Algebra and trigonometry 10th edition larson solutions manual 1 by alex.heart421

Algebra and Trigonometry 10th Edition Larson

1337271179 9781337271172

Full download at: Solution Manual: https://testbankpack.com/p/solution-manual-for-algebra-and-trigonometry10th-edition-larson-1337271179-9781337271172/

C H A P T E R 7

Section7.1 UsingFundamentalIdentities 584 Section7.2 VerifyingTrigonometricIdentities....................................................591 Section7.3 SolvingTrigonometricEquations......................................................597 Section7.4 Sumand DifferenceFormulas 612 Section7.5 Multiple-Angleand Product-to-SumFormulas................................628 Review Exercises........................................................................................................640 ProblemSolving 651 PracticeTest 658

Analytic Trigonometry

C H A P T E R 7 Analytic Trigonometry Section 7.1 Using Fundamental Identities

1. tan u 2. csc u 9. sin θ 3 = − , cosθ 4 > 0  θ is in Quadrant IV  3  9 7 3. cot u cosθ = 1   =   3 1 = 16 4 4. csc u tan θ = sin θ = 4 = 3 3 7 5. 1 cosθ 7 7 7 4 6. sin u secθ 1 1 4 = = = = 4 7 7. sec x 5 , tan x < 0  x is in Quadrant II cosθ 7 7 7 4 2 cot θ 1 1 7 = = = cos x = 1 = 1 = 2 tan θ 3 3 7 sec x 5 5 2 cscθ 1 1 4 = = = sin x = 1  2  =  1 4 = 21 25 5 sin θ 3 3 4 2 tan x = sin x 21 = 5 = 21 10. cos θ = , sin θ 3 < 0  θ is in Quadrant IV 2 cos x 2 2 sinθ = 1  2  = 1 4 = 5 5 csc x = 1 = 5 = 5 21     9 3 5 sin x 21 21 tanθ = sinθ = 2 = 5 cot x = 1 = 2 = 2 21 cosθ 2 2 3 tan x 21 21 1 1 3 8. csc x 7 , tan x > 0  x is in Quadrant III. secθ = cosθ = 2 = 2 3 6 cotθ = 1 = 1 = 2 2 5 sin x = 1 = 1 = 6 tanθ 5 5 5 csc x 7 7 6 cscθ 2 = 1 = 1 = 3 3 5

584 © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.  2 7   cos x = 1  6   6 = 1 36 = 13 49 7 sinθ 5 5 5 3 tan x = sin x cos x = 7 = 6 = 13 13 6 13 13 7 sec x = 1 = 1 = 7 = 7 13 cos x 13 13 13 7 cot x = 1 = 1 = 13 tan x 6 6 13

Section 7.1 Using Fundamental Identities 585

11. tan x = 2 , cos x > 0  x is in Quadrant I. 15. cos x(1 + tan2 x) = cos x(sec2 x) 3  1  cot x = 1 = 1 = 3 = cos x cos2 x  tan x 2 2 3  2  4 13 = 1 cos x = sec x sec x = 1 +   =  3   3  1 + = 9 3 9 13 Matches (f). cos x 1 1 csc x = 1 +   = 1 + = 16. cot x sec x = = = csc x  2  4 2 sin x cos x sin x sin x = 1 = 1 = 2 = 2 13 Matches (a). csc x 13 13 13 sec2 x 1 tan2 x sin2 x 1 2 17. = = = sec2 x cos x = 1 = 1 = 3 = 3 13 sin2 x sin2 x cos2 x sin2 x sec x 13 13 13 3 Matches (e). cos2  (π 2) x  sin2 x sin x cot x = 1 = 1 = 3 18. cos x = cos x = sin x = tan x sin x cos x tan x 2 2 3 Matches (d). 12. cot x = 7 , sin x < 0  x is in Quadrant III. 4 19. tanθ cotθ  1  tanθ   = tanθ  tan x = 1 = 1 = 4 secθ 1 cot x 7 7 4  4  16 65 cosθ = 1 1 sec x = 1 +   =  7  2     = 1 + = 49 7 1 + 49 = 65  π  cosθ = cosθ  4  16 4 20. cos x sec x = sin x sec x sin x = 1 = 1 = 4 4 65  2   1  csc x 65 65 65 4 = sin x cos x  cos x = 1 = 1 = 7 7 65 = tan x sec x 65 65 65 7 21. tan2 x tan2 x sin2 x = tan2 x(1 sin2 x) = tan2 x cos2 x 2 13. sec x cos x =  1  cos x   = sin x ⋅ cos 2 x cos 2 x = 1 Matches (c). 22.

© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2 = s i n 2 x sin2 x sec2 x sin2 x = sin2 x(sec2 x 1) 14. cot2 x csc 2 x = (csc 2 x 1) csc 2 x = 1 = sin2 x tan2 x Matches (b). 23. sec x 1 = (sec x + 1)(sec x 1) sec x 1 sec x 1 = sec x + 1 24. cos x 2 = cos x 2 cos2 x 4 (cos x + 2)(cos x 2) = 1 cos x + 2

31. cot2 x + csc x 1 = (csc 2 x 1) + csc x 1

1 sin x

(csc x 1)(csc x + 2) csc x 1 cot x

csc2 x + csc x 2

cos x 32. sin2 x + 3 cos x + 3 = (1 cos2 x) + 3 cos x + 3

cos2 x + 3 cos x + 4

(cos2

© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. = 2 586 Chapter 7 Analytic Trigonometry 25. 1 2 cos x + cos x = (1 cos x) 26. sec 4 x tan4 x = (sec 2 x + tan2 x)(sec 2 x tan2 x) 2 4 2 2 ( 2 2 )() = (sin2 x) = sec x + tan x 1 27. 28. = sin4 x cot3 x + cot2 x + cot x + 1 = cot2 x(cot x + 1) + (cot x + 1) = (cot x + 1)(cot2 x + 1) = (cot x + 1)csc 2 x sec3 x sec2 x sec x + 1 = sec2 x(sec x 1) (sec x 1) = (sec2 x 1)(sec x 1) = tan2 x(sec x 1) = sec2 x + tan2 x 29. 3 sin2 x 5 sin x 2 = (3 sin x + 1)(sin x 2) 38. cot u sin u + tan u cos u = cos u(sin u)+ sin u (cos u) 30. 6 cos 2 x + 5 cos x 6 = (3 cos x 2)(2 cos x + 3) sin u = cos u + sin u cos u 2 2 2

39.

2 2 2 =

x 3 cos x 4) = (cos x + 1)(cos x 4) 40. cos2 y 1 sin y 1 sin2 y = 1 sin y = (1 + sin y)(1 sin y) 1 sin y = 1 + sin y 33. tanθ cscθ = sinθ ⋅ 1 = 1 = secθ 41. (sin x + cos x)2 = sin2 x + 2 sin x cos x + cos2 x = (sin2 x + cos2 x) + 2 sin x cos x cos θ sinθ cosθ = 1 + 2 sin x cos x 34. tan( x) cos x = tan x cos x sin x cos x cos x = sin x 42. (2 csc x + 2)(2 csc x 2) = 4 csc 2 x 4 = 4(csc2 x 1) = 4 cot2 x 35. sin φ(cscφ sinφ) = (sin φ) 1 sin2 φ 43. 1 + 1 = 1 cos x + 1 + cos x sinφ 1 + cos x 1 cos x (1 + cos x)(1 cos x) = 1 sin2 φ = cos2 φ = 2 1 cos2 x 36. cos x(sec x cos x) = cos x  1 cos x  = 2  cos x  sin2 x   = 1 cos2 x = sin2 x 44. = 2 csc2 x 1 1 = sec x 1 (sec x + 1)

cos x = cos

x tan

(cos

x) sin x

sin

© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 37. sin β tan β + cos β = (sin β) sin β + cos β cos β sec x + 1 sec x 1 (sec x + 1)(sec x 1) = sec x 1 sec x 1 sin2 β = + cos β cos2 β cos β sec2 x 1 = 2 sin2 β + cos2 β = cos β tan2 x  1  = 2 2  = 1  tan x  cos β = sec β = 2 cot2 x

Section 7.1 Using Fundamental Identities 587

45. cos x cos x = cos x(1 sin x) cos x(1 + sin x) 1 + sin x 1 sin x (1 + sin x)(1 sin x) = cos x sin x cos x cos x sin x cos x (1 + sin x)(1 sin x) = 2sin x cos x 1 sin2 x = 2sin x cos x cos 2 x = 2sin x cos x = 2 tan x

sin x 1 + cos x sin x + = 1 cos x sin x(1 cos x) + sin x(1 + cos x) (1 + cos x)(1 cos x) sin x sin x cos x + sin x + sin x cos x = (1 + cos x)(1 cos x) 2 sin x = 1 cos 2 x

sin x = sin

cos x + 1 + sin x = cos x + (1 + sin x)

1 + sin x cos x cos x(1 + sin x)

(

x tan x)

51. y = tan x + 1 1 sec x + csc x sin x + 1 cos x = 2 + 2 sin x cos x(1 + sin x)

cos2 x + 1 + 2 sin x + sin2 x = cos x(1 + sin x)

= 1 + 1 cos x sin x sin x + cos x = cos x(1 + sin x)

2(1+ sin x)

= cos x sin x + cos x = 2 cos x = 2 sec x

sin x cos x  sin x + cos x  sin x cos x  =  cos x  sin x + cos x 

   = sin x 2 (1 + cos y)(1 cos y) = 1 cos y = 1 + cos y

Cengage Learning. All

not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2018

Rights Reserved. May

47.

50.

tan

tan x 2 2 tan

) = 1 = 5

46.

= sin x = 2 csc x

sec2 x tan x = tan

x sec

5 ⋅ tan x sec x = 5(tan x sec x)

x tan x = 1 = cot x

sec x tan x sec x tan

x sec

(tan x sec x

sec

48.

49. sin2 y 1 cos y 1 cos2 y = 1 cos y 2π 2π 2

588 Chapter 7 Analytic Trigonometry

52. y1 = 1  1   cos x = tan x sin x cos x  5 1  1  1 cos x cos x = sin x  cos x  sin x cos x sin x 2π 2π   1 cos2 x sin2 x sin x = = = = tan x 5 sin x cos x sin x cos x cos x 53. Let x = 3 cos θ 55. Let x = 2 sec θ 9 x2 = = = 9 (3 cosθ)2 9 9 cos2 θ 9(1 cos2 θ) x2 4 = = = (2 secθ)2 4 4(sec2 θ 1) 4 tan2 θ = 9 sin2 θ = 3 sin θ = 2 tan θ 54. Let x = 7 sin θ 56. Let 3x = 5 tan θ 2 2 49 x2 = 49 (7 sin θ)2 9x + 25 = (3x) + 25 = 49 49 sin2 θ = 49(1 sin2 θ) = 49 cos2 θ = (5 tan θ)2 + 25 = 25 tan2 θ + 25 = 25(tan2 θ + 1) = 7 cosθ = 25 sec2 θ 57. Let x = 2 sin θ 4 x2 = 2 4 (2 sin θ)2 = 2 4 4 sin2 θ = 2 = 5 secθ 4(1 sin2 θ) = 2 4 cos2 θ = 2 2 cosθ = 2 cosθ = 2 2 2 sin θ = 1 cos2 θ =  2  2 1   = ±  2  2

Section 7.1 Using Fundamental Identities 589

58. x = 10 cosθ  tan t  5 3 = 5 3 = 100 x2 100 (10 cosθ)2 63. ln tan t ln(1 cos2 t) = ln  1 cos2 t  = ln tan t 5 3 = 5 3 = 100(1 cos2 θ) 100 sin2 θ sin2 t = ln sin t cos t 1 sin2 t 5 3 = 10 sin θ = ln 1 cos t sin t sin θ = 5 3 = 3 10 2 2 = ln sec t csc t cosθ = 1 sin2 θ =  3  1 1   = 64. ln(cos2 t) + ln(1 + tan2 t) = lncos2 t(1 + tan2 t)  2  2 = ln cos2 t sec2 t 59. sin θ = 1 cos2 θ  1  = lncos 2 t ⋅  Let y1 = sin x and y2 = 1 cos2 x, 0 ≤ x ≤ 2π  cos t  y1 = y2 for 0 ≤ x ≤ π = ln(1) = 0 So, sin θ = 2 1 cos2 θ y2 for 0 ≤ θ ≤ π 65. μW cosθ μ = W sin θ W sin θ = W cosθ = tan θ 0 2π y1 66. sec x tan x sin x = 1 sin x sin x cos x 2 cos x sin x = sin x cos2 x 60. secθ = 1 + tan2 θ sin x sin x cos 2 x Let y = 1 and y = 1 cos x 2 1 + tan2 x, 0 ≤ x ≤ 2π. = cos 2 x sin x(1 cos 2 x) y1 = y for 0 ≤ x < π and 3π 2 2 2 < x ≤ 2π = cos 2 x sin x sin2 x = So, secθ = 1 + tan2 θ for 0 ≤ θ < π 2 and cos2 x = sin x tan2 x 3π < θ 2 < 2π 4 y2 0 2π y1 67. True. tan u = sin u cos u 61. ln sin x + ln cot x 4 = ln sin x cot x = ln sin x ⋅ cos x sin x = ln cos x

False. A cofunction identity can be used to transform a tangent function so that it can be represented by a cotangent function.

© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. cot u = cos u sin u s e c u = 1 cos u csc u = 1 sin u 62. ln cos x ln sin x = ln cos x sin x = ln cot x 68.

69. As x → π , tan x → ∞ and cot x → 0. 2

Chapter 7 Analytic Trigonometry 70. As x → π + , sin x → 0 and csc x = 1 → ∞ . sin x 72. Let u = a tan θ, then 71. cos( θ) ≠ cosθ cos( θ) = cosθ The correct identity is sin θ = cos( θ) sin θ cosθ a2 + u2 = = = = a 2 + (a tan θ)2 a2 + a2 tan2 θ a2(1 + tan2 θ) a2 sec2 θ = tan θ = a secθ. 73. Because sin2 θ + cos2 θ = 1, then cos 2 θ = 1 sin2 θ cos θ = ± 1 sin θ tan θ = sin θ = sin θ cot θ cos θ ± = cos θ = ± sin θ 1 sin2 θ 1 sin2 θ sin θ sec θ = 1 = 1 csc θ cos θ 1 = sin θ ± 1 sin2 θ 74. To derive sin2 θ + cos2 θ = 1, let sin θ = a and cos θ = b a2 + b2 a2 + b2 2 2  a   b  a b2 So, sin2 θ + cos2 θ =   +   = +  a2 + b2   a2 + b2  a2 + b2 a 2 + b2 = a2 + b2 a2 + b2 = 1. To derive 1 + tan2 θ = sec2 θ, let tan θ = a and sec θ b a 2 + b2 = b So, 1 + tan2 θ 2      b  a 2 = 1 + = b2 b2 + a2 b2 2 2  a2 + b2   a2 + b2  =   =    b2   b      = sec2 θ To derive 1 + cot2 θ = csc2 θ, let cot θ = b and csc θ a a 2 + b2 = . a

590

© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. = 1 + b So, 1 + cot2 θ 2      a  = 1 + b2 a2 2 a2 + b2  a2 + b2  = =   a 2  a 2    2  a2 + b2  =   = csc2 θ  a    Answers will vary.

Section 7.2 Verifying Trigonometric Identities

Verifying Trigonometric Identities

75. sec θ(1 + tan θ) sec θ + csc θ   cos θ  cos θ  = 1 + 1 cos θ sin θ cos θ + sin θ = cos2 θ sin θ + cos θ sin θ cos θ =  sin θ + cos θ  sin θ cos θ   cos2 θ  sin θ + cos θ     = sin θ cos θ

Section 7.2

591

1. identity 2. conditional equation 3. tan u 16. cos (π 2) x  sin (π 2) x   π  sin x = = tan x cos x  1  4. cot u 17. sin t csc 2 t  = sin t sec t = sin t cos t  5. sin u sin t = cos t = tan t 6. cot2 u 7. csc u 18. sec2 y cot2 π 2 y  = sec2 y tan2 y = 1   8. sec u 19. 1 + 1 = cot x + tan x 9. tan t cot t = sin t cos t cos t = 1 sin t tan x cot x tan x cot x = cot x + tan x 1 10. tan x cot x cos x = 1 = sec x cos x 20. = tan x + cot x 1 1 = csc x sin x 11. (1 + sin α)(1 sin α) = 1 sin2 α = cos2 α sin x csc x sin x csc x csc x sin x = 12. cos2 β sin2 β = cos2 β (1 cos2 β) = 2

1 = csc x sin x 13. 14. cos2 β sin2 β sin2 α sin4 α = (1 sin2 β) sin2 β = 1 2 sin2 β = sin2 α(1 sin2 α) = (1 cos2 α)(cos2 α) = cos2 α cos4 α 21. 1 + sin θ cosθ + cosθ = 1 + sin θ = = (1 + sin θ)2 + cos 2 θ cosθ(1 + sin θ) 1 + 2 sin θ + sin2 θ + cos2 θ cosθ(1 + sin θ) 2 + 2 sin θ cosθ(1 + sin θ) 2(1+ sin θ) 15. tan π  2 θ  tan θ = cot θ tan θ = cosθ(1 + sin θ)

cos

β 1

© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.    1  =   tan θ  tan θ  = 1 = 2 cosθ = 2 secθ

cosθ cot θ cosθ cot θ (1 sin θ) 

1 sin θ

cosθ  cosθ sin θ 

sin θ sin θ

25. 26. 27.

sec y cos y =   cos y = 1  cos y  cot2 y(sec2 y 1) = cot2 y tan2 y = 1 tan = ( ) = sin θ tan θ 2 θ sin θ cosθ tan θ = sin θ(1 sin θ) secθ 1 cosθ = 1 sin θ sin θ(1 sin θ) = 1 sin θ

23. 1 1 + = cos x + 1 cos x 1 =

cos x 1 + cos x + 1 (cos x + 1)(cos x 1)

2 cos x cos2 x 1

28. cot3 t csc t cot t cot2t = csc t cot t(csc2 t 1) = csc t cos t csc2 t 1 sin t

= 1 sin t

= cos t sin t csc2 sin t t 1) = sin2 x = cos t(csc2 t 1)

2 cos x

= 2 csc x cot x cos x cos x = cos x(1 tan x) cos x

1 tan x 1 tan x

= cos x tan x 1 tan x cos x(sin x cos x) cos x = ⋅ 1 (sin x cos x)

sec2 β = tan β

= 2 1 cos x 1 1 + tan2 β sin x sin x 29. tan β + tan β = tan β 24.

sin x cos x

= secθ 1 ⋅ secθ secθ(secθ 1) = = secθ

t cos

= = =

sec x cos x = cos x

cos

Cengage

2018

Learning.

 ( ) (

592 Chapter 7 Analytic Trigonometry 

 22. 1 sin θ 1 = =

+ sin θ ⋅ 1 sin θ cos2 θ sin θ + sin2 θ

cscθ

1 31.

32. 

x 

 

= sin x cos x cos x sin x sin x cos x 

cos x

30. secθ

1 cosθ 1 (1 secθ) secθ secθ 1 cot2 t cos2 t sin

t 1 sin

33.

csc t 1 sin t sin t sin t cos x

x =

sin

cos x + sin x tan x = cos x + sin x

cos x

© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. c o s 2 x + s i n 2 x cos x sin2 x = cos x = cos x = sin x sin x = 1 cos x = sec x cos x = sin x tan x

cot x cot y

42. cos x cos y + sin x sin y = (cos x cos y)(cos x + cos y) + (sin x sin y)(sin x + sin y)

sin x + sin y cos x + cos y (sin x + sin y)(cos x + cos y) cos2 x cos2 y + sin2 x sin2 y

sin x + sin y)(cos x + cos y)

cos2 x + sin2 x) (cos2 y + sin2 y)

sin x + sin y)(cos x + cos y)

34. cot x tan x = cos x sin x sin x cos x cos2 x sin2 x = sin x cos x 1 sin2 x sin2 x = sin x cos x 1 2 sin2 x = sin x cos x 1 1 2 sin2 x  =   cos x sin x  1  1 2 sin2 x  = cos x  sin x sin x  = sec x(csc x 2 sin x) cot x cos x sin x cos2 x 1 sin2 x 1 sin2 x 35. = = = = = csc x sin x 36. sec x csc( x) sec( x) 1 cos x 1 sin( x) = 1 cos( x) cos( x) sin x sin x sin x sin x = sin( x) 37. = cos x sin x = cot x sin1 2 x cos x sin5 2 x cos x = sin1 2 x cos x(1 sin2 x) = sin1 2 x cos x cos 2 x = cos 3 x sin x 38. sec 6 x(sec x tan x) sec 4 x(sec x tan x) = sec 4 x(sec x tan x)(sec 2 x 1) = sec 4 x(sec x tan x) tan2 x = sec 5 x tan3 x 39. (1 + sin y) 1 + sin( y)  = (1 + sin y)(1 sin y) = 1 sin2 y 41. 1 + sin θ = 1 sin θ 1 + sin θ 1 sin θ 1 + sin θ 1 + sin θ 40. = cos2 y 1 + 1 tan x + tan y = cot x cot y

(1 +

θ)2 = 1 sin2 θ (1 + sin θ)2 = 1 tan

cos 2 θ cot

1 + sin θ = = cot

cot

cosθ

Section 7.2 Verifying Trigonometric Identities 593

sin

x tan y 1 1 1 cot x cot y

x cot y

y + cot x

x cot y 1

43. = 0

(

© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. cot( x) ≠ cot x 44. The first line claims that sec( θ ) = sec θ and The correct substitution is cot( x) = cot x sin( θ) = sin θ The correct substitutions are 1 + cot( x) = cot x cot x = 0 tan x sec( θ) = sec θ and sin( θ) = sin θ

45. (a) 3 2π 2π 1 Identity (b) Identity (c) 1 + cot2 x cos2 x = csc2 x cos2 x = 1 cos2 x = cot2 x sin2 x 3 46. (a) (b) 2π 2π 1 Identity Identity (c) csc x(csc x sin x) + sin x cos x + cot x = csc2 x csc x sin x + 1 cos x + cot x sin x sin x = csc2 x 1 + 1 cot x + cot x = csc2 x 47. (a) 5 y2 y1 2π 2π 1 Not an identity (b) Not an identity (c) 2 + cos2 x 3 cos4 x = (1 cos2 x)(2 + 3 cos2 x) = sin2 x(2 + 3 cos2 x) ≠ sin2 x(3 + 2 cos2 x) 48. (a) π 5 y1 y2 π (b) 5 Not an identity sin4 x sin2 x Not an identity (c) tan4 x + tan2 x 3 = + 3 cos4 x 1 cos2 x  sin4 x  = cos2 x  cos2 x + sin2 x 3  1  sin4 x + sin2 x cos2 x  =  cos 2 x cos2 x  3   1  sin2 x  =   sin x + cos x 3 cos2 x cos2 x  1  sin2 x  = cos 2 x  cos 2 x 1 3

Chapter 7 Analytic Trigonometry

© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. .  cos2 x ( ) Section 7.2 Verifying Trigonometric Identities 595 49. (a) 3 50. (a) 3 y1 2π 2π 2 2 y2 3 5 (b) Identity (b) Not an identity (c) Identity 1 + cos x (1 + cos x)(1 cos x) = (c) Not an identity cot α is the reciprocal of cscα + 1 sin x sin x(1 cos x) cscα + 1 cot α 1 cos2 x = sin x(1 cos x) sin2 x = sin x(1 cos x) sin x = 51. They will only be equivalent at isolated points in their respective domains So, not an identity. tan3 x sec2 x tan3 x = tan3 x(sec2 x 1) = tan3 x tan2 x 5 1 cos x = tan x 2 4 52. (tan2 x + tan4 x) sec2 x =  sin x sin x  1 +  cos 4 x cos 2 x   1  = sin2 x + sin4 x   cos4 x cos2 x  1  sin2 x cos2 x + sin4 x  =   cos4 x cos2 x  1  sin2 x cos 2 x + sin2 x  =   cos4 x cos2 x    1  sin2 x =  1 = sec4 x tan2 x cos4 x cos2 x  53. (sin2 x sin4 x) cos x = sin2 x(1 sin2 x) cos x = sin2 x cos2 x cos x = sin2 x cos3 x 54. sin4 x + cos4 x = sin2 x sin2 x + cos4 x = (1 cos2 x)(1 cos2 x) + cos4 x = 1 2 cos2 x + cos4 x + cos4 x = 1 2 cos2 x + 2 cos4 x 55. sin2 25° + sin2 65° = sin2 25° + cos2 (90° 65°) = sin2 25° + cos2 25° = 1 56. tan2 63° + cot2 16° sec2 74° csc2 27° = tan2 63° + cot2 16° csc2(90° 74°) sec2(90° 27°) = tan2 63° + cot2 16° csc2 16° sec2 63°

© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. = ( t a n 2 6 3 ° s e c 2 6 3 ° ) + ( c o t 2 1 6 ° c s c 2 1 6 ° ) = 1 + ( 1 ) = 2

63. False tan x2 = tan(x ⋅ x) and

2 x = (tan x)(tan x), tan x

64. True. Cosine is an even function,

© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 1 596 Chapter 7 Analytic Trigonometry 57. Let θ = sin 1 x  sin θ = x = x 60. Let θ = cos 1 x + 1  cosθ = x + 1 1 2 2 1 x 2 4 (x + 1)2 θ 1 x2 θ x + 1 From the diagram, From the diagram, tan(sin 1 x) = tan θ = x .  1 x + 1 4 (x + 1)2 1 x2 tancos 2  = tan θ = x 1 . 58. Let θ = sin 1 x  sin θ = x = x   + 1 cos x 1 61. cos x csc x cot x = cos x sin x sin x  1  = cos x1   sin2 x  x θ 1 x2 = cos x(1 csc 2 x) = cos x(csc2 x 1) = cos x cot2 x From the diagram, cos(sin 1 x) = cosθ 1 x 2 = = 1 1 x2 62. (a) h sin(90° θ) sin θ = h cosθ = h cot θ sin θ 59. Let θ = sin 1 x 1  sin θ = x 1 . (b) 15° 30° 45° 60° 75° 90° s 18.66 8.66 5 2.89 1.34 0 4 4 4 x 1 (c) Maximum:

Minimum:

(d) Noon θ

15°

90°

16 (x 1)2 From the diagram, tansin 1 x 1 = tan θ x 1 = tan

≠ tan2 x    π   π   4  16 (x 1)2 cosθ  = cos  θ   2   = cos  π  2  θ   2    = sin θ

65. False. For the equation to be an identity, it must be true for all values of θ in the domain

publicly

or posted to a

accessible website, in whole or in part.

Section 7.3 Solving Trigonometric Equations 597

Section 7.3 Solving Trigonometric Equations

The equation is not an identity because it is only true when cos

= 0 or cos θ = 1 So, one angle for which

This equation is not an identity because it is only true

when tan

= 0 So, one angle for which the equation

66. If sin θ = a , sec θ = c , and 68. tan θ = sec2 θ 1 c b True identity: tan θ = ± sec 2 θ 1 a2 + b2 = c2  a2  c  = c2 b2 , then tan θ = sec2 θ 1 is not true for π 2 < θ < π sec2 θ 1   1 =   or 3π 2 < θ < 2π So, the equation is not true for sec2 θ  c      c2 θ 69. = 3π 4. 1 cosθ = sin θ b2 1 = c 2 (1 cosθ) = (sin θ) 2 2 b2 1 2 cosθ + cos θ = sin θ c2 b2 = b2 c2 1 2 cosθ + cos2 θ 2 cos2 θ 2 cosθ = 1 cos2 θ = 0 b2 c 2 b2 b2 = b2 c2 c2 b2 2 cosθ(

= 0

cosθ 1)

the equation

π = c2 a 2 70. = c2 1 + tan θ (1 + tan θ)2 2 = secθ = (secθ)2  a  2 2 =   1 + 2 tan θ + tan θ = sec θ  c  = sin2 θ 1 + 2 tan θ + tan2 θ 2 tan θ = 1 + tan2 θ = 0 67. Because sin2 θ = 1 cos2 θ, then tan θ = 0 sin θ = ± 1 cos 2 θ; sin θ ≠ 1 cos 2 θ if θ

is not true is

lies in Quadrant III or IV.

One such angle is

= 7

4 is not true is π 6

1. isolate 6. sec x 2 = 0 2. general 3. quadratic (a) x = π 3 sec π 2 = 1 2 4. extraneous 5. ta n x

© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 3 = 0 3 cos(π 3) = 1 2 = 2 2 = 0 1 2 (a) x = π 3 tan π 3 = 3 3 = 0 (b) x = 5π 3 sec 5π 2 = 1 2 (b) 3 x = 4π 3 3 cos(5π 3) = 1 2 = 2 2 = 0 1 2 tan 4π 3 = 3 3 3 = 0

598 Chapter 7 Analytic Trigonometry

7. 3 tan2 2x 1 = 0 10. csc4 x 4 csc2 x = 0 (a) x = π 12 2 3 tan 2 π  1 = 3 tan2 π 1 (a) x = π 6 csc 4 π 4 csc2 π 1 4 =  12  6 4 2   2  1  3  1 6 6 sin (π 6) 1 4 = sin (π 6)  3  (1 2)4 (1 2)2 (b) = 0 x = 5π 12 2 (b) x = 5π 6 = 16 16 = 0 3tan 2 5π  1 = 3 tan2 5π 1 csc4 5π 4 csc 5π 1 4     12  6 = 4 2   2  1  = 3  1 6 6 sin (5π 6) 1 4 = sin (5π 6)  3  (1 2)4 (1 2)2 = 0 8. 2 cos2 4x 1 = 0 11. 3 csc x 2 = 0 = 16 16 = 0 (a) x = π 16   π  π 4  1 = 2 cos 1 3 csc x = 2 csc x = 2 3  16  4 2   x = π + 2nπ 3 = 2 2  1  2  (b) x = 3π    1  = 2  1 = 1 1 = 0   12. tan x + or x = 2π 3 3 = 0 + 2nπ 16   3π  3π 4  1 = 2 cos 1 tan x = 3 x = 2π + nπ   16  4 2  2  = 2  1 3 13. cos x + 1 = cos x  2  2 cos x + 1 = 0   =  1  1 = 0  2  cos x 1 2 x = 2π + 2nπ or x = 4π + 2nπ

© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2 9. 2 sin2 x sin x 1 = 0 3 3 (a) x = π 2 2 sin2 π sin π 1 = 2(1)2 1 1 14. 3 sin x + 1 = sin x 2 sin x + 1 = 0 1 2 2 sin x = = 0 7π (b) x = 7π 6 2 sin2 7π sin 7π 1 2 1 = 2  1   1 x = + 2nπ or 6 x = 11π + 2nπ 6     6 6  2   2  = 1 + 1 1 2 2 = 0

Solving Trigonometric

15. 3 sec2 x 4 = 0 20. (2 sin2 x 1)(tan2 x 3) = 0 sec2 x = 4 3 2 sin2 x 1 = 0 or tan2 x = 3 sec x = ± 2 3 sin2 x = 1 2 tan x = ± 3 x = π 6 + nπ sin x = ± 1 2 x = π 3 + nπ or x = 5π + nπ sin x = ± 2 x = 2π + nπ 16. 6 3 cot2 x 1 = 0 cot2 x = 1 x = π 4 x = 3π 2 3 + 2nπ + 2nπ 3 cot x = ± 1 3 4 x = 5π 4 x = 7π + 2nπ + 2nπ x = π 3 + nπ 4 or x = 2π 3 + nπ 21. cos3 x cos x = 0 cos x(cos2 x 1) = 0 17. 4 cos2 x 1 = 0 cos x = 0 or cos2 x 1 = 0 cos2 x = 1 4 cos x 1 x = π + nπ 2 cos x = ±1 x = nπ = ± 2 Both of these answers can be represented as x = nπ x = π + nπ or x = 2π 2 + nπ 18. 3 3 2 4 sin2 x = 0 sin2 x = 1 2 22. sec2 x 1 = 0 sec2 x = 1 sec x = ±1 x = nπ sin x = ± 1 2 = ± 2 2 23. 3 tan3 x = tan x 3 x = π + 2nπ 3 tan x tan x = 0 4 x = 3π + 2nπ tan x(3 tan2 x 1) = 0 4 x = 5π 4 x = 7π + 2nπ + 2nπ tan x = 0 x = nπ or 3 tan2 x 1 = 0 tan x = ± x = π 3 3 + nπ, 5π + nπ

Section 7.3

Equations 599

© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 4 These answers can be represented as x = π + nπ 24. 6 6 sec x csc x = 2 csc x 19. sin x(sin x + 1) = 0 4 2 sec x csc x 2 csc x = 0 csc x(sec x 2) = 0 sin x = 0 or sin x = 1 csc x = 0 or sec x 2 = 0 x = nπ x = 3π 2 + 2nπ No solution sec x = 2 x = π + 2nπ, 5π + 2nπ 3 3

600 Chapter 7 Analytic Trigonometry

25. 2 cos2 x + cos x 1 = 0 (2 cos x 1)(cos x + 1) = 0 2 cos x 1 = 0 or cos x + 1 = 0 cos x = 1 cos x = 1 2 x = π + 2nπ, 5π + 2nπ x = π + 2nπ 3 3 26. 2 sin2 x + 3 sin x + 1 = 0 (2 sin x + 1)(sin x + 1) = 0 2 sin x + 1 = 0 or sin x + 1 = 0 sin x 1 2 sin x = 1 x = 3π + 2nπ x = 7π + 2nπ, 11π + 2nπ 2 6 6 27. sec2 x sec x = 2 sec2 x sec x 2 = 0 (sec x 2)(sec x + 1) = 0 sec x 2 = 0 sec x = 2 or sec x + 1 = 0 sec x = 1 x = π + 2nπ, 5π + 2nπ x = π + 2nπ 3 3 28. csc2 x + csc x = 2 csc2 x + csc x 2 = 0 (csc x + 2)(csc x 1) = 0 csc x + 2 = 0 csc x = 2 or csc x 1 = 0 csc x = 1 x = 7π + 2nπ, 11π + 2nπ x = π + 2nπ 6 6 2 29. sin x 2 = cos x 2 sin x = cos x 30. cos x + sin x tan x = 2 cos x + sin x  sin x  = 2 sin x = 1 cos x  cos x  cos2 x + sin2 x tan x = 1 x = tan 1 1 x = π , 5π 4 4 cos x = 2 1 = 2 cos x 1 cos x = 2 x = π , 5π 3 3

Section 7.3 Solving Trigonometric Equations 601

31. 2 sin2 x = 2 + cos x 2 2 cos2 x = 2 + cos x 2 cos2 x + cos x = 0 36. 3 sec x 4 cos x = 0 3 4 cos x = 0 cos x cos x(2 cos x + 1) = 0 3 4 cos2 x = 0 cos x cos x = 0 or 2 cos x + 1 = 0 3 4 cos2 x = 0 x = π , 3π 2 cos x = 1 3 2 2 1 cos x = 2 cos2 x = 4 3 32. tan2 x = sec x 1 sec2 x 1 = sec x 1 x = 2π , 4π 3 3 37. cos x = ± 2 x = π , 5π , 7π , 11π 6 6 6 6 csc x + cot x = 1 sec2 x sec x = 0 (csc x + cot x)2 = 12 33. sec x(sec x 1) = 0 sec x = 0 or sec x 1 = 0 No Solutions sec x = 1 x = 0 sin2 x = 3 cos2 x csc2 x + 2 csc x cot x + cot2 x = 1 cot2 x + 1 + 2 csc x cot x + cot2 x = 1 2 cot2 x + 2 csc x cot x = 0 2 cot x(cot x + csc x) = 0 2 cot x = 0 or cot x + csc x = 0 x π , 3π cos x 1 sin2 x 3 cos2 x = 0 = = 2 2 sin x sin x sin2 x 3(1 sin2 x) = 0 4 sin2 x = 3 sin x = ± 3  3π   2 is extraneous   cos x = 1 x = π (π is extraneous) 2 x = π , 2π , 4π , 5π 3 3 3 3 34. 2 sec2 x + tan2 x 3 = 0 38. x = π 2 is the only solution. sec x + tan x = 1 1 sin x + = 1 cos x cos x 2(tan2 x + 1) + tan2 x 3 = 0 1 + sin x = cos x 3 tan2 x 1 = 0 (1 + sin x)2 = cos2 x tan x = ± 3 3 1 + 2 sin x + sin2 x = cos2 x 1 + 2 sin x + sin2 x = 1 sin2 x x = π , 5π , 7π , 11π 6 6 6 6 2 sin2 x + 2 sin x = 0 35. 2 sin x + csc x = 0 2 sin x + 1 = 0 2 sin x(sin x + 1) = 0 sin x = 0 or sin x + 1 = 0 sin x 2 sin2 x + 1 = 0

© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.  sin2 x 1  No solution x = 0, π (π is extraneous ) sin x = 1 x = 3π 2 = 2 x = 0 is the only solution.  3π   2 is extraneous  

602 Chapter 7 Analytic Trigonometry

39. 2 cos 2x 1 = 0 45. 3 tan x 3 = 0 cos 2x = 1 2 2 tan x = 3 2x = π + 2nπ or 2x = 5π 2 3 + 2nπ 3 3 x = π + nπ  x = π + 2nπ x = π + nπ x = 5π + nπ 2 6 3 6 6 x 46. tan + 3 = 0 40. 2 sin 2x + 3 = 0 2 x sin 2x = 3 2 tan 2 x = 3 = 2π + nπ  x = 4π + 2nπ 2x = 4π + 2nπ or 2x = 5π + 2nπ 2 3 3 3 3 x = 2π + nπ x = 5π + nπ 47. y = sin π x + 1 3 6 2 π x  + = 41. tan 3x 1 = 0 tan 3x = 1 sin  1 0   π x  3x = π + nπ sin  = 1   4 x = π + nπ π x 2 = 3π 2 + 2nπ 12 3 42. sec 4x 2 = 0 sec 4x = 2 cos 4x = 1 2 48. x = 3 + 4n For 2 < x < 4, the intercepts are 1 and 3 y = sin π x + cos π x sin π x + cos π x = 0 sin π x = cosπ x 4x = π + 2nπ or 4x = 5π + 2nπ π 3 3 π x = + nπ 4 x = π + nπ x = 5π + nπ 1 12 2 12 2 x = + n 4 43. 2 cos x = 2 = 0 For 1 x 3, the intercepts are 1 , 3 , 7 , 11 2 cos x = 2 < < 4 4 4 4 2 2 49. 5 sin x + 2 = 0 8 x = π + 2nπ or x = 7π + 2nπ 2 4 2 4

© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. x = π + 4nπ x = 7π + 4nπ 0 2π 2 2 44. 2 sin x = 2 3 = 0 5 x ≈ 3.553 and x ≈ 5.872 sin x = 3 50. 2 tan x + 7 = 0 2 2 15 x = 4π + 2nπ or x 5π = + 2nπ 2 3 2 3 x = 8π + 4nπ x = 10π + 4nπ 0 2π 3 3 5 x ≈ 1 849 and x ≈ 4 991

Solving

51. sin x 3 cos x = 0 5 55. sec2 x 3 = 0 5 0 2π 0 2π 5 x ≈ 1.249 and x ≈ 4 391 52. sin x + 4 cos x = 0 5 56. 4 x ≈ 0.955, x ≈ 2.186, x ≈ 4.097 and x ≈ 5.328 csc2 x 5 = 0 5 0 2π 0 2π 5 x ≈ 1 816 and x ≈ 4 957 5 x ≈ 0 464, x ≈ 2 678, x = 3 605 and x ≈ 5.820 53. cos x = x 4 57. 2 tan2 x = 15 6 0 2π 0 2π 8 x ≈ 0.739 54. tan x = csc x 18 x ≈ 1.221, x ≈ 1.921, x ≈ 4.362 and x ≈ 5.062 10 58. 6 sin2 x = 5 6 0 2π 0 2π 10 x ≈ 0.905 and x ≈ 5 379 18 x ≈ 1 150, x ≈ 1.991, x ≈ 4 292 and x ≈ 5.133 59. tan2 x + tan x 12 = 0 (tan x + 4)(tan x 3) = 0 tan x + 4 = 0 or tan x 3 = 0 tan x = 4 tan x = 3 x = arctan( 4) + nπ x = arctan 3 + nπ 60. tan2 x tan x 2 = 0 (tan x + 1)(tan x 2) = 0 tan x + 1 = 0 or tan x 2 = 0 tan x = 1 tan x = 2 x = 3π 4 + nπ x = arