oblique triangle, you need to know the measure of at least one side and any two other parts of the triangle—either two sides, two angles, or one angle and one side. This breaks down into the following four cases. 1. Two angles and any side (AAS or ASA) 2. Two sides and an angle opposite one of them (SSA) 3. Three sides (SSS) 4. Two sides and their included angle (SAS) The first two cases can be solved using the Law of Sines, whereas the last two |
cases require the Law of Cosines (see Section 6.2). Law of Sines If ABC is a triangle with sides a, b, and c, then Hideo Kurihara/Getty Images a sin A b sin B c sin is acute. A is obtuse. The HM mathSpace® CD-ROM and Eduspace® for this text contain additional resources related to the concepts discussed in this chapter. The Law of Sines can also be written in the reciprocal form sin A a sin B b sin C c. For a proof of the Law of Sines, see Proofs in Mathematics on page 489. 333202_0601.qxd 12/5/05 10:40 AM Page 431 C b = 27.4 ft 102.3° a 28.7° c A FIGURE 6.2 When solving triangles, a careful sketch is useful as a quick test for the feasibility of an answer. Remember that the longest side lies opposite the largest angle, and the shortest side lies opposite the smallest angle. C b a 8° 43° Section 6.1 Law of Sines 431 Example 1 Given Two Angles and One Side—AAS For the triangle in Figure 6.2, the remaining angle and sides. C 102.3, B 28.7, and b 27.4 feet. Find B Solution The third angle of the triangle is A 180 B C 180 28.7 102.3 49.0. By the Law of Sines, you have a sin A b sin B c sin C. Using b 27.4 a b sin B and produces sin A 27.4 sin 28.7 sin 49.0 43.06 feet c b sin B sin C 27.4 sin 28.7 sin 102.3 55.75 feet. Now try Exercise 1. Example 2 Given Two Angles and One Side—ASA A pole tilts toward the sun at an angle from the vertical, and it casts a 22-foot shadow. The angle of elevation from the tip of the shadow to the top of the pole is How tall is the pole? 43. 8 Solution From Figure 6.3, note that is C 180 A B A 43 and B 90 8 98. So, the third angle 180 43 98 39. By the Law of Sines, you have a sin A c sin C. B c = 22 ft A Because c 22 FIGURE 6.3 a c sin C feet, the length of the pole is sin A 22 sin |
43 23.84 feet. sin 39 Now try Exercise 35. For practice, try reworking Example 2 for a pole that tilts away from the sun under the same conditions. 333202_0601.qxd 12/5/05 10:40 AM Page 432 432 Chapter 6 Additional Topics in Trigonometry The Ambiguous Case (SSA) In Examples 1 and 2 you saw that two angles and one side determine a unique triangle. However, if two sides and one opposite angle are given, three possible situations can occur: (1) no such triangle exists, (2) one such triangle exists, or (3) two distinct triangles may satisfy the conditions. The Ambiguous Case (SSA) Consider a triangle in which you are given a, b, and A. h b sin A A is acute. A is acute. A is acute. A is acute. A is obtuse. A is obtuse. b h a b ah None One One Two a ≤ b None a b A a > b One Sketch Necessary condition Triangles possible C b = 12 in. 42° A c One solution: FIGURE 6.4 a > b Example 3 Single-Solution Case—SSA a = 22 in. For the triangle in Figure 6.4, the remaining side and angles. a 22 inches, b 12 inches, and A 42. Find Solution By the Law of Sines, you have B sin A a sin B b sin B bsin A a sin B 12sin 42 22 B 21.41. Reciprocal form Multiply each side by b. Substitute for A, a, and b. B is acute. Now, you can determine that C 180 42 21.41 116.59. Then, the remaining side is c sin C a sin A c a sin A sin C 22 sin 42 sin 116.59 29.40 inches. Now try Exercise 19. 333202_0601.qxd 12/5/05 10:40 AM Page 433 a = 15 b = 25 h 85° A No solution: FIGURE 6.5 a < h Section 6.1 Law of Sines 433 Example 4 No-Solution Case—SSA Show that there is no triangle for which a 15, b 25, and A 85. Solution Begin by making the sketch shown in Figure 6.5. From this figure it appears that no triangle is formed. You can verify this using the Law of Sines. sin A a sin B b sin |
B bsin A a sin B 25sin 85 15 1.660 > 1 This contradicts the fact that and sides a 15 b 25 sin B ≤ 1. and an angle of Reciprocal form Multiply each side by b. So, no triangle can be formed having A 85. Now try Exercise 21. Example 5 Two-Solution Case—SSA Find two triangles for which a 12 meters, b 31 meters, and A 20.5. Solution By the Law of Sines, you have sin A a sin B b sin B bsin A a 31sin 20.5 12 0.9047. Reciprocal form B2 64.8, 180 64.8 115.2 you obtain between 0 There are two angles and 64.8 and B1 B1 180 whose sine is 0.9047. For C 180 20.5 64.8 94.7 c a sin C 12 sin A 115.2, B2 you obtain C 180 20.5 115.2 44.3 c a sin C 12 sin A sin 20.5 sin 20.5 For sin 94.7 34.15 meters. sin 44.3 23.93 meters. The resulting triangles are shown in Figure 6.6. b = 31 m a = 12 m 20.5° 64.8° B1 A A FIGURE 6.6 Now try Exercise 23. b = 31 m 20.5° 115.2° B2 a = 12 m 333202_0601.qxd 12/5/05 10:40 AM Page 434 434 Chapter 6 Additional Topics in Trigonometry Area of an Oblique Triangle 180 A To see how to obtain the height of the obtuse triangle in Figure 6.7, notice the use of the reference angle and the difference formula for sine, as follows. h b sin180 A bsin 180 cos A cos 180 sin A b0 cos A 1 sin A b sin A b = 52 m 102° C a = 90 m FIGURE 6.8 The procedure used to prove the Law of Sines leads to a simple formula for the area of an oblique triangle. Referring to Figure 6.7, note that each triangle has a height of h b sin A. Consequently, the area of each triangle is cb sin A 1 2 baseheight 1 2 bc sin A. Area 1 2 By similar arguments, you can develop the formulas Area 1 2 ab sin C 1 2 ac sin B is acute FIGURE 6 |
explaining how to use the Law of Sines to solve each triangle. Is there an easier way to solve these triangles? a. AAS B 50° c = 20 b. ASA B 50° a = 10 C A C A 333202_0601.qxd 12/5/05 10:40 AM Page 436 436 Chapter 6 Additional Topics in Trigonometry 6.1 Exercises The HM mathSpace® CD-ROM and Eduspace® for this text contain step-by-step solutions to all odd-numbered exercises. They also provide Tutorial Exercises for additional help. VOCABULARY CHECK: Fill in the blanks. 1. An ________ triangle is a triangle that has no right angle. 2. For triangle ABC, the Law of Sines is given by 3. The area of an oblique triangle is given by a sin A 2 bc sin A 1 1 ________ c sin C. 2ab sin C ________. PREREQUISITE SKILLS REVIEW: Practice and review algebra skills needed for this section at www.Eduspace.com. In Exercises 1–18, use the Law of Sines to solve the triangle. Round your answers to two decimal places. C c C 105° b 30° b c = 20 C a = 20 45° a 40° B B 14. 15. 16. 17. 18. A 100, A 110 15, C 85 20, A 55, B 28, a 125, c 10 a 48, a 35, b 16 c 50 B 42, C 104, c 3 4 a 35 8 In Exercises 19–24, use the Law of Sines to solve (if possible) the triangle. If two solutions exist, find both. Round your answers to two decimal places. 19. 20. 21. 22. 23. 24. A 110, A 110, A 76, A 76, A 58, A 58, a 125, a 125, a 18, a 34, a 11.4, a 4.5, b 100 b 200 b 20 b 21 b 12.8 b 12.8 b 25° C 135° b a = 3.5 35° B B c a 10° c = 45 b 5 c 10 In Exercises 25–28, find values for b such that the triangle has (a) one solution, (b) two solutions, and (c) no solution. 25. 26. 27. 28. A |
36, A 60, A 10, A 88, a 5 a 10 a 10.8 a 315.6 In Exercises 29–34, find the area of the triangle having the indicated angle and sides. a 8, a 9, A 36, A 60, A 102.4, A 24.3, A 83 20, A 5 40, B 15 30, B 2 45, C 145, C 16.7, C 54.6, a 21.6 c 2.68 C 54.6, c 18.1 b 4.8 B 8 15, a 4.5, b 6.2, b 4, c 14 b 6.8 c 5.8 29. 30. 31. 32. 33. 34. C 120, B 130, A 43 45, A 5 15, B 72 30, C 84 30, a 4, a 62, b 6 c 20 b 57, b 4.5, a 105, a 16, c 85 c 22 c 64 b 20 A A A A 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 333202_0601.qxd 12/5/05 10:40 AM Page 437 Section 6.1 Law of Sines 437 35. Height Because of prevailing winds, a tree grew so that it was leaning 4 from the vertical. At a point 35 meters from the tree, the angle of elevation to the top of the tree is 23 (see figure). Find the height of the tree. h 39. Bridge Design A bridge is to be built across a small lake from a gazebo to a dock (see figure). The bearing from the gazebo to the dock is S W. From a tree 100 meters from the gazebo, the bearings to the gazebo and the dock are S 74 E, respectively. Find the distance from the gazebo to the dock. E and S 41 28 h 94° Tree 74° 28° 100 m W N S E Gazebo 41° 23° 35 m 36. Height A flagpole at a right angle to the horizontal is located on a slope that makes an angle of with the horizontal. The flagpole’s shadow is 16 meters long and points directly up the slope. The angle of elevation from the tip of the shadow to the sun is 20. 12 (a) Draw a triangle that represents the problem. Show the |
known quantities on the triangle and use a variable to indicate the height of the flagpole. (b) Write an equation involving the unknown quantity. (c) Find the height of the flagpole. 37. Angle of Elevation A 10-meter telephone pole casts a 17-meter shadow directly down a slope when the angle of the angle of elevation of the sun is elevation of the ground. (see figure). Find 42, A 10 m C 42° 42° − θ 1 7 m θ B 316 38. Flight Path A plane flies 500 kilometers with a bearing of from Naples to Elgin (see figure). The plane then flies 720 kilometers from Elgin to Canton. Find the bearing of the flight from Elgin to Canton. W N S E Elgin N 720 km 500 km Canton Not drawn to scale 44° Naples Dock 40. Railroad Track Design The circular arc of a railroad curve has a chord of length 3000 feet and a central angle of 40. (a) Draw a diagram that visually represents the problem. Show the known quantities on the diagram and use the s variables and to represent the radius of the arc and the length of the arc, respectively. r r (b) Find the radius of the circular arc. s (c) Find the length of the circular arc. 41. Glide Path A pilot has just started on the glide path for landing at an airport with a runway of length 9000 feet. The angles of depression from the plane to the ends of the runway are 18.8. 17.5 and (a) Draw a diagram that visually represents the problem. (b) Find the air distance the plane must travel until touching down on the near end of the runway. (c) Find the ground distance the plane must travel until touching down. (d) Find the altitude of the plane when the pilot begins the descent. 42. Locating a Fire The bearing from the Pine Knob fire tower to the Colt Station fire tower is N E, and the two towers are 30 kilometers apart. A fire spotted by rangers in N 80 E each tower has a bearing of from Pine Knob and S 70 E from Colt Station (see figure). Find the distance of the fire from each tower. 65 N S W 80° 65° E Colt Station 30 km 70° Fire Pine Knob Not drawn to scale 333202_0601.qxd 12/5/05 10:40 AM Page 438 438 Chapter 6 Additional Topics in Trigonometry 43. Distance A |
boat is sailing due east parallel to the shoreline at a speed of 10 miles per hour. At a given time, the bearing to the lighthouse is S E, and 15 minutes later the bearing is S E (see figure). The lighthouse is located at the shoreline. What is the distance from the boat to the shoreline? 63 70 70° d 63° W N S E Model It 44. Shadow Length The Leaning Tower of Pisa in Italy is characterized by its tilt. The tower leans because it was built on a layer of unstable soil—clay, sand, and water. The tower is approximately 58.36 meters tall from its foundation (see figure). The top of the tower leans about 5.45 meters off center. 5.45 m β α 58.36 m Synthesis True or False? the statement is true or false. Justify your answer. In Exercises 45 and 46, determine whether 45. If a triangle contains an obtuse angle, then it must be oblique. 46. Two angles and one side of a triangle do not necessarily determine a unique triangle. 47. Graphical and Numerical Analysis In the figure, and are positive angles. (a) Write. (b) Use a graphing utility to graph the function. Determine as a function of its domain and range. (c) Use the result of part (a) to write as a function of. (d) Use a graphing utility to graph the function in part (c). c Determine its domain and range. (e) Complete the table. What can you infer? 0.4 0.8 1.2 1.6 2.0 2.4 2.8 c 20 cm θ 2 8 cm θ 30 cm FIGURE FOR 48 γ 9 β 18 α c FIGURE FOR 47 θ d Not drawn to scale 48. Graphical Analysis (a) Find the angle of lean of the tower. (b) Write as a function of d and where, is the angle of elevation to the sun. (c) Use the Law of Sines to write an equation for the length of the shadow cast by the tower. d (d) Use a graphing utility to complete the table. (a) Write the area of the shaded region in the figure as a A function of. (b) Use a graphing utility to graph the area function. (c) Determine the domain of the area function. Explain how the area of the |
region and the domain of the function would change if the eight-centimeter line segment were decreased in length. 10 20 30 40 50 60 Skills Review d In Exercises 49–52, use the fundamental trigonometric identities to simplify the expression. 49. 51. sin x cot x 1 sin2 2 x 50. 52. 1 cot2 2 tan x cos x sec x x 333202_0602.qxd 12/5/05 10:41 AM Page 439 6.2 Law of Cosines Section 6.2 Law of Cosines 439 What you should learn • Use the Law of Cosines to solve oblique triangles (SSS or SAS). • Use the Law of Cosines to model and solve real-life problems. • Use Heron’s Area Formula to find the area of a triangle. Why you should learn it You can use the Law of Cosines to solve real-life problems involving oblique triangles. For instance, in Exercise 31 on page 444, you can use the Law of Cosines to approximate the length of a marsh. Introduction Two cases remain in the list of conditions needed to solve an oblique triangle— SSS and SAS. If you are given three sides (SSS), or two sides and their included angle (SAS), none of the ratios in the Law of Sines would be complete. In such cases, you can use the Law of Cosines. Law of Cosines Standard Form a2 b2 c 2 2bc cos A b2 a 2 c 2 2ac cos B c 2 a 2 b2 2ab cos C Alternative Form cos A b2 c 2 a 2 2bc cos B a 2 c 2 b2 2ac cos C a 2 b2 c 2 2ab For a proof of the Law of Cosines, see Proofs in Mathematics on page 490. Example 1 Three Sides of a Triangle—SSS Find the three angles of the triangle in Figure 6.11. B c = 14 ft b = 19 ft A a = 8 ft C FIGURE 6.11 © Roger Ressmeyer/Corbis Solution It is a good idea first to find the angle opposite the longest side—side case. Using the alternative form of the Law of Cosines, you find that b in this cos B a 2 c 2 b2 2ac 82 142 192 2814 0.45089. Because B is negative, you know that cos B 116.80. At this point, it is simpler to use |
the Law of Sines to determine sin A asin B b is an obtuse angle given by A. 8sin 116.80 0.37583 19 B B Because obtuse angle. So, A A 22.08 is obtuse, must be acute, because a triangle can have, at most, one and C 180 22.08 116.80 41.12. Now try Exercise 1. 333202_0602.qxd 12/5/05 10:41 AM Page 440 440 Chapter 6 Additional Topics in Trigonometry Exploration What familiar formula do you obtain when you use the third form of the Law of Cosines c2 a2 b2 2ab cos C C 90? What is and you let the relationship between the Law of Cosines and this formula? Do you see why it was wise to find the largest angle first in Example 1? Knowing the cosine of an angle, you can determine whether the angle is acute or obtuse. That is, cos > 0 cos < 0 0 < < 90 90 < < 180. Obtuse Acute for for So, in Example 1, once you found that angle was obtuse, you knew that angles A C and were both acute. If the largest angle is acute, the remaining two angles are acute also. B Example 2 Two Sides and the Included Angle—SAS Find the remaining angles and side of the triangle in Figure 6.12. C b = 15 cm FIGURE 6.12 a 115° A c = 10 cm B Solution Use the Law of Cosines to find the unknown side a in the figure. a2 b2 c2 2bc cos A a2 152 102 21510 cos 115 a2 451.79 a 21.26 a 21.26 Because the reciprocal form of the Law of Sines to solve for centimeters, you now know the ratio B. sin Aa and you can use When solving an oblique triangle given three sides, you use the alternative form of the Law of Cosines to solve for an angle. When solving an oblique triangle given two sides and their included angle, you use the standard form of the Law of Cosines to solve for an unknown. sin A a sin B b sin B bsin A a 15sin 115 21.26 0.63945 So, B arcsin 0.63945 39.75 and C 180 115 39.75 25.25. Now try Exercise 3. 333202_0602.qxd |
12/5/05 10:41 AM Page 441 Section 6.2 Law of Cosines 441 Applications Example 3 An Application of the Law of Cosines The pitcher’s mound on a women’s softball field is 43 feet from home plate and the distance between the bases is 60 feet, as shown in Figure 6.13. (The pitcher’s mound is not halfway between home plate and second base.) How far is the pitcher’s mound from first base? Solution In triangle p 60. H 45 HPF, bisects the right angle at Using the Law of Cosines for this SAS case, you have (line HP H ), f 43, and 60 ft 60 ft P h F f = 43 ft 60 ft 45° p = 60 ft H FIGURE 6.13 h2 f 2 p2 2fp cos H 432 602 24360 cos 45º 1800.3 So, the approximate distance from the pitcher’s mound to first base is h 1800.3 42.43 feet. Now try Exercise 31. Example 4 An Application of the Law of Cosines A ship travels 60 miles due east, then adjusts its course northward, as shown in Figure 6.14. After traveling 80 miles in that direction, the ship is 139 miles from its point of departure. Describe the bearing from point to point C. B N S E W A FIGURE 6.14 c = 60 mi Solution You have of Cosines, you have a 80, b 139, and c 60; so, using the alternative form of the Law cos B a 2 c 2 b2 2ac 802 602 1392 28060 0.97094. B arccos0.97094 166.15, So, north from point B to point C is and thus the bearing measured from due or 166.15 90 76.15, N 76.15 E. Now try Exercise 37. 333202_0602.qxd 12/5/05 10:41 AM Page 442 442 Chapter 6 Additional Topics in Trigonometry Historical Note Heron of Alexandria (c. 100 B.C.) was a Greek geometer and inventor. His works describe how to find the areas of triangles, quadrilaterals, regular polygons having 3 to 12 sides, and circles as well as the surface areas and volumes of three-dimensional objects. Heron’s Area Formula The Law of Cosines can be used to establish the following formula for the area of |
a triangle. This formula is called Heron’s Area Formula after the Greek mathematician Heron (c. 100 B.C.). Heron’s Area Formula Given any triangle with sides of lengths triangle is Area ss as bs c where s a b c2. a, b, and c, the area of the For a proof of Heron’s Area Formula, see Proofs in Mathematics on page 491. Example 5 Using Heron’s Area Formula Find the area of a triangle having sides of lengths and c 72 meters. a 43 meters, b 53 meters, Solution Because s a b c2 1682 84, Heron’s Area Formula yields Area ss as bs c 84413112 1131.89 square meters. Now try Exercise 47. You have now studied three different formulas for the area of a triangle. Standard Formula Area 1 Area 1 2 ab sin C 1 Oblique Triangle Heron’s Area Formula Area ss as bs c 2 bh 2 bc sin A 1 2 ac sin B W RITING ABOUT MATHEMATICS The Area of a Triangle Use the most appropriate formula to find the area of each triangle below. Show your work and give your reasons for choosing each formula. 2 ft 50° a. c. 4 ft 2 ft 4 ft b. d. 2 ft 3 ft 4 ft 3 ft 4 ft 5 ft 333202_0602.qxd 12/5/05 10:41 AM Page 443 Section 6.2 Law of Cosines 443 6.2 Exercises VOCABULARY CHECK: Fill in the blanks. 1. If you are given three sides of a triangle, you would use the Law of ________ to find the three angles of the triangle. 2. The standard form of the Law of Cosines for cos B a2 c2 b2 2ac is ________. 3. The Law of Cosines can be used to establish a formula for finding the area of a triangle called ________ ________ Formula. PREREQUISITE SKILLS REVIEW: Practice and review algebra skills needed for this section at www.Eduspace.com. In Exercises 1–16, use the Law of Cosines to solve the triangle. Round your answers to two decimal places. 1. 3. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 2. 4 105° b = 4 |
.5 A a = 10 c B C b = 10 A c = 15 a = 7 B b = 15 30° A C a c = 30 B c 20 c 72 b 14, b 25, a 11, a 55, a 75.4, a 1.42, A 135, A 55, B 10 35, B 75 20, B 125 40, C 15 15, C 43, C 103, c 52 c 1.25 b 52, b 0.75, b 4, b 3, c 9 c 10 c 30 c 9.5 c 32 b 2.15 a 40, a 6.2, a 32, a 6.25, b 7 a 4 9, 9 a 3 b 3 8, 4 In Exercises 17–22, complete the table by solving the parallelogram shown in the figure. (The lengths of the d. ) diagonals are given by and c φ θ a c d b a 5 17. 18. 25 19. 10 20. 40 21. 15 22. b 8 35 14 60 25 c d 45 120 20 80 25 20 35 50 In Exercises 23–28, use Heron’s Area Formula to find the area of the triangle. 23. 24. 25. 26. 27. 28. c 10 c 9 a 5, a 12, a 2.5, a 75.4, a 12.32, a 3.05, b 7, b 15, b 10.2, b 52, b 8.46, b 0.75, c 9 c 52 c 15.05 c 2.45 B,A, 29. Navigation A boat race runs along a triangular course C. marked by buoys and The race starts with the boats headed west for 3700 meters. The other two sides of the course lie to the north of the first side, and their lengths are 1700 meters and 3000 meters. Draw a figure that gives a visual representation of the problem, and find the bearings for the last two legs of the race. 30. Navigation A plane flies 810 miles from Franklin to Then it flies 648 miles Centerville with a bearing of 32. from Centerville to Rosemount with a bearing of Draw a figure that visually represents the problem, and find the straight-line distance and bearing from Franklin to Rosemount. 75. 333202_0602.qxd 12/5/05 10:41 AM Page 444 444 Chapter 6 Additional Topics in Trigonometry 31 |
. Surveying To approximate the length of a marsh, a B, then (see figure). surveyor walks 250 meters from point turns Approximate the length and walks 220 meters to point AC of the marsh. to point C 75 A B 75° 220 m 250 m C A 32. Surveying A triangular parcel of land has 115 meters of frontage, and the other boundaries have lengths of 76 meters and 92 meters. What angles does the frontage make with the two other boundaries? 33. Surveying A triangular parcel of ground has sides of lengths 725 feet, 650 feet, and 575 feet. Find the measure of the largest angle. 34. Streetlight Design Determine the angle in the design of the streetlight shown in the figure. 3 θ 1 4 2 2 35. Distance Two ships leave a port at 9 A.M. One 53 travels at a bearing of N W at 12 miles per hour, and the other travels at a bearing of S W at 16 miles per hour. Approximate how far apart they are at noon that day. 67 36. Length A 100-foot vertical tower is to be erected on the 6 side of a hill that makes a angle with the horizontal (see figure). Find the length of each of the two guy wires that will be anchored 75 feet uphill and downhill from the base of the tower. 37. Navigation On a map, Orlando is 178 millimeters due south of Niagara Falls, Denver is 273 millimeters from Orlando, and Denver is 235 millimeters from Niagara Falls (see figure). 235 mm 235 mm Niagara Falls Niagara Falls Denver Denver 273 mm 273 mm 178 mm 178 mm Orlando Orlando (a) Find the bearing of Denver from Orlando. (b) Find the bearing of Denver from Niagara Falls. 38. Navigation On a map, Minneapolis is 165 millimeters due west of Albany, Phoenix is 216 millimeters from Minneapolis, and Phoenix is 368 millimeters from Albany (see figure). Minneapolis Minneapolis 165 mm 165 mm Albany Albany 216 mm 216 mm Phoenix Phoenix 368 mm 368 mm (a) Find the bearing of Minneapolis from Phoenix. (b) Find the bearing of Albany from Phoenix. 39. Baseball On a baseball diamond with 90-foot sides, the pitcher’s mound is 60.5 feet from home plate. How far is it from the pitcher’s mound to third base? 40. Baseball The baseball player in center field is playing approximately 330 feet from the television camera that is behind home plate. A batter hits a fly ball that goes to |
the wall 420 feet from the camera (see figure). The camera turns to follow the play. Approximately how far does the center fielder have to run to make the catch? 8 100 ft 6° 75 f t 75 f t 330 ft 8° 420 ft 333202_0602.qxd 12/8/05 9:22 AM Page 445 Section 6.2 Law of Cosines 445 41. Aircraft Tracking To determine the distance between two aircraft, a tracking station continuously determines the distance to each aircraft and the angle between them (see figure). Determine the distance between the planes when A 42, a c 20 miles, and b 35 miles. A a C b B c A FIGURE FOR 41 42. Aircraft Tracking Use the figure for Exercise 41 to a between the planes when c 20 miles, and determine the distance A 11, b 20 43. Trusses Q is the midpoint of the line segment in the truss rafter shown in the figure. What are the lengths of the RS? line segments miles. and PQ, QS, PR R Q 10 45. Paper Manufacturing In a process with continuous paper, the paper passes across three rollers of radii 3 inches, 4 inches, and 6 inches (see figure). The centers of the three-inch and six-inch rollers are inches apart, and the length of the arc in contact with the paper on the four-inch roller is inches. Complete the table. d s 3 in. s θ d 4 in. 6 in. d (inches) 9 10 12 13 14 15 16 (degrees) s (inches) 46. Awning Design A retractable awning above a patio door from the exterior wall at a height lowers at an angle of of 10 feet above the ground (see figure). No direct sunlight is to enter the door when the angle of elevation of the sun is greater than What is the length of the awning? 70. 50 x P 8 S 8 8 8 50° x Sun’s rays Model It 44. Engine Design An engine has a seven-inch connecting rod fastened to a crank (see figure). 10 ft 70° 1.5 in. 7 in. θ x (a) Use the Law of Cosines to write an equation giving the relationship between and x (b) Write x. as a function of x. ) yields positive values of. (Select the sign that 47. Geometry The lengths of the sides of a triangular parcel of land |
are approximately 200 feet, 500 feet, and 600 feet. Approximate the area of the parcel. 48. Geometry A parking lot has the shape of a parallelogram (see figure). The lengths of two adjacent sides are 70 meters and 100 meters. The angle between the two sides is 70 What is the area of the parking lot?. (c) Use a graphing utility to graph the function in part (b). (d) Use the graph in part (c) to determine the maximum distance the piston moves in one cycle. 70 m 70° 100 m 333202_0602.qxd 12/5/05 10:41 AM Page 446 446 Chapter 6 Additional Topics in Trigonometry 49. Geometry You want to buy a triangular lot measuring 510 yards by 840 yards by 1120 yards. The price of the land is $2000 per acre. How much does the land cost? (Hint: 1 acre 4840 square yards) 50. Geometry You want to buy a triangular lot measuring 1350 feet by 1860 feet by 2490 feet. The price of the land is $2200 per acre. How much does the land cost? (Hint: 1 acre 43,560 square feet) 57. Proof Use the Law of Cosines to prove that bc 1 cos. 58. Proof Use the Law of Cosines to prove that a b c 2 bc 1 cos A a b c 1 2 2. Synthesis Skills Review True or False? statement is true or false. Justify your answer. In Exercises 51–53, determine whether the In Exercises 59– 64, evaluate the expression without using a calculator. 51. In Heron’s Area Formula, s is the average of the lengths of the three sides of the triangle. 52. In addition to SSS and SAS, the Law of Cosines can be used to solve triangles with SSA conditions. 53. A triangle with side lengths of 10 centimeters, 16 centimeters, and 5 centimeters can be solved using the Law of Cosines. 54. Circumscribed and Inscribed Circles Let r be the radii of the circumscribed and inscribed circles of a ABC, triangle s a b c 2 respectively (see figure), and let and a) Prove that (b) Prove that b c 2R a sin A r s as bs c sin C sin B.. s Circumscribed and Inscribed Circles 56, use the results of Exercise 54. In Exerc |
ises 55 and 55. Given a triangle with a 25, b 55, and c 72 find the areas of (a) the triangle, (b) the circumscribed circle, and (c) the inscribed circle. 56. Find the length of the largest circular running track that can be built on a triangular piece of property with sides of lengths 200 feet, 250 feet, and 325 feet. 59. 60. 61. 62. 63. 64. arcsin1 arccos 0 arctan 3 arctan3 arcsin 2 3 2 arccos 3 In Exercises 65– 68, write an algebraic expression that is equivalent to the expression. 65. 66. 67. 68. secarcsin 2x tanarccos 3x cotarctanx 2 x 1 cosarcsin 2 In Exercises 69–72, use trigonometric substitution to write, the algebraic equation as a trigonometric function of where Then find /2 < < /2. sec csc. and 69. 70. 71. 72. 5 25 x2, 2 4 x2, 3 x2 9, 12 36 x2, x 5 sin x 2 cos x 3 sec x 6 tan In Exercises 73 and 74, write the sum or difference as a product. 73. cos 5 6 74. sinx cos 3 sinx 2 2 333202_0603.qxd 12/5/05 10:42 AM Page 447 6.3 Vectors in the Plane Section 6.3 Vectors in the Plane 447 What you should learn • Represent vectors as directed line segments. • Write the component forms of vectors. • Perform basic vector operations and represent them graphically. • Write vectors as linear combinations of unit vectors. • Find the direction angles of vectors. • Use vectors to model and solve real-life problems. Why you should learn it You can use vectors to model and solve real-life problems involving magnitude and direction. For instance, in Exercise 84 on page 459, you can use vectors to determine the true direction of a commercial jet. Bill Bachman/Photo Researchers, Inc. Introduction Quantities such as force and velocity involve both magnitude and direction and cannot be completely characterized by a single real number. To represent such a quantity, you can use a directed line segment, as shown in Figure 6.15. The PQ Its magnitude directed line segment (or length) is denoted by and terminal point and can be found using the Distance |
Formula. has initial point \ PQ Q. P \ Terminal point Q PQ P Initial point FIGURE 6.15 FIGURE 6.16 Two directed line segments that have the same magnitude and direction are equivalent. For example, the directed line segments in Figure 6.16 are all equivalent. The set of all directed line segments that are equivalent to the directed line segment Vectors are denoted by lowercase, boldface letters such as v is a vector v in the plane, written w. PQ, and vu, PQ. \ \ Example 1 Vector Representation by Directed Line Segments u Let be represented by the directed line segment from v and let S 4, 4, P 0, 0 be represented by the directed line segment from as shown in Figure 6.17. Show that u v. Q to R 1, 2 3, 2, to y 5 4 3 2 1 P (0, 0) (4, 4) v S (3, 2) Q (1, 2) R u 1 2 3 4 x FIGURE 6.17 Solution From the Distance Formula, it follows that PQ \ 3 0 2 2 02 13 \ 4 12 4 22 13 PQ RS \ \ and RS have the same magnitude. Moreover, both line segments have the same direction because they are both directed toward the upper right on lines having a slope of So, have the same magnitude and direction, and it follows that 2 3. u v. and PQ RS \ \ Now try Exercise 1. 333202_0603.qxd 12/8/05 9:23 AM Page 448 448 Chapter 6 Additional Topics in Trigonometry Component Form of a Vector The directed line segment whose initial point is the origin is often the most convenient representative of a set of equivalent directed line segments. This representative of the vector v is in standard position. 0, 0 A vector whose initial point is the origin. v1, v2 can be uniquely represented by This is the component form of a the coordinates of its terminal point vector v, written as v v1, v2 v1 The coordinates the terminal point lie at the origin, 0 0, 0. and v2. are the components of v v. If both the initial point and is the zero vector and is denoted by P p1, p2 and termi- Component Form of a Vector The component form of the vector with initial point nal point Q q1, q2 PQ \ q1 p1, q |
2 is given by v1, v2 p2 v The magnitude (or length) of is given by 2 q2 p2 is a unit vector. Moreover, v q1 v 1, p1 v v. 2 v1 2 v2 2. If vector 0. v 0 if and only if v is the zero u2 and v v1, v2 u u1, u2 Two vectors v2. u PQ For instance, in Example 1, the vector \ 3 0, 2 0 3, 2 R 1, 2 S 4, 4 and the vector v RS \ v from 4 1, 4 2 to 3, 2. is are equal if and only if u to P 0, 0 from v1 u1 Q 3, 2 and is Example 2 Finding the Component Form of a Vector Find the component form and magnitude of the vector 4, 7 and terminal point 1, 5. v that has initial point and let Q 1, 5 q1, q2, as shown in Figure v v1, v2 are v1 Solution P 4, 7 p1, p2 Let 6.18. Then, the components of p1 q1 p2 q2 v2 v 5, 12 v 52 122 169 13. So, 1 4 5 5 7 12. and the magnitude of v is Te c h n o l o g y You can graph vectors with a graphing utility by graphing directed line segments. Consult the user’s guide for your graphing utility for specific instructions. Q = (−1, 5) y 6 2 −8 −6 −4 −2 2 4 6 x −2 −4 −6 −8 v P = (4, −7) FIGURE 6.18 Now try Exercise 9. 333202_0603.qxd 12/5/05 10:42 AM Page 449 v 1 2 v 2v −v − v3 2 Vector Operations Section 6.3 Vectors in the Plane 449 The two basic vector operations are scalar multiplication and vector addition. In operations with vectors, numbers are usually referred to as scalars. In this text, v scalars will always be real numbers. Geometrically, the product of a vector and k has the a scalar v same direction as, as shown in Figure 6.19. times as long as kv has the direction opposite that of k is negative, is the vector that is is positive, v, and if v. If kv k |
k To add two vectors geometrically, position them (without changing their lengths or directions) so that the initial point of one coincides with the terminal is formed by joining the initial point of the secpoint of the other. The sum ond vector with the terminal point of the first vector, as shown in Figure 6.20. This technique is called the parallelogram law for vector addition because the, often called the resultant of vector addition, is the diagonal of a vector parallelogram having and as its adjacent sides FIGURE 6.20 y u v + v u x x and u u1, u2 Definitions of Vector Addition and Scalar Multiplication v v1, v2 Let number). Then the sum of and u v2 k times. k be vectors and let be a scalar (a real v v1, u2 and the scalar multiple of u v u1 is the vector is the vector Scalar multiple ku ku1, u2 ku1, ku2 Sum u The negative of v 1v v v1, v2 is v1, v2 u and the difference of and u v u v u1 v1, u2 Negative v is Add v. See Figure 8.21.. Difference v2 u v To represent same initial point. The difference v geometrically, you can use directed line segments with the is the vector from the terminal point of as shown in Figure 6.21. to the terminal point of, which is equal to u v, u v u FIGURE 6.19 y −v u − v u v u + (−v) u v u v FIGURE 6.21 x 333202_0603.qxd 12/5/05 10:42 AM Page 450 450 Chapter 6 Additional Topics in Trigonometry The component definitions of vector addition and scalar multiplication are illustrated in Example 3. In this example, notice that each of the vector operations can be interpreted geometrically. Example 3 Vector Operations Let v 2, 5 w 3, 4, and w v and find each of the following vectors. v 2w c. a. 2 v b. Solution a. Because v 2, 5, you have 2v 22, 5 22, 25 4, 10. A sketch of 2 is shown in Figure 6.22. v b. The difference of w w v 3 2, 4 5 and is v 5, 1. w v A sketch of vector difference is shown in |
Figure 6.23. Note that the figure shows the w v as the sum w v. c. The sum of and 2 is w v 2w 2, 5 23, 4 v 2, 5 23, 24 2, 5 6, 8 2 6, 5 8 4, 13. v 2w A sketch of is shown in Figure 6.24. y y (3, 4) w −v 4 3 2 1 − ( 4, 10) − ( 2, 5) 10 8 2v 6 4 v −8 −6 −4 −2 2 x −1 FIGURE 6.22 FIGURE 6.23 x w − v 3 4 5 (5, −1) Now try Exercise 21. (4, 13) y 2w v + 2w 14 12 10 8 v (−2, 5) 2 4 6 8 x −6 −4 −2 FIGURE 6.24 333202_0603.qxd 12/5/05 10:42 AM Page 451 Section 6.3 Vectors in the Plane 451 Vector addition and scalar multiplication share many of the properties of ordinary arithmetic. c d w 1. vu,, and Properties of Vector Addition and Scalar Multiplication Let be vectors and let and be scalars. Then the following properties are true. u v v u u 0 u cdu cdu cu v cu cv du cu du 1u u, 0u 0 8. 6. 2. 4. 5. 3. 7. 9. cv c v Property 9 can be stated as follows: the magnitude of the vector cv is the absolute value of c times the magnitude of v. Unit Vectors In many applications of vectors, it is useful to find a unit vector that has the same direction as a given nonzero vector To do this, you can divide by its magnitude to obtain v. v u unit vector v v 1 vv. Unit vector in direction of v Note that same direction as The vector is a scalar multiple of The vector has a magnitude of 1 and the v. is called a unit vector in the direction of v. v. u u u Example 4 Finding a Unit Vector Find a unit vector in the direction of magnitude of 1. v 2, 5 and verify that the result has a Solution The unit vector in the direction of v is v v 2, 5 22 52 1 29 2 29 2, 5., 5 29 This vector has a magnitude of 1 because 5 29 2 29 4 2 2 |
29 25 29 29 29 1. Now try Exercise 31 Historical Note William Rowan Hamilton (1805–1865), an Irish mathematician, did some of the earliest work with vectors. Hamilton spent many years developing a system of vector-like quantities called quaternions. Although Hamilton was convinced of the benefits of quaternions, the operations he defined did not produce good models for physical phenomena. It wasn’t until the latter half of the nineteenth century that the Scottish physicist James Maxwell (1831–1879) restructured Hamilton’s quaternions in a form useful for representing physical quantities such as force, velocity, and acceleration. 333202_0603.qxd 12/5/05 10:42 AM Page 452 Chapter 6 Additional Topics in Trigonometry The unit vectors 1, 0 and 0, 1 are called the standard unit vectors and are denoted by i 1, 0 and j 0, 1 452 y 2 1 j = 〈0, 1〉 i = 〈1, 0〉 1 2 x FIGURE 6.25 (−1, 3) −8 −6 −4 −2 y 8 6 4 −2 −4 −6 4 6 2 u (2, −5) as shown in Figure 6.25. (Note that the lowercase letter is written in boldface to distinguish it from the imaginary number ) These vectors can be used to represent any vector i 1. as follows., v v1, v2 i 0, 1 1, 0 v2 v v1, v2 v1 v1i v2 j v1 and The scalars respectively. The vector sum v2 are called the horizontal and vertical components of v, v1i v2 j is called a linear combination of the vectors and. Any vector in the plane can be written as a linear combination of the standard unit vectors and j. j i i Example 5 Writing a Linear Combination of Unit Vectors u Let be the vector with initial point as a linear combination of the standard unit vectors and and terminal point j. i 2, 5 1, 3. Write u Solution Begin by writing the component form of the vector u. x u 1 2, 3 5 3, 8 3i 8j FIGURE 6.26 This result is shown graphically in Figure 6.26. Now try Exercise 43. Example 6 Vector Operations Let u 3i 8j and let v 2i j. Find 2u 3v. Solution You could |
solve this problem by converting to component form. This, however, is not necessary. It is just as easy to perform the operations in unit vector form. and u v 2u 3v 23i 8j 32i j 6i 16j 6i 3j 12i 19j Now try Exercise 49. 333202_0603.qxd 12/8/05 9:26 AM Page 453 y 1 x y (, ) y = sin θ u θ −1 x = cos θ 1 x −1 FIGURE 6.27 u 1 y 3 2 1 (3, 3) u x x θ = 45° 1 2 3 FIGURE 6.28 y 1 306.87° 1 2 3 4 v −1 −1 −2 −3 −4 (3, −4) FIGURE 6.29 Section 6.3 Vectors in the Plane 453 Direction Angles u is a unit vector such that If the positive -axis to u x, the terminal point of is the angle (measured counterclockwise) from lies on the unit circle and you have u u x, y cos, sin cos i sin j Suppose that as shown in Figure 6.27. The angle u u. is any vector that makes an angle with the positive -axis, it has the same direction as u is the direction angle of the vector v ai bj. is a unit vector with direction angle and you can write If x v v cos, sin v cos i v sin j. it follows that the direction is determined from v ai bj v cos i v sin j, Because v for angle tan sin cos v sin v cos Quotient identity Multiply numerator and denominator by v. b a. Simplify. Example 7 Finding Direction Angles of Vectors Find the direction angle of each vector. a. b. u 3i 3j v 3i 4j Solution a. The direction angle is tan b a 3 3 1. So, 45, as shown in Figure 6.28. b. The direction angle is 4 tan b a 3 v 3i 4j Moreover, because its reference angle is. arctan 4 53.13 53.13. 3 360 53.13 306.87, So, it follows that Now try Exercise 55. lies in Quadrant IV, lies in Quadrant IV and as shown in Figure 6.29. 333202_0603.qxd 12/5/05 10:42 |
AM Page 454 454 Chapter 6 Additional Topics in Trigonometry Applications of Vectors − 100 −75 −50 210° −50 −75 100 FIGURE 6.30 W 15° B C D 15° A FIGURE 6.31 y Example 8 Finding the Component Form of a Vector Find the component form of the vector that represents the velocity of an airplane descending at a speed of 100 miles per hour at an angle 30 below the horizontal, as shown in Figure 6.30. x Solution The velocity vector has a magnitude of 100 and a direction angle of v v v cos i v sin j 100cos 210i 100sin 210j 100 i 1001 2 3 2 j 210. 503 i 50j 503, 50 You can check that has a magnitude of 100, as follows. v v 5032 502 7500 2500 10,000 100 Now try Exercise 77. Example 9 Using Vectors to Determine Weight A force of 600 pounds is required to pull a boat and trailer up a ramp inclined at 15 from the horizontal. Find the combined weight of the boat and trailer. Solution Based on Figure 6.31, you can make the following observations. BA BC AC \ force of gravity combined weight of boat and trailer \ force against ramp \ force required to move boat up ramp 600 pounds BWD and ABC are similar. So, angle ABC, is 15 and By construction, triangles so in triangle sin 15 ABC you have AC \ 600 \ BA \ BA \ 600 sin 15 2318. BA Consequently, the combined weight is approximately 2318 pounds. (In Figure 6.31, note that is parallel to the ramp.) AC \ Now try Exercise 81. 333202_0603.qxd 12/5/05 10:42 AM Page 455 Recall from Section 4.8 that in air navigation, bearings are measured in degrees clockwise from north. Section 6.3 Vectors in the Plane 455 Example 10 Using Vectors to Find Speed and Direction 330 An airplane is traveling at a speed of 500 miles per hour with a bearing of at a fixed altitude with a negligible wind velocity as shown in Figure 6.32(a). When the airplane reaches a certain point, it encounters a wind with a velocity of 70 miles per hour in the direction as shown in Figure 6.32(b).What are the resultant speed and direction of the airplane? N 45 E, y y v1 120° (a) FIGURE 6.32 v2 d W i n v v1 θ x ( |
b) xx Solution Using Figure 6.32, the velocity of the airplane (alone) is v1 500cos 120, sin 120 250, 2503 and the velocity of the wind is 70cos 45, sin 45 352, 352. v2 So, the velocity of the airplane (in the wind) is v2 v v1 250 352, 2503 352 200.5, 482.5 and the resultant speed of the airplane is v 200.52 482.52 miles per hour. 522.5 Finally, if is the direction angle of the flight path, you have tan 482.5 200.5 2.4065 which implies that 180 arctan2.4065 So, the true direction of the airplane is Now try Exercise 83. 180 67.4 337.4. 112.6. 333202_0603.qxd 12/5/05 10:42 AM Page 456 456 Chapter 6 Additional Topics in Trigonometry 6.3 Exercises VOCABULARY CHECK: Fill in the blanks. 1. A ________ ________ ________ can be used to represent a quantity that involves both magnitude and direction. 2. The directed line segment \ PQ has ________ point P 3. The ________ of the directed line segment \ PQ and ________ point PQ. is denoted by Q. 4. The set of all directed line segments that are equivalent to a given directed line segment \ PQ is a ________ v in the plane. 5. The directed line segment whose initial point is the origin is said to be in ________ ________. 6. A vector that has a magnitude of 1 is called a ________ ________. 7. The two basic vector operations are scalar ________ and vector ________. 8. The vector u v 9. The vector sum v2 scalars and v1 is called the ________ of vector addition. v1i v2 j are called the ________ and ________ components of is called a ________ ________ of the vectors and i v, j, and the respectively. PREREQUISITE SKILLS REVIEW: Practice and review algebra skills needed for this section at www.Eduspace.com. In Exercises 1 and 2, show that u v. 1. y 2. 6 4 2 (0, 0) −2 −2 (6, 5) u (2, 4) (4, 1) 4 6 v 2 |
x y 4 (0, 4) u −2 −4 v (3, 3) x 2 4 −4 (−3, −4) (0, −5) In Exercises 3–14, find the component form and the magnitude of the vector v. 3. y 4 3 2 1 5. (3, 2) x 3 4 v 1 2 y (−1, 4) 5 v 3 2 1 (2, 2) −3 −2 −1 21 3 x 4. 6. y 1 −1 −2 −3 x −4 −3 −2 v − − ( 4, 2) (3, 5) y 6 4 2 v −4 −2 (−1, −1) x 2 4 7. y 8. y 4 3 2 1 (3, 3) v −2 −1 −2 −3 1 2 4 5 (3, −2) x −5 (−4, −1) 4 3 2 1 −2 −3 −4 −5 x 4 v (3, −1) Initial Point 1, 5 1, 11 3, 5 3, 11 1, 3 2, 7 9. 10. 11. 12. 13. 14. Terminal Point 15, 12 9, 3 5, 1 9, 40 8, 9 5, 17 In Exercises 15–20, use the figure to sketch a graph of the specified vector. To print an enlarged copy of the graph, go to the website, www.mathgraphs.com. y u v x v 16. 5 u v 18. 20. v 1 2u 15. 17. 19. v u v u 2v 333202_0603.qxd 12/5/05 10:42 AM Page 457 In Exercises 21–28, find (a) 2u 3v., (b) Then sketch the resultant vector. u v Section 6.3 Vectors in the Plane 457 u v, and (c) In Exercises 53–56, find the magnitude and direction angle of the vector v. v 1, 3 v 4, 0 v 0, 0 v 2, 1 v 2i 3j 21. 22. 23. 24. 25. 26. 27. 28. u 2, 1, u 2, 3, u 5, 3, u 0, 0, u i j, u 2i j, u 2i, v j u 3j, v 2i v i 2j In Exercises |
29–38, find a unit vector in the direction of the given vector. 29. 31. 33. 35. 37. u 3, 0 v 2, 2 v 6i 2j w 4j w i 2j 30. 32. 34. 36. 38. u 0, 2 v 5, 12 v i j w 6i w 7j 3i In Exercises 39– 42, find the vector v with the given magnitude and the same direction as u. Magnitude v 5 v 6 v 9 v 10 39. 40. 41. 42. Direction u 3, 3 u 3, 3 u 2, 5 u 10, 0 In Exercises 43–46, the initial and terminal points of a vector are given. Write a linear combination of the standard unit vectors i and j. Initial Point 3, 1 0, 2 1, 5 6, 4 43. 44. 45. 46. Terminal Point 4, 5 3, 6 2, 3 0, 1 In Exercises 47–52, find the component form of and sketch the specified vector operations geometrically, w i 2j. where u 2i j and v 47. 48. 49. 50. 51. 52. v 3 2u v 3 4 w v u 2w v u w v 1 2 v u 2w 3u w 53. 54. 55. 56. v 3cos 60i sin 60j v 8cos 135i sin 135j v 6i 6j v 5i 4j In Exercises 57–64, find the component form of v given its magnitude and the angle it makes with the positive -axis. Sketch v. x Magnitude 32 v 43 v 2 v 3 57. 58. 59. 60. 61. 62. 63. 64. Angle 0 45 150 45 150 90 v in the direction v in the direction i 3j 3i 4j In Exercises 65–68, find the component form of the sum of u and v with direction angles and v. u 65. 66. 67. 68. Magnitude u 5 v 5 u 4 v 4 u 20 v 50 u 50 v 30 Angle 0 90 60 90 45 180 30 110 u v u v u v u v In Exercises 69 and 70, use the Law of Cosines to find the angle between the vectors. ( Assume 0 ≤ ≤ 180. ) 69. 70. v i j, v i 2j, w 2i 2j w 2i j Result |
ant Force In Exercises 71 and 72, find the angle between the forces given the magnitude of their resultant. (Hint: Write force 1 as a vector in the direction of the positive -axis and force 2 as a vector at an angle with the positive -axis.) x x Force 1 71. 45 pounds Force 2 60 pounds Resultant Force 90 pounds 72. 3000 pounds 1000 pounds 3750 pounds 333202_0603.qxd 12/5/05 10:42 AM Page 458 458 Chapter 6 Additional Topics in Trigonometry 73. Resultant Force Forces with magnitudes of 125 newtons and 300 newtons act on a hook (see figure). The angle between the two forces is Find the direction and magnitude of the resultant of these forces. 45. y 125 newtons 45° 300 newtons x 74. Resultant Force Forces with magnitudes of 2000 newtons and 900 newtons act on a machine part at angles of respectively, with the -axis (see figure). Find the direction and magnitude of the resultant of these forces. 45, and 30 x 2000 newtons 30° −45° x 900 newtons 75. Resultant Force Three forces with magnitudes of 75 pounds, 100 pounds, and 125 pounds act on an object at angles of respectively, with the positive x -axis. Find the direction and magnitude of the resultant of these forces. 120, 45, 30, and 76. Resultant Force Three forces with magnitudes of 70 pounds, 40 pounds, and 60 pounds act on an object at angles of respectively, with the x positive -axis. Find the direction and magnitude of the resultant of these forces. 30, 135, 45, 4 and 77. Velocity A ball is thrown with an initial velocity of 70 feet per second, at an angle of with the horizontal (see figure). Find the vertical and horizontal components of the velocity. 35 70 ft sec 35˚ 78. Velocity A gun with a muzzle velocity of 1200 feet per second is fired at an angle of with the horizontal. Find the vertical and horizontal components of the velocity. 6 Cable Tension In Exercises 79 and 80, use the figure to determine the tension in each cable supporting the load. 79. A 50° 30° B 80. 10 in. 20 in. C 2000 lb A 24 in. B C 5000 lb 81. Tow Line Tension A loaded barge is being towed by two tugboats, and the magnitude of the resultant is 6000 pounds directed along the axis of the barge (see figure |
). Find the tension in the tow lines if they each make an angle with the axis of the barge. 18 18° 18° 82. Rope Tension To carry a 100-pound cylindrical weight, two people lift on the ends of short ropes that are tied to an eyelet on the top center of the cylinder. Each rope makes a 20 angle with the vertical. Draw a figure that gives a visual representation of the problem, and find the tension in the ropes. 148, 83. Navigation An airplane is flying in the direction of with an airspeed of 875 kilometers per hour. Because of the wind, its groundspeed and direction are 140, respectively (see figure). 800 kilometers per hour and Find the direction and speed of the wind. y 140° 148° N S E W x d Win 800 kilometers per hour 875 kilometers per hour 333202_0603.qxd 12/5/05 10:42 AM Page 459 Model It 84. Navigation A commercial jet is flying from Miami to Seattle. The jet’s velocity with respect to the air is 580 The wind, at the miles per hour, and its bearing is altitude of the plane, is blowing from the southwest with a velocity of 60 miles per hour. 332. (a) Draw a figure that gives a visual representation of the problem. (b) Write the velocity of the wind as a vector in component form. (c) Write the velocity of the jet relative to the air in component form. (d) What is the speed of the jet with respect to the ground? (e) What is the true direction of the jet? 85. Work A heavy implement is pulled 30 feet across a floor, using a force of 100 pounds. The force is exerted at an above the horizontal (see figure). Find the angle of F work done. (Use the formula for work, is the component of the force in the direction of motion and D is the distance.) W FD, where 50 100 lb 50° 30 ft Tension 45° u 1 lb FIGURE FOR 85 FIGURE FOR 86 86. Rope Tension A tetherball weighing 1 pound is pulled outward from the pole by a horizontal force until the rope angle with the pole (see figure). Determine the makes a resulting tension in the rope and the magnitude of 45 u. u Synthesis True or False? statement is true or false. Justify your answer. In Exercises 87 and 88, decide whether the 87. If and v have the |
same magnitude and direction, then u u v. Section 6.3 Vectors in the Plane 459 (b) If the resultant of the forces is make a conjecture 0, about the angle between the forces. (c) Can the magnitude of the resultant be greater than the sum of the magnitudes of the two forces? Explain. 90. Graphical Reasoning Consider two forces F1 10, 0 and F2 F2 F1 (a) Find 5cos, sin. as a function of. (b) Use a graphing utility to graph the function in part (a) for 0 ≤ < 2. (c) Use the graph in part (b) to determine the range of the function. What is its maximum, and for what value of does it occur? What is its minimum, and for what value of does it occur? (d) Explain why the magnitude of the resultant is never 0. cos i sin j is a unit vector for 91. Proof Prove that any value of. 92. Technology Write a program for your graphing utility that graphs two vectors and their difference given the vectors in component form. In Exercises 93 and 94, use the program in Exercise 92 to find the difference of the vectors shown in the figure. 93. y 94. y 8 6 4 2 (1, 6) (4, 5) (9, 4) (5, 2) 2 4 6 8 125 (−20, 70) x (−100, 0) −50 (80, 80) (10, 60) x 50 Skills Review In Exercises 95–98, use the trigonometric substitution to write the algebraic expression as a trigonometric function of where 0 < < /2., 95. 96. 97. 98. x 2 64, 64 x 2, x 2 36, x 2 253, x 8 sec x 8 sin x 6 tan x 5 sec 88. If u ai bj is a unit vector, then a 2 b2 1. In Exercises 99–102, solve the equation. 89. Think About It Consider two forces of equal magnitude acting on a point. 99. 100. (a) If the magnitude of the resultant is the sum of the magnitudes of the two forces, make a conjecture about the angle between the forces. cos xcos x 1 0 sin x2 sin x 2 0 3 sec x sin x 23 sin x 0 101. 102. cos x csc x cos x2 0 333202 |
_0604.qxd 12/5/05 10:44 AM Page 460 460 Chapter 6 Additional Topics in Trigonometry 6.4 Vectors and Dot Products What you should learn • Find the dot product of two vectors and use the Properties of the Dot Product. • Find the angle between two vectors and determine whether two vectors are orthogonal. • Write a vector as the sum of two vector components. • Use vectors to find the work done by a force. Why you should learn it You can use the dot product of two vectors to solve real-life problems involving two vector quantities. For instance, in Exercise 68 on page 468, you can use the dot product to find the force necessary to keep a sport utility vehicle from rolling down a hill. Edward Ewert The Dot Product of Two Vectors So far you have studied two vector operations—vector addition and multiplication by a scalar—each of which yields another vector. In this section, you will study a third vector operation, the dot product. This product yields a scalar, rather than a vector. Definition of the Dot Product The dot product of and u v u1v1 u u1, u2 u2v2. v v1, v2 is be vectors in the plane or in space and let be a scalar. c Properties of the Dot Product Let 1. 2. 3. vu w,, and. 5. cu v cu v u cv For proofs of the properties of the dot product, see Proofs in Mathematics on page 492. Example 1 Finding Dot Products Find each dot product. 4, 5 2, 3 a. b. 2, 1 1, 2 c. 0, 3 4, 2 Solution a. 4, 5 2, 3 42 53 8 15 23 b. c. 2, 1 1, 2 21 12 2 2 0 0, 3 4, 2 04 32 0 6 6 Now try Exercise 1. In Example 1, be sure you see that the dot product of two vectors is a scalar (a real number), not a vector. Moreover, notice that the dot product can be positive, zero, or negative. 333202_0604.qxd 12/5/05 10:44 AM Page 461 Section 6.4 Vectors and Dot Products 461 Example 2 Using Properties of Dot Products Let a. u 1, 3, u vw v 2, 4, u 2v b. and w 1, 2. Find each dot |
product. Solution Begin by finding the dot product of and u v. u v 1, 3 2, 4 12 34 14 u vw 141, 2 14, 28 u 2v 2u v 214 28 a. b. Notice that the product in part (a) is a vector, whereas the product in part (b) is a scalar. Can you see why? Now try Exercise 11. Example 3 Dot Product and Magnitude u The dot product of with itself is 5. What is the magnitude of? u Solution Because u 2 u u u u u 5. and u u 5, it follows that Now try Exercise 19. The Angle Between Two Vectors The angle between two nonzero vectors is the angle between their respective standard position vectors, as shown in Figure 6.33. This angle can be found using the dot product. (Note that the angle between the zero vector and another vector is not defined.), 0 ≤ ≤, Angle Between Two Vectors If is the angle between two nonzero vectors and u v, then cos u v u v. For a proof of the angle between two vectors, see Proofs in Mathematics on page 492. − v u θ u v Origin FIGURE 6.33 333202_0604.qxd 12/8/05 9:34 AM Page 462 462 Chapter 6 Additional Topics in Trigonometry Example 4 Finding the Angle Between Two Vectors Find the angle between u 4, 3 and v 3, 5. v = 〈3, 5〉 Solution cos u v u v 4, 3 3, 5 4, 3 3, 5 u = 〈4, 3〉 27 534 θ This implies that the angle between the two vectors is FIGURE 6.34 arccos 27 534 22.2 x as shown in Figure 6.34. Now try Exercise 29. Rewriting the expression for the angle between two vectors in the form u v u v cos Alternative form of dot product produces an alternative way to calculate the dot product. From this form, you can will always are always positive, see that because have the same sign. Figure 6.35 shows the five possible orientations of two vectors. u v cos u and and v θ u θ u u v v cos 1 Opposite Direction FIGURE 6.35 < < 2 1 < cos < 0 Obtuse Angle θ v 2 cos 0 90 Angle u θ v 0 < |
< 2 0 < cos < 1 Acute Angle v u 0 cos 1 Same Direction Definition of Orthogonal Vectors The vectors and are orthogonal if u v 0. u v The terms orthogonal and perpendicular mean essentially the same thing—meeting at right angles. Even though the angle between the zero vector and another vector is not defined, it is convenient to extend the definition of orthogonality to include the zero vector. In other words, the zero vector is orthogonal to every vector u, because 0 u 0. 333202_0604.qxd 12/5/05 10:44 AM Page 463 Te c h n o l o g y Example 5 Determining Orthogonal Vectors Section 6.4 Vectors and Dot Products 463 The graphing utility program Finding the Angle Between Two Vectors, found on our website college.hmco.com, graphs two v c, d vectors in standard position and finds the measure of the angle between them. Use the program to verify the solutions for Examples 4 and 5. u a, b and Are the vectors u 2, 3 and v 6, 4 orthogonal? Solution Begin by finding the dot product of the two vectors. 0 u v 2, 3 6, 4 26 34 Because the dot product is 0, the two vectors are orthogonal (see Figure 6.36). y 4 3 2 1 −1 −2 −3 v = 〈6, 42, −3〉 FIGURE 6.36 Now try Exercise 47. Finding Vector Components Consider a boat on an inclined ramp, as shown in Figure 6.37. The force You have already seen applications in which two vectors are added to produce a resultant vector. Many applications in physics and engineering pose the reverse problem—decomposing a given vector into the sum of two vector components. F due to gravity pulls the boat down the ramp and against the ramp. These two orthogonal forces, w2. F w1 are vector components of Vector components of. That is, and w2, w1 F F The negative of component rolling down the ramp, whereas withstand against the ramp. A procedure for finding following page. represents the force needed to keep the boat from represents the force that the tires must is shown on the and w2 w1 w2 w1 w1 w2 F FIGURE 6.37 333202_0604.qxd 12/5/05 10:44 AM Page |
464 464 Chapter 6 Additional Topics in Trigonometry Definition of Vector Components Let and be nonzero vectors such that u v u w1 w1 w2 w2 w1 u projvu. w2 and are orthogonal and w1 where v, as shown in Figure 6.38. The vectors nents of is parallel to (or a scalar multiple of) w1 are called vector compo- w2 and is the projection of onto and is denoted by u. The vector v w1 The vector is given by w2 u w1. w2 u w1 vθ is acute. FIGURE 6.38 w2 θ v u w1 is obtuse. From the definition of vector components, you can see that it is easy to find v. To find the once you have found the projection of onto the component projection, you can use the dot product, as follows. w2 u w2 u w1 u v cv w2 cv w2 v cv v w2 v cv2 0 w1 is a scalar multiple of v. Take dot product of each side with v. w2 and are orthogonal. v So, and c u v v2 projv u cv u v v2 v. w1 Projection of u onto v Let and be nonzero vectors. The projection of onto u u v v is projvu u v v 2v. 333202_0604.qxd 12/5/05 10:44 AM Page 465 v = 〈6, 2〉 Example 6 Decomposing a Vector into Components Section 6.4 Vectors and Dot Products 465 Find the projection of onto two orthogonal vectors, one of which is u 3, 5 v 6, 2. projvu. Then write as the sum of u 1 2 3 4 5 6 w2 u = 〈3, −5〉 x Solution The projection of onto u v is projvu u v v2 v 8 40 6, 2 6 5 2 5, w1 as shown in Figure 6.39. The other component, 9 5 3, 5 6 5 u w1 w2, is., 27 5 w2 2 5, y w1 −1 2 1 −1 −2 −3 −4 −5 FIGURE 6.39 So, u w1 w2 6 5 9 5 2 5, 27 5, 3, 5 |
. Now try Exercise 53. Example 7 Finding a Force A 200-pound cart sits on a ramp inclined at 30, as shown in Figure 6.40. What force is required to keep the cart from rolling down the ramp? Solution Because the force due to gravity is vertical and downward, you can represent the gravitational force by the vector F 200j. Force due to gravity To find the force required to keep the cart from rolling down the ramp, project in the direction of the ramp, as follows. onto a unit vector v F v cos 30i sin 30j 3 2 i 1 2 j Unit vector along ramp Therefore, the projection of onto F v is v 30° w1 F FIGURE 6.40 w1 projvF F v v2 v F vv 2001 v 2 1003 j. i 1 2 2 The magnitude of this force is 100, and so a force of 100 pounds is required to keep the cart from rolling down the ramp. Now try Exercise 67. 333202_0604.qxd 12/5/05 10:44 AM Page 466 466 Chapter 6 Additional Topics in Trigonometry Work The work object is given by W done by a constant force F acting along the line of motion of an W magnitude of forcedistance \ F PQ W proj PQ as shown in Figure 6.41. If the constant force W motion, as shown in Figure 6.42, the work \ F PQ cos F PQ \ F PQ proj PQ \ \ \. Projection form for work F cos F Alternative form of dot product is not directed along the line of F done by the force is given by F P F θ F projPQ P Q Q Force acts along the line of motion. FIGURE 6.41 Force acts at angle with the line of motion. FIGURE 6.42 This notion of work is summarized in the following definition. done by a constant force F as its point of application moves is given by either of the following. Definition of Work The work W along the vector W projPQ W F PQ PQ \ \F PQ 1. 2. \ \ Projection form Dot product form Example 8 Finding Work To close a sliding door, a person pulls on a rope with a constant force of 50 pounds at a constant angle of 60, as shown in Figure 6.43. Find the work done in moving the door 12 feet to its closed position. Solution Using a projection, you can calculate the |
work as follows. Projection form for work W proj PQ \ \F PQ cos 60F PQ \ 1 2 5012 300 foot-pounds So, the work done is 300 foot-pounds. You can verify this result by finding the vectors and calculating their dot product. and PQ F \ 12 ft P projPQF 60° Q F 12 ft FIGURE 6.43 Now try Exercise 69. 333202_0604.qxd 12/5/05 10:44 AM Page 467 Section 6.4 Vectors and Dot Products 467 6.4 Exercises VOCABULARY CHECK: Fill in the blanks. 1. The ________ ________ of two vectors yields a scalar, rather than a vector. 2. If is the angle between two nonzero vectors and cos ________. 3. The vectors and are ________ if u v 4. The projection of onto u v is given by 5. The work W W ________ is given by done by a constant force or v, then u u v 0. projvu ________. F W ________. as its point of application moves along the vector \ PQ PREREQUISITE SKILLS REVIEW: Practice and review algebra skills needed for this section at www.Eduspace.com. In Exercises 1–8, find the dot product of and u v. 1. 3. 5. 7. u 6, 1 v 2, 3 u 4, 1 v 2, 3 u 4i 2j v i j u 3i 2j v 2i 3j 2. 4. 6. 8. u 5, 12 v 3, 2 u 2, 5 v 1, 2 u 3i 4j v 7i 2j u i 2j v 2i j 32. u 2i 3j v 4i 3j 31. 33. 34. u 5i 5j v 6i 6j i sin u cos j 3 3 i sin3 v cos3 j 4 4 i sin u cos j 4 4 i sin v cos j 2 2 In Exercises 35–38, graph the vectors and find the degree measure of the angle between the vectors. 35. 37. u 3i 4j v 7i 5j u 5i 5j v 8i 8j 36. 38. u 6i 3j v 4i 4j u 2i 3j v 8i 3j In Exercises |
39–42, use vectors to find the interior angles of the triangle with the given vertices. 39. 41. 1, 2, 3, 4, 2, 5 3, 0, 2, 2, 0, 6) 40. 42. 3, 4, 3, 5, 1, 7, 1, 9, 8, 2 7, 9 where is the angle between In Exercises 9–18, use the vectors and whether the result is a vector or a scalar. u <2, 2>, v <3, 4>, to find the indicated quantity. State w <1, 2> u u u vv 3w vu w 1 u v u w 9. 11. 13. 15. 17. 10. 12. 14. 16. 18. 3u v v uw u 2vw 2 u v u w v In Exercises 19–24, use the dot product to find the magnitude of u. 19. 21. 23. u 5, 12 u 20i 25j u 6j 20. 22. 24. u 2, 4 u 12i 16j u 21i In Exercises 25 –34, find the angle between the vectors. 43. 25. 27. 29. u 1, 0 v 0, 2 u 3i 4j v 2j u 2i j v 6i 4j 26. 28. 30. u 3, 2 v 4, 0 u 2i 3j v i 2j u 6i 3j v 8i 4j u v, In Exercises 43–46, find u and v. u 4, v 10, 2 3 44. u 100, v 250, 6 45. u 9, v 36, 3 4 46. u 4, v 12, 3 333202_0604.qxd 12/5/05 10:44 AM Page 468 468 Chapter 6 Additional Topics in Trigonometry In Exercises 47–52, determine whether and are orthogonal, parallel, or neither. v u 47. 49. 51. u 12, 30 v 1 2, 5 4 3i j u 1 4 v 5i 6j u 2i 2j v i j 48. 50. 52. u 3, 15 v 1, 5 u i v 2i 2j u cos, sin v sin, cos In Exercises 53–56, find the projection of onto. Then write as the sum of two orthogonal vectors, one of which |
is u projv u. v u 53. 55. u 2, 2 v 6, 1 u 0, 3 v 2, 15 54. 56. u 4, 2 v 1, 2 u 3, 2 v 4, 1 In Exercises 57 and 58, use the graph to determine mentally the projection of onto. (The coordinates of the terminal points of the vectors in standard position are given.) Use the formula for the projection of onto to verify your result. u u v v 57. y 58. y 6 (−2, 3) u −2 −2 (6, 4) v x 2 4 6 (6, 4) v x 2 6 4 (2, −3) u 4 2 −2 −2 −4 In Exercises 59–62, find two vectors in opposite directions (There are many that are orthogonal to the vector correct answers.) u. 59. 60. 61. 62. u 3, 5 u 8 3j Work In Exercises 63 and 64, find the work done in if the magnitude and moving a particle from Q v. direction of the force are given by to P 63. 64. P 0, 0, Q 4, 7, v 1, 4 P 1, 3, Q 3, 5, v 2i 3j 65. Revenue The vector u 1650, 3200 gives the numbers of units of two types of baking pans produced by a company. The vector gives the prices (in dollars) of the two types of pans, respectively. v 15.25, 10.50 (a) Find the dot product u v and interpret the result in the context of the problem. (b) Identify the vector operation used to increase the prices by 5%. 66. Revenue The vector u 3240, 2450 gives the numbers of hamburgers and hot dogs, respectively, sold at a fast-food stand in one month. The vector gives the prices (in dollars) of the food items. v 1.75, 1.25 (a) Find the dot product u v and interpret the result in the context of the problem. (b) Identify the vector operation used to increase the prices by 2.5%. Model It 67. Braking Load A truck with a gross weight of 30,000 (see figure). Assume pounds is parked on a slope of that the only force to overcome is the force of gravity. d d° Weight = 30,000 lb (a) Find the force required to |
keep the truck from rolling down the hill in terms of the slope d. (b) Use a graphing utility to complete the table 10 d Force d Force (c) Find the force perpendicular to the hill when d 5. 68. Braking Load A sport utility vehicle with a gross weight of 5400 pounds is parked on a slope of Assume that the only force to overcome is the force of gravity. Find the force required to keep the vehicle from rolling down the hill. Find the force perpendicular to the hill. 10. 333202_0604.qxd 12/8/05 10:08 AM Page 469 69. Work Determine the work done by a person lifting a 25-kilogram (245-newton) bag of sugar. Synthesis Section 6.4 Vectors and Dot Products 469 70. Work Determine the work done by a crane lifting a 2400-pound car 5 feet. 71. Work A force of 45 pounds exerted at an angle of 30 above the horizontal is required to slide a table across a floor (see figure). The table is dragged 20 feet. Determine the work done in sliding the table. 45 lb 30° 20 ft 72. Work A tractor pulls a log 800 meters, and the tension in the cable connecting the tractor and log is approximately 1600 kilograms (15,691 newtons). The direction of the 35 above the horizontal. Approximate the work force is done in pulling the log. 73. Work One of the events in a local strongman contest is to pull a cement block 100 feet. One competitor pulls the block by exerting a force of 250 pounds on a rope attached to the block at an angle of with the horizontal (see figure). Find the work done in pulling the block. 30 30˚ 100 ft Not drawn to scale 74. Work A toy wagon is pulled by exerting a force of 25 pounds on a handle that makes a angle with the horizontal (see figure). Find the work done in pulling the wagon 50 feet. 20 20° True or False? the statement is true or false. Justify your answer. In Exercises 75 and 76, determine whether 75. The work done by a constant force line of motion of an object is represented by a vector. acting along the W F 76. A sliding door moves along the line of vector If a force is applied to the door along a vector that is orthogonal to then no work is done. PQ PQ., \ \ 77. Think About It |
What is known about v between two nonzero vectors condition? and u, the angle, under each (a) u v 0 u v > 0 78. Think About It What can be said about the vectors and u v < 0 (b) (c) u v under each condition? (a) The projection of onto equals u v u. (b) The projection of onto equals u v 0. 79. Proof Use vectors to prove that the diagonals of a rhombus are perpendicular. 80. Proof Prove the following. u v 2 u2 v 2 2u v Skills Review In Exercises 81–84, perform the operation and write the result in standard form. 81. 82. 83. 84. 42 24 18 112 3 8 12 96 In Exercises 85–88, find all solutions of the equation in the interval [0, 2. 85. 86. 87. 88. sin 2x 3 sin x 0 sin 2x 2 cos x 0 2 tan x tan 2x cos 2x 3 sin x 2 find the exact value of the and sin u 12 13 (Both and are in Quadrant IV.) v u 89. In Exercises 89–92, trigonometric function given that cos v 24 25. sinu v sinu v cosv u 91. 92. tanu v 90. 333202_0605.qxd 12/5/05 10:45 AM Page 470 470 Chapter 6 Additional Topics in Trigonometry 6.5 Trigonometric Form of a Complex Number What you should learn • Plot complex numbers in the complex plane and find absolute values of complex numbers. • Write the trigonometric forms of complex numbers. • Multiply and divide complex numbers written in trigonometric form. • Use DeMoivre’s Theorem to find powers of complex numbers. n numbers. • Find th roots of complex Why you should learn it You can use the trigonometric form of a complex number to perform operations with complex numbers. For instance, in Exercises 105–112 on page 480, you can use the trigonometric forms of complex numbers to help you solve polynomial equations. The Complex Plane Just as real numbers can be represented by points on the real number line, you can represent a complex number z a bi a, b in a coordinate plane (the complex plane). The horizontal axis as the point is called the real axis and the vertical axis is called the imaginary axis, as shown in Figure 6 |
.44. Imaginary axis 3 2 1 (3, 1) or 3 + i 1 2 3 Real axis −3 −2 −1 (−2, −1) or −2 − i −1 −2 FIGURE 6.44 The absolute value of the complex number between the origin 0, 0 and the point a, b. a bi is defined as the distance Definition of the Absolute Value of a Complex Number The absolute value of the complex number z a bi is a bi a2 b2. If the complex number a bi is a real number (that is, if b 0 ), then this definition agrees with that given for the absolute value of a real number a 0i a2 02 a. Example 1 Finding the Absolute Value of a Complex Number Plot z 2 5i and find its absolute value. Solution The number is plotted in Figure 6.45. It has an absolute value of Imaginary axis 5 4 3 (−2, 5) 29 −4 −3 −2 −1 1 2 3 4 Real axis z 22 52 29. FIGURE 6.45 Now try Exercise 3. 333202_0605.qxd 12/5/05 10:45 AM Page 471 Section 6.5 Trigonometric Form of a Complex Number 471 Imaginary axis Trigonometric Form of a Complex Number (a, b) b r a θ Real axis In Section 2.4, you learned how to add, subtract, multiply, and divide complex numbers. To work effectively with powers and roots of complex numbers, it is helpful to write complex numbers in trigonometric form. In Figure 6.46, consider the nonzero complex number By letting be the angle from the positive real axis (measured counterclockwise) to the line segment connecting the origin and the point you can write a bi. a, b, a r cos and b r sin where r a2 b2. Consequently, you have a bi r cos r sin i FIGURE 6.46 from which you can obtain the trigonometric form of a complex number. Trigonometric Form of a Complex Number The trigonometric form of the complex number z a bi is z rcos i sin where number a r cos, b r sin, r a2 b2, tan ba. and is called an argument of z. The r is the modulus of z, and The trigonometric form of a complex number is also called the polar form. the trigonometric form of a is restricted to the interval Because |
there are infinitely many choices for complex number is not unique. Normally, 0 ≤ < 2, although on occasion it is convenient to use < 0., Example 2 Writing a Complex Number in Trigonometric Form Write the complex number z 2 23i in trigonometric form. Solution The absolute value of z is r 2 23i 22 232 16 4 Imaginary axis and the reference angle is given by −3 −2 π 4 3 Real axis 1 z = 4 z = −2 − 2 3 i FIGURE 6.47 −2 −3 − 4 tan b a 23 2 tan3 3 Because choose to be 3. and because 3 43. z 2 23i So, the trigonometric form is lies in Quadrant III, you z rcos i sin 4cos i sin 4 3 4. 3 See Figure 6.47. Now try Exercise 13. 333202_0605.qxd 12/5/05 10:45 AM Page 472 472 Chapter 6 Additional Topics in Trigonometry Example 3 Writing a Complex Number in Standard Form Write the complex number in standard form a bi. z 8cos i sin 3 3 Te c h n o l o g y A graphing utility can be used to convert a complex number in trigonometric (or polar) form to standard form. For specific keystrokes, see the user’s manual for your graphing utility. Solution Because cos3 1 2 sin3 32, you can write and i sin i 3 3 2 3 z 8cos 221 2 6i. 2 Now try Exercise 35. Multiplication and Division of Complex Numbers The trigonometric form adapts nicely to multiplication and division of complex numbers. Suppose you are given two complex numbers cos 1 i sin 1 and z2 r2 cos 2 i sin 2. z1 r1 The product of r1r2 r1r2 z1z2 is given by z1 and cos 1 cos z 2 i sin 1 1 cos 2 cos i sin 2 isin sin 1 sin 2 2 1 cos 2 cos 1 sin 2. Using the sum and difference formulas for cosine and sine, you can rewrite this equation as z1z2 r1r2 cos 1 2 i sin 1 2. This establishes the first part of the following rule. The second part is left for you to verify (see Exercise 117). Product and Quotient of Two Complex Numbers z1 and Let numbers. i sin 1 i sin 2 cos 1 |
cos 2 r2 r1 z2 be complex z1z2 z1 z2 r1r2 r1 r2 cos 1 2 i sin 1 2 cos 1 2 i sin 1 2, z 2 0 Product Quotient Note that this rule says that to multiply two complex numbers you multiply moduli and add arguments, whereas to divide two complex numbers you divide moduli and subtract arguments. 333202_0605.qxd 12/5/05 10:45 AM Page 473 Section 6.5 Trigonometric Form of a Complex Number 473 Example 4 Multiplying Complex Numbers Find the product 2cos z1 z1z2 2 3 Solution of the complex numbers. i sin 2 3 8cos z 2 11 6 i sin 11 6 Te c h n o l o g y z1z2 Some graphing utilities can multiply and divide complex numbers in trigonometric form. If you have access to such a graphing utility, use z1 it to find in Examples 4 and 5. and z1z2 z2 2cos 2 3 16cos i sin i sin 8cos 2 3 11 i sin 6 11 6 11 2 6 3 11 6 Multiply moduli and add arguments. 2 3 5 2 5 2 2 16cos i sin 16cos 2 i sin 160 i1 16i You can check this result by first converting the complex numbers to the standard z1 forms and then multiplying algebraically, as in Section 2.4. 1 3i 43 4i and z2 z1z2 1 3i43 4i 43 4i 12i 43 16i Now try Exercise 47. Example 5 Dividing Complex Numbers Find the quotient z 2 24cos 300 i sin 300 z1 of the complex numbers. 8cos 75 i sin 75 z 2 z1 Solution z1 z2 24cos 300 i sin 300 8cos 75 i sin 75 24 8 cos300 75 i sin300 75 Divide moduli and subtract arguments. 3cos 225 i sin 225 3 i 2 2 2 2 32 2 i 32 2 Now try Exercise 53. 333202_0605.qxd 12/5/05 10:45 AM Page 474 474 Chapter 6 Additional Topics in Trigonometry Powers of Complex Numbers The trigonometric form of a complex number is used to raise a complex number to a power. To accomplish this, consider repeated use of the multiplication rule. z rcos i sin z 2 rcos i sin rcos i sin r 2cos 2 i sin |
2 z3 r 2cos 2 i sin 2rcos i sin r 3cos 3 i sin 3 z4 r 4cos 4 i sin 4 z5 r5cos 5 i sin 5... This pattern leads to DeMoivre’s Theorem, which is named after the French mathematician Abraham DeMoivre (1667–1754). DeMoivre’s Theorem z rcos i sin If then is a complex number and n is a positive integer, zn rcos i sin n rncos n i sin n. Example 6 Finding Powers of a Complex Number Use DeMoivre’s Theorem to find 1 3i12. Solution First convert the complex number to trigonometric form using r 12 32 2 and arctan 3 1 2. 3 So, the trigonometric form is z 1 3i 2cos 2 3 i sin 2. 3 Then, by DeMoivre’s Theorem, you have Historical Note Abraham DeMoivre (1667–1754) is remembered for his work in probability theory and DeMoivre’s Theorem. His book The Doctrine of Chances (published in 1718) includes the theory of recurring series and the theory of partial fractions. 1 3i12 2cos 12 2 3 i sin 212cos 2 3 122 3 4096cos 8 i sin 8 40961 0 4096. i sin 122 3 Now try Exercise 75. 333202_0605.qxd 12/5/05 10:45 AM Page 475 Section 6.5 Trigonometric Form of a Complex Number 475 Roots of Complex Numbers Recall that a consequence of the Fundamental Theorem of Algebra is that a polynomial equation of degree has solutions in the complex number system. has six solutions, and in this particular case you can find So, the equation the six solutions by factoring and using the Quadratic Formula. x6 1 n n x 6 1 x3 1x3 1 x 1x 2 x 1x 1x 2 x 1 0 Consequently, the solutions are x ±1, x 1 ± 3i 2, and x 1 ± 3 i 2. Each of these numbers is a sixth root of 1. In general, the nth root of a complex number is defined as follows. Definition of the nth Root of a Complex Number The complex number is an nth root of the complex number u a bi z if z un a bin. n Exploration The th roots of a |
complex number are useful for solving some polynomial equations. For instance, explain how you can use DeMoivre’s Theorem to solve the polynomial equation x4 16 0. 16 [Hint: Write as 16cos i sin. ] To find a formula for an th root of a complex number, let be an th root n n u of where z, u scos i sin and z rcos i sin. By DeMoivre’s Theorem and the fact that un z, you have sn cos n i sin n rcos i sin. Taking the absolute value of each side of this equation, it follows that Substituting back into the previous equation and dividing by you get r, sn r. cos n i sin n cos i sin. So, it follows that cos n cos and sin n sin. Because both sine and cosine have a period of solutions if and only if the angles differ by a multiple of must exist an integer such that k 2, these last two equations have Consequently, there 2. n 2k 2k n. By substituting this value of stated on the following page. into the trigonometric form of you get the result u, 333202_0605.qxd 12/5/05 10:45 AM Page 476 476 Chapter 6 Additional Topics in Trigonometry Finding nth Roots of a Complex Number For a positive integer n exactly distinct th roots given by 2k n n nrcos 2k n the complex number i sin n, z rcos i sin has where k 0, 1, 2,..., n 1. When exceeds k n 1, the roots begin to repeat. For instance, if k n, the angle 2n n n 2 is coterminal with n, which is also obtained when n k 0. z nr, The formula for the th roots of a complex number has a nice geometrical n th roots of all interpretation, as shown in Figure 6.48. Note that because the nr have the same magnitude with center at they all lie on a circle of radius the origin. Furthermore, because successive th roots have arguments that differ by 2n, You have already found the sixth roots of 1 by factoring and by using the Quadratic Formula. Example 7 shows how you can solve the same problem with the formula for th roots. roots are equally spaced around the circle. the n n n z Example 7 Finding the nth Roots of |
a Real Number Find all the sixth roots of 1. Solution First write 1 in the trigonometric form root formula, with 61cos r 1, 0 2k 6 0 2k 6 i sin n 6 and 1 1cos 0 i sin 0. the roots have the form k 3 cos i sin k 3 Then, by the th n. So, for 1, 2, 3, 4, and 5, the sixth roots are as follows. (See Figure 6.49.) k 0, cos 0 i sin 0 1 cos i sin 3 1 2 3 2 i Increment by 2 n 2 6 3 3 2 3 cos 2 3 i sin 1 2 cos i sin sin i sin 3 2 i 3 2 i 3 2 i cos cos Now try Exercise 97. Imaginary axis rn π 2 n π 2 n Real axis FIGURE 6.48 Imaginary axis − 1 + 0i −1 1 + 0i 1 Real axis − − 1 2 3 2 i FIGURE 6.49 − 1 2 3 2 i 333202_0605.qxd 12/5/05 10:45 AM Page 477 Section 6.5 Trigonometric Form of a Complex Number 477 In Figure 8.49, notice that the roots obtained in Example 7 all have a magnitude of 1 and are equally spaced around the unit circle. Also notice that the complex roots occur in conjugate pairs, as discussed in Section 2.5. The distinct th roots of 1 are called the nth roots of unity. n n Example 8 Finding the nth Roots of a Complex Number Find the three cube roots of z 2 2i. Solution z Because lies in Quadrant II, the trigonometric form of z is z 2 2i 8 cos 135 i sin 135. arctan22 135 By the formula for th roots, the cube roots have the form 68 cos Finally, for 68cos n 135 360k 3 k 0, 1, and 2, 135 3600 3 68cos 135 3601 3 1 + i 68cos 135 3602 3 Real axis 1 2 See Figure 6.50. i sin 135º 360k 3. you obtain the roots i sin 135 3600 3 2cos 45 i sin 45 i sin 135 3601 3 1 i 2cos 165 i sin 165 i sin 135 3602 3 1.3660 0.3660i 2cos 285 i sin 285 0.3660 1.3660i. 0.3660 − 1.3660i Now try Exercise |
103. W RITING ABOUT MATHEMATICS A Famous Mathematical Formula The famous formula Imaginary axis −1.3660 + 0.3660i 1 −1 −2 −2 FIGURE 6.50 is Note in Example 8 that the z absolute value of r 2 2i 22 22 8 and the angle tan b a 2 2 1. is given by ea bi e acos b i sin b is called Euler’s Formula, after the Swiss mathematician Leonhard Euler (1707–1783). Although the interpretation of this formula is beyond the scope of this text, we decided to include it because it gives rise to one of the most wonderful equations in mathematics. ei 1 0 This elegant equation relates the five most famous numbers in mathematics—0, 1, e,, and —in a single equation. Show how Euler’s Formula can be used to derive this equation. i 333202_0605.qxd 12/8/05 10:09 AM Page 478 478 Chapter 6 Additional Topics in Trigonometry 6.5 Exercises VOCABULARY CHECK: Fill in the blanks. 1. The ________ ________ of a complex number and the point a, b. 2. The ________ ________ of a complex number is the ________ of and where z r 3. ________ Theorem states that if zn r ncos n i sin n. then a bi is the distance between the origin 0, 0 z a bi is given by z r cos i sin, is the ________ of z r cos i sin z. is a complex number and n is a positive integer, 4. The complex number u a bi is an ________ ________ of the complex number z if z un a bin. PREREQUISITE SKILLS REVIEW: Practice and review algebra skills needed for this section at www.Eduspace.com. In Exercises 1–6, plot the complex number and find its absolute value. 1. 3. 5. 7i 4 4i 6 7i 2. 4. 6. 7 5 12i 8 3i In Exercises 7–10, write the complex number in trigonometric form. 7. Imaginary axis 8. Imaginary axi s z = 3i 4 3 2 1 −2 −1 1 2 Real axis z = 2− 4 2 −6 −4 −2 2 −4 Real axis 9. Imaginary axis |
Real axis 3 = 3 − i z −3 10. Imaginary axis 3 i = 1 + 3 − z 25. 27. 29. 3 i 5 2i 8 53 i 26. 28. 30. 1 3i 8 3i 9 210 i In Exercises 31– 40, represent the complex number graphically, and find the standard form of the number. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 3cos 120 i sin 120 5cos 135 i sin 135 cos 300 i sin 300 3 2 cos 225 i sin 225 1 4 3 3 3.75cos 4 4 5 5 12 12 6cos i sin i sin 2 i sin 8cos 2 7cos 0 i sin 0 3cos18 45 i sin18 45 6cos230º 30 i sin230º 30 −3 −2 −1 Real axis In Exercises 41– 44, use a graphing utility to represent the complex number in standard form. In Exercises 11–30, represent the complex number graphically, and find the trigonometric form of the number. 11. 13. 15. 17. 19. 3 3i 3 i 21 3 i 5i 7 4i 21. 7 23. 3 3 i 12. 14. 16. 18. 20. 2 2i 4 43 i 3 i 5 2 4i 3 i 22. 4 24. 22 i 9 i sin 5cos 3cos 165.5 i sin 165.5 9cos 58º i sin 58º 9 41. 43. 44. 42. 10cos 2 5 i sin 2 5 In Exercises 45 and 46, represent the powers graphically. Describe the pattern. z, z2, z3, and z 4 45. z 2 2 1 i 46. z 1 2 1 3 i 333202_0605.qxd 12/8/05 10:09 AM Page 479 Section 6.5 Trigonometric Form of a Complex Number 479 In Exercises 47–58, perform the operation and leave the result in trigonometric form. 6cos 4 4cos i sin 12 3 i sin 4 cos 60 i sin 60 12 3 4 2cos 47. i sin 4 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 3 3 3 cos 3 i sin 4 cos 140 i sin 1402 5 3 0.5cos 100 i sin 100 0.8cos 300 i sin 300 0. |
45cos 310i sin 310 0.60cos 200 i sin 200 cos 5 i sin 5cos 20 i sin 20 cos 50 i sin 50 cos 20 i sin 20 2cos 120 i sin 120 4cos 40 i sin 40 cos53 i sin53 cos i sin 5cos 4.3 i sin 4.3 4cos 2.1 i sin 2.1 12cos 52 i sin 52 3cos 110 i sin 110 6cos 40 i sin 40 7cos 100 i sin 100 In Exercises 59–66, (a) write the trigonometric forms of the complex numbers, (b) perform the indicated operation using the trigonometric forms, and (c) perform the indicated operation using the standard forms, and check your result with that of part (b). 60. 62. 64. 3 i1 i 41 3 i 1 3 i 6 3i 59. 61. 63. 65. 66. 2 2i1 i 2i1 i 3 4i 1 3 i 5 2 3i 4i 4 2i In Exercises 67–70, sketch the graph of all complex numbers z satisfying the given condition. 67. 68. 69. 70. z 2 z 3 6 5 4 71. 73. In Exercises 71–88, use DeMoivre’s Theorem to find the indicated power of the complex number. Write the result in standard form. 1 i5 1 i10 23 i7 5cos 20 i sin 203 3cos 150 i sin 1504 cos 2 2i6 3 2i8 41 3 i3 i sin 77. 75. 79. 74. 76. 72. 78. 2 2 4 i sin 12 4 8 2cos 5cos 3.2 i sin 3.24 cos 0 i sin 020 3 2i5 5 4i3 3cos 15 i sin 154 2cos 10 i sin 108 5 2cos 10 6 2cos i sin i sin 10 8 8 80. 81. 82. 83. 84. 85. 86. 87. 88. In Exercises 89–104, (a) use the theorem on page 476 to find the indicated roots of the complex number, (b) represent each of the roots graphically, and (c) write each of the roots in standard form. 89. Square roots of 90. Square roots of 91. Cube roots of 5cos 120 i sin 120 16cos 60 i sin 60 8cos i sin 2 3 5 6 2 3 5 6 92. Fifth roots |
of 93. Square roots of 94. Fourth roots of 95. Cube roots of 96. Cube roots of i sin 32cos 25i 625i 1 3 i 125 2 421 i 97. Fourth roots of 16 98. Fourth roots of i 99. Fifth roots of 1 100. Cube roots of 1000 125 4 102. Fourth roots of 101. Cube roots of 103. Fifth roots of 1281 i 104. Sixth roots of 64i 333202_0605.qxd 12/8/05 10:09 AM Page 480 480 Chapter 6 Additional Topics in Trigonometry In Exercises 105–112, use the theorem on page 476 to find all the solutions of the equation and represent the solutions graphically. 105. 106. 107. 108. 109. 110. 111. 112. x 4 i 0 x3 1 0 x5 243 0 x3 27 0 x 4 16i 0 x6 64i 0 x3 1 i 0 x 4 1 i 0 Synthesis True or False? the statement is true or false. Justify your answer. In Exercises 113–116, determine whether 113. Although the square of the complex number is given by the absolute value of the complex number bi bi2 b2, z a bi is defined as a bi a2 b2. 114. Geometrically, the th roots of any complex number are all equally spaced around the unit circle centered at the origin. n z Graphical Reasoning graph of the roots of a complex number. In Exercises 123 and 124, use the (a) Write each of the roots in trigonometric form. (b) Identify the complex number whose roots are given. (c) Use a graphing utility to verify the results of part (b). 123. Imaginary axis 30° 2 2 30° 2 1 −1 Real axis 124. Imaginary axis 45° 45° 3 3 3 45° 3 45° Real axis Skills Review In Exercises 125–130, solve the right triangle shown in the figure. Round your answers to two decimal places. 115. The product of two complex numbers z1 r1 cos 1 i sin 1 and z2 r2 cos 2 i sin 2 0 116. By DeMoivre’s Theorem, is zero only when r1. and/or 0. r2 4 6 i8 cos32 i sin86. z1 117. Given two complex numbers 0, i sin 2 cos 2 z2 z2, cos i sin r |
1 1 1 show that r2 r1 r2 z1 z 2 cos 1 2 i sin 1 2. B a C 125. 127. 129. and a 8 b 112.6 A 22, A 30, A 42 15, c 11.2 c b 126. 128. 130. A a 33.5 b 211.2 B 66, B 6, B 81 30, c 6.8 Harmonic Motion In Exercises 131–134, for the simple harmonic motion described by the trigonometric function, find the maximum displacement and the least positive value of for which t d 0. 118. Show that conjugate of z r cos i sin z r cos i sin. is the complex 119. Use the trigonometric forms of and z z in Exercise 118 to find (a) zz and (b) zz, z 0. 120. Show that the negative of z r cos i sin. 1 3 i 121. Show that 1 2 2141 i is a sixth root of 1. 2. is a fourth root of 122. Show that z r cos i sin 131. d 16 cos t 4 133. d 1 16 sin t 5 4 132. 134. d 1 8 d 1 12 cos 12t sin 60t is In Exercises 135 and 136, write the product as a sum or difference. 135. 6 sin 8 cos 3 136. 2 cos 5 sin 2 333202_060R.qxd 12/5/05 10:48 AM Page 481 6 Chapter Summary What did you learn? Section 6.1 Use the Law of Sines to solve oblique triangles (AAS, ASA, or SSA) (p. 430, 432). Find areas of oblique triangles (p. 434). Use the Law of Sines to model and solve real-life problems (p. 435). Section 6.2 Use the Law of Cosines to solve oblique triangles (SSS or SAS) (p. 439). Use the Law of Cosines to model and solve real-life problems (p. 441). Use Heron's Area Formula to find areas of triangles (p. 442). Section 6.3 Represent vectors as directed line segments (p. 447). Write the component forms of vectors (p. 448). Perform basic vector operations and represent vectors graphically (p. 449). Write vectors as linear combinations of unit vectors (p. 451). Find the direction angles of vectors (p. 453 |
). Use vectors to model and solve real-life problems (p. 454). Section 6.4 Find the dot product of two vectors and use the properties of the dot product (p. 460). Find the angle between two vectors and determine whether two vectors are orthogonal (p. 461). Write vectors as sums of two vector components (p. 463). Use vectors to find the work done by a force (p. 466). Section 6.5 Plot complex numbers in the complex plane and find absolute values of complex numbers (p. 470). Write the trigonometric forms of complex numbers (p. 471). Multiply and divide complex numbers written in trigonometric form (p. 472). Use DeMoivre’s Theorem to find powers of complex numbers (p. 474) Find nth roots of complex numbers (p. 475). Chapter Summary 481 Review Exercises 1–12 13–16 17–20 21–28 29–32 33–36 37, 38 39–44 45–56 57–62 63–68 69–72 73–80 81–88 89–92 93–96 97–100 101–104 105, 106 107–110 111–118 333202_060R.qxd 12/5/05 10:48 AM Page 482 482 Chapter 6 Additional Topics in Trigonometry 6 Review Exercises 6.1 In Exercises 1–12, use the Law of Sines to solve (if possible) the triangle. If two solutions exist, find both. Round your answers to two decimal places. 1. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. B 2. 71° a = 8 A C c 35° b A B c 22° 121° a = 17 C b B 72, B 10, A 16, A 95, A 24, B 64, B 150, B 150, A 75, B 25, C 82, C 20, B 98, B 45, C 48, C 36, b 30, a 10, a 51.2, a 6.2, b 54 c 33 c 8.4 c 104.8 b 27.5 a 367 c 10 b 3 b 33.7 b 4 In Exercises 13–16, find the area of the triangle having the indicated angle and sides. c 7 c 8 13. 14. 15. 16. A 27, B 80, C |
123, A 11, b 5, a 4, a 16, b 22, b 5 c 21 17. Height From a certain distance, the angle of elevation to 17. At a point 50 meters closer to Approximate the the top of a building is the building, the angle of elevation is height of the building. 31. 18. Geometry Find the length of the side w of the parallelogram. w 140° 12 16 19. Height A tree stands on a hillside of slope from the horizontal. From a point 75 feet down the hill, the angle of elevation to the top of the tree is (see figure). Find the height of the tree. 45 28 5 ft 7 45° 28° FIGURE FOR 19 20. River Width A surveyor finds that a tree on the opposite bank of a river, flowing due east, has a bearing of N E from a certain point and a bearing of N W from a point 400 feet downstream. Find the width of the river. 22 30 15 In Exercises 21–28, use the Law of Cosines to solve 6.2 the triangle. Round your answers to two decimal places. 21. 22. 23. 24. 25. 26. 27. 28. c 10 a 5, a 80, a 2.5, a 16.4, B 110, B 150, C 43, A 62, b 8, b 60, b 5.0, b 8.8, a 4, a 10, a 22.5, b 11.34, c 100 c 4.5 c 12.2 c 4 c 20 b 31.4 c 19.52 29. Geometry The lengths of the diagonals of a parallelogram are 10 feet and 16 feet. Find the lengths of the sides of the 28. parallelogram if the diagonals intersect at an angle of 30. Geometry The lengths of the diagonals of a parallelogram are 30 meters and 40 meters. Find the lengths of the sides of the parallelogram if the diagonals intersect at an angle of 34. 31. Surveying To approximate the length of a marsh, a B. Then C and walks 300 meters to point (see surveyor walks 425 meters from point 65 the surveyor turns figure). Approximate the length of the marsh. to point AC A B 65° 300 m 425 m C A 333202_060R.qxd 12/5/05 10:48 AM Page 483 32. |
Navigation Two planes leave Raleigh-Durham Airport at approximately the same time. One is flying 425 miles per, and the other is flying 530 miles hour at a bearing of 67. per hour at a bearing of Draw a figure that gives a visual representation of the problem and determine the distance between the planes after they have flown for 2 hours. 355 In Exercises 33–36, use Heron’s Area Formula to find the area of the triangle. 33. 34. 35. 36. a 4, a 15, a 12.3, a 38.1, b 5, b 8, c 7 c 10 b 15.8, b 26.7, c 3.7 c 19.4 6.3 In Exercises 37 and 38, show that u v. 37. y 6 4 − ( 2, 1) (4, 6) u (6, 3) v −2 −2 (0, 2)− x 6 38. y (−3, 2) u 4 2 −4 (−1, −4) (1, 4) v x 2 4 (3, − 2) In Exercises 39– 44, find the component form of the vector v satisfying the conditions. 39. y 6 4 2 (−5, 4) v −4 −2 x (2, −1) 40. y 6 4 2 −2 ( 7 26, ) v (0, 1) 2 4 6 x 41. Initial point: terminal point: terminal point: 7, 3 15, 9 43. 42. Initial point: v 8, v 1 2, 44. 0, 10; 1, 5; 120 225 In Exercises 45–52, find (a) (d) 2v 5u. u v, (b) u v, (c) 3u, and 45. 46. 47. 48. 49. u 1, 3, v 3, 6 u 4, 5, v 0, 1 u 5, 2, v 4, 4 u 1, 8, v 3, 2 u 2i j, v 5i 3j Review Exercises 483 50. 51. 52. u 7i 3j, v 4i j u 4i, v i 6j u 6j, v i j and In Exercises 53–56, find the component form of sketch the specified vector operations geometrically, where v 1 i 3j. u 6i 5j and w 53. 54. 55 |
. 56. w 2u v w 4u 5v w 3v w 1 2v In Exercises 57– 60, write vector as a linear combination of the standard unit vectors and u j. i 57. 58. 59. u 3, 4 u 6, 8 u has initial point 60. u has initial point 3, 4 2, 7 and terminal point and terminal point 9, 8. 5, 9. In Exercises 61 and 62, write the vector vcos i sin j. v in the form 61. 62. v 10i 10j v 4i j In Exercises 63–68, find the magnitude and the direction angle of the vector v. 63. 64. 65. 66. 67. 68. v 7cos 60i sin 60j v 3cos 150i sin 150j v 5i 4j v 4i 7j v 3i 3j v 8i j 69. Resultant Force Forces with magnitudes of 85 pounds and 50 pounds act on a single point. The angle between the forces is Describe the resultant force. 15. 70. Rope Tension A 180-pound weight is supported by two ropes, as shown in the figure. Find the tension in each rope. 30° 30° 180 lb 333202_060R.qxd 12/8/05 10:10 AM Page 484 484 Chapter 6 Additional Topics in Trigonometry 71. Navigation An airplane has an airspeed of 430 miles per The wind velocity is 35 miles per E. Find the resultant speed hour at a bearing of 30 hour in the direction of N and direction of the airplane. 135. 72. Navigation An airplane has an airspeed of 724 kilometers per hour at a bearing of The wind velocity is 32 kilometers per hour from the west. Find the resultant speed and direction of the airplane. 30. 6.4 In Exercises 73–76, find the dot product of u. and v. 73. 75. u 6, 7 v 3, 9 u 3i 7j v 11i 5j 74. 76. u 7, 12 v 4, 14 u 7i 2j v 16i 12j In Exercises 77– 80, use the vectors and to find the indicated quantity. State whether u <3, 4> the result is a vector or a scalar. 77. v <2, 1> 2u u v2 uu v 3u v 78. 80. 79. In Exercises 81– 84 |
, find the angle between the vectors. u cos i sin v cos i sin cos 45i sin 45j v cos 300i sin 300j u 22, 4, u 3, 3, v 4, 33 v 2, 1 81. 82. 83. 84. In Exercises 85–88, determine whether orthogonal, parallel, or neither. u and v are 85. 87. u 3, 8 v 8, 3 u i v i 2j 86. 88. u 1 4, 1 2 v 2, 4 u 2i j v 3i 6j In Exercises 89–92, find the projection of onto Then write as the sum of two orthogonal vectors, one of which is u projvu. v. u Work In Exercises 93 and 94, find the work done in Q if the magnitude and moving a particle from v. direction of the force are given by to P 93. 94. P 5, 3, Q 8, 9, v 2, 7 P 2, 9, Q 12, 8, v 3i 6j 95. Work Determine the work done by a crane lifting an 18,000-pound truck 48 inches. 96. Work A mover exerts a horizontal force of 25 pounds on a crate as it is pushed up a ramp that is 12 feet long and inclined at an angle of above the horizontal. Find the work done in pushing the crate. 20 In Exercises 97–100, plot the complex number and 6.5 find its absolute value. 97. 99. 100. 7i 5 3i 10 4i 98. 6i In Exercises 101–104, write the complex number in trigonometric form. 101. 103. 5 5i 33 3i 102. 104. 5 12i 7 In Exercises 105 and 106, (a) write the two complex (b) use the numbers 0. z2 where trigonometric forms to find in trigonometric form, and z1/ z2, and 105. 106. z1 z1 23 2i, 31 i, z2 z2 z1z2 10i 23 i In Exercises 107–110, use DeMoivre’s Theorem to find the indicated power of the complex number. Write the result in standard form. 5cos i sin 107. 4 12 4 15 5 i sin 12 4 2cos 15 2 3i 6 1 i 8 108. 109. 110. In Exerc |
ises 111–114, (a) use the theorem on page 476 to find the indicated roots of the complex number, (b) represent each of the roots graphically, and (c) write each of the roots in standard form. v 8, 2 111. Sixth roots of 89. 90. 91. 92. u 4, 3, u 5, 6, u 2, 7, u 3, 5, v 10, 0 v 1, 1 v 5, 2 729i 256i 112. Fourth roots of 113. Cube roots of 8 114. Fifth roots of 1024 333202_060R.qxd 12/8/05 10:10 AM Page 485 In Exercises 115–118, use the theorem on page 476 to find all solutions of the equation and represent the solutions graphically. 129. Give a geometric description of the scalar multiple ku of the vector u, for k > 0 and for k < 0. 130. Give a geometric description of the sum of the vectors u Review Exercises 485 115. 116. 117. 118. x4 81 0 x5 32 0 x3 8i 0 x3 1x2 1 0 Synthesis and v. Graphical Reasoning graph of the roots of a complex number. In Exercises 131 and 132, use the (a) Write each of the roots in trigonometric form. (b) Identify the complex number whose roots are given. (c) Use a graphing utility to verify the results of part (b). True or False? the statement is true or false. Justify your answer. In Exercises 119–123, determine whether 131. Imaginary axis 119. The Law of Sines is true if one of the angles in the triangle is a right angle. 120. When the Law of Sines is used, the solution is always unique. 121. If u is a unit vector in the direction of v, then v v u. 122. If v a i bj 0, then a b. 123. x 3 i is a solution of the equation x2 8i 0. 124. State the Law of Sines from memory. 125. State the Law of Cosines from memory. 126. What characterizes a vector in the plane? 127. Which vectors in the figure appear to be equivalent? y 2 4 −2 −2 4 60° 60° 4 132. Imaginary axis 3 30° 4 60° 4 4 60° 3 30 |
° 4 Real axis Real axis C B A x D E 133. The figure shows z1 and Describe z2. z1z2 and z1 z2. Imaginary axis z2 1 θ z1 θ −1 1 Real axis u 128. The vectors and have the same magnitudes in the two figures. In which figure will the magnitude of the sum be greater? Give a reason for your answer. v (a) y (b) y v u v x u x 134. One of the fourth roots of a complex number z is shown in the figure. (a) How many roots are not shown? (b) Describe the other roots. Imaginary axis 1 −1 1 z 30° Real axis 333202_060R.qxd 12/8/05 10:11 AM Page 486 486 Chapter 6 Additional Topics in Trigonometry 6 Chapter Test Take this test as you would take a test in class. When you are finished, check your work against the answers given in the back of the book. In Exercises 1–6, use the information to solve the triangle. If two solutions exist, find both solutions. Round your answers to two decimal places. 1. 3. 5. A 24, A 24, B 100, B 68, a 11.2, a 15, a 12.2 b 13.4 b 23 2. 4. 6. B 104, a 4.0, C 123, C 33, a 18.1 b 7.3, a 41, c 12.4 b 57 7. A triangular parcel of land has borders of lengths 60 meters, 70 meters, and 82 meters. Find the area of the parcel of land. 240 mi C 37° B 8. An airplane flies 370 miles from point to point with a bearing of B 240 miles from point and bearing from point A B to point with a bearing of A C to point C. 24. It then flies (see figure). Find the distance 37 In Exercises 9 and 10, find the component form of the vector conditions. v satisfying the given 370 mi 10. Magnitude of v: 9. Initial point of v: 3, 7; v 12; terminal point of v: 11, 16 direction of v: u 3, 5 24° A FIGURE FOR 8 In Exercises 11–13, its graph. u <3, 5> and v <7, 1>. Find the resultant vector and sketch 11. u v 12 |
. u v 13. 5u 3v 14. Find a unit vector in the direction of u 4, 3. 15. Forces with magnitudes of 250 pounds and 130 pounds act on an object at angles of, respectively, with the -axis. Find the direction and magnitude of the x and 45 60 resultant of these forces. 16. Find the angle between the vectors u 1, 5 and v 3, 2. 17. Are the vectors 18. Find the projection of orthogonal vectors. u 6, 10 and u 6, 7 v 2, 3 onto orthogonal? v 5, 1. Then write as the sum of two u 19. A 500-pound motorcycle is headed up a hill inclined at What force is required to 12. keep the motorcycle from rolling down the hill when stopped at a red light? 20. Write the complex number 21. Write the complex number z 5 5i z 6cos 120 i sin 120 in trigonometric form. in standard form. In Exercises 22 and 23, use DeMoivre’s Theorem to find the indicated power of the complex number. Write the result in standard form. 8 3cos 3 3i6 i sin 23. 22. 7 6 7 6 24. Find the fourth roots of 2561 3 i. 25. Find all solutions of the equation x3 27i 0 and represent the solutions graphically. 333202_060R.qxd 12/5/05 10:48 AM Page 487 6 Cumulative Test for Chapter 4–6 Cumulative Test for Chapters 4–6 487 y 4 −3 −4 FIGURE FOR 7 Take this test to review the material from earlier chapters. When you are finished, check your work against the answers given in the back of the book. 1. Consider the angle 120. (a) Sketch the angle in standard position. (b) Determine a coterminal angle in the interval 0, 360. (c) Convert the angle to radian measure. (d) Find the reference angle. (e) Find the exact values of the six trigonometric functions of. 2. Convert the angle 2.35 radians to degrees. Round the answer to one decimal place. 3. Find cos if tan 4 3 and sin < 0. x 1 3 In Exercises 4 –6, sketch the graph of the function. (Include two full periods.) 4. f x 3 2 sin x 5. gx 1 2 tanx 2 6 |
. hx secx 7. Find a, b, graph in the figure. c and such that the graph of the function hx a cosbx c matches the 8. Sketch the graph of the function f x 1 2x sin x over the interval 3 ≤ x ≤ 3. In Exercises 9 and 10, find the exact value of the expression without using a calculator. 9. tanarctan 6.7 10. tanarcsin 3 5 11. Write an algebraic expression equivalent to 12. Use the fundamental identities to simplify: sinarccos 2x. cos 2 x csc x. 13. Subtract and simplify: sin 1 cos cos sin 1. In Exercises 14 –16, verify the identity. 14. 15. 16. cot2 sec2 1 1 sinx y sinx y sin2 x sin2 y sin2 x cos2 x 1 8 1 cos 4x In Exercises 17 and 18, find all solutions of the equation in the interval [0, 2. 17. 18. 2 cos2 cos 0 3 tan cot 0 19. Use the Quadratic Formula to solve the equation in the interval 0, 2: sin2 x 2 sin x 1 0. sin u 12 13, cos v 3 5, and angles u and v are both in Quadrant I, find 20. Given that tanu v. tan 1 2, 21. If find the exact value of tan2. 333202_060R.qxd 12/5/05 10:48 AM Page 488 488 Chapter 6 Additional Topics in Trigonometry C b a A c B FIGURE FOR 25–28 22. If tan 4 3, find the exact value of sin. 2 23. Write the product 5 sin 3 4 cos 7 4 as a sum or difference. 24. Write cos 8x cos 4x as a product. In Exercises 25–28, use the information to solve the triangle shown in the figure. Round your answers to two decimal places. 25. 26. 27. 28. A 30, a 9, b 8 A 30, b 8, c 10 A 30, a 4, C 90, b 8, c 9 b 10 29. Two sides of a triangle have lengths 7 inches and 12 inches. Their included angle measures 60. Find the area of the triangle. 30. Find the area of a triangle with sides of lengths 11 inches, 16 inches, and 17 inches. 31. Write the |
vector u 3, 5 as a linear combination of the standard unit vectors i and j. 32. Find a unit vector in the direction of v i j. u 3i 4j for 33. Find u v 34. Find the projection of orthogonal vectors. and u 8, 2 v i 2j. onto v 1, 5. Then write u as the sum of two 35. Write the complex number 36. Find the product of 2 2i in trigonometric form. 4cos 30 i sin 306cos 120 i sin 120. Write the answer in standard form. 37. Find the three cube roots of 1. 38. Find all the solutions of the equation x5 243 0. 39. A ceiling fan with 21-inch blades makes 63 revolutions per minute. Find the angular speed of the fan in radians per minute. Find the linear speed of the tips of the blades in inches per minute. 40. Find the area of the sector of a circle with a radius of 8 yards and a central angle of 5 feet 114. 41. From a point 200 feet from a flagpole, the angles of elevation to the bottom and top respectively. Approximate the height of the flag to the 16 45 18, and of the flag are nearest foot. 12 feet FIGURE FOR 42 42. To determine the angle of elevation of a star in the sky, you get the star in your line of vision with the backboard of a basketball hoop that is 5 feet higher than your eyes (see figure). Your horizontal distance from the backboard is 12 feet. What is the angle of elevation of the star? 43. Write a model for a particle in simple harmonic motion with a displacement of 4 inches and a period of 8 seconds. 44. An airplane’s velocity with respect to the air is 500 kilometers per hour, with a The wind at the altitude of the plane has a velocity of 50 kilometers E. What is the true direction of the plane, and what bearing of 60 per hour with a bearing of N is its speed relative to the ground? 30. 45. A force of 85 pounds exerted at an angle of above the horizontal is required to slide an object across a floor. The object is dragged 10 feet. Determine the work done in sliding the object. 60 333202_060R.qxd 12/5/05 10:48 AM Page 489 Proofs in Mathematics Law of Tangents Besides the Law of Sines and the Law of Cosines, there is also a Law |
of Tangents, which was developed by Francois Vi`ete (1540–1603). The Law of Tangents follows from the Law of Sines and the sum-to-product formulas for sine and is defined as follows. a b a b tanA B2 tanA B2 The Law of Tangents can be used to solve a triangle when two sides and the included angle are given (SAS). Before calculators were invented, the Law of Tangents was used to solve the SAS case instead of the Law of Cosines, because computation with a table of tangent values was easier. C b a A c B A is acute. C b a A c B A is obtuse. Law of Sines If ABC (p. 430) is a triangle with sides a, b, and c, then a sin A b sin B c sin is acute. A is obtuse. Proof Let h be the altitude of either triangle found in the figure above. Then you have sin A h b sin B h a or or h b sin A h a sin B. Equating these two values of h, you have a sin B b sin A or a sin A b sin B. and sin B 0 sin A 0 0 or B (extended in the obtuse triangle), as shown at the left. Then you have In a similar manner, construct an altitude from vertex because no angle of a triangle can have a to 180. Note that measure of AC side sin A h c sin C h a or or h c sin A h a sin C. Equating these two values of h, you have a sin C c sin A or a sin A c sin C. By the Transitive Property of Equality you know that a sin A b sin B c sin C. So, the Law of Sines is established. 489 333202_060R.qxd 12/5/05 10:48 AM Page 490 Law of Cosines (p. 439) Standard Form a2 b2 c2 2bc cos A b2 a2 c2 2ac cos B c2 a2 b2 2ab cos C Alternative Form cos A b2 c2 a2 2bc cos B a2 c2 b2 2ac cos C a2 b2 c2 2ab Proof To prove the first formula, consider the top triangle at the left, which has three has acute angles. Note that vertex x, y, coordinates is the distance C from vertex has coordinates y |
b sin A. it follows that Furthermore, Because c, 0. and C B a x b cos A where B, to vertex a x c2 y 02 a2 x c2 y 02 a2 b cos A c2 b sin A2 a2 b2 cos2 A 2bc cos A c2 b2 sin2 A a2 b2sin2 A cos2 A c2 2bc cos A a2 b2 c2 2bc cos A. To prove the second formula, consider the bottom triangle at the left, which also A C has three acute angles. Note that vertex has coordinates is the has coordinates distance from vertex y a sin B. x a cos B x, y, C it follows that Furthermore, Because c, 0. and A, b where to vertex b x c2 y 02 b2 x c2 y 02 b2 a cos B c2 a sin B2 b2 a2 cos2 B 2ac cos B c2 a2 sin2 B b2 a2sin2 B cos2 B c2 2ac cos B b2 a2 c2 2ac cos B. A similar argument is used to establish the third formula. Distance Formula Square each side. Substitute for x and y. Expand. Factor out b2. sin2 A cos2 A 1 Distance Formula Square each side. Substitute for x and y. Expand. Factor out a2. sin2 B cos2 = (, 0c, 0) 490 333202_060R.qxd 12/5/05 10:48 AM Page 491 a, b, and c, the area of the triangle is Heron’s Area Formula Given any triangle with sides of lengths Area ss as bs c (p. 442) where s a b c 2. Proof From Section 6.1, you know that bc sin A b2c2 sin2 A Area 1 2 Area2 1 4 Area 1 1 1 4 4 b2c2 sin2 A b2c21 cos2 A Formula for the area of an oblique triangle Square each side. Take the square root of each side. Pythagorean Identity bc1 cos A1 2 bc1 cos A. Factor. 2 Using the Law of Cosines, you can show that bc1 cos and 1 2 bc1 cos A a b c 2 a b c 2. Letting s a b c2, these two equations can be rewritten as and 1 2 1 2 bc1 cos A ss a |
bc1 cos A s bs c. By substituting into the last formula for area, you can conclude that Area ss as bs c. 491 333202_060R.qxd 12/8/05 9:34 AM Page 492 be vectors in the plane or in space and let be a scalar. c 1. Properties of the Dot Product Let v,u, w and. 5. cu v cu v u cv (p. 460) 2. 4. 0 v 0 v v v2 0 0, 0, and let c be a scalar. u2w2 u v u w w w1, w2 v2u2, v u w2 u2w2 w2 v2 u2v2 u1w1 22 2 v2 v2 Proof Let 1. 2. 3. 4. 5., u u1, u2 v v1, v2 u v u1v1 u2v2 0 v2 0 v 0 v1 u v w u v1, v1u1 0 w1, v2 u2 v1 u1 u1v1 u1v1 w1 u1w1 u2v2 v v v1 cu v cu1, u2 2 v2 2 v1 v1, v2 u2v2 cu2 v2 v1, v2 cu1v1 cu1 v1 cu1, cu2 cu v Angle Between Two Vectors (p. 461) If is the angle between two nonzero vectors and u v, then cos u v u v. Proof Consider the triangle determined by vectors u, v, and figure. By the Law of Cosines, you can write v u, as shown in the v u2 u2 v2 2u vcos v u v u u2 v2 2u v cos v u v v u u u2 v2 2u v cos v v u v v u u u u2 v2 2u v cos v2 2u v u2 u2 v2 2u v cos cos u v u v. u − v u θ Origin v 492 333202_060R.qxd 12/5/05 10:48 AM Page 493 P.S. Problem Solving This collection of thought-provoking and challenging exercises further explores and expands upon concepts learned in this chapter. 1. In the figure, a beam of light is directed at the blue mirror |
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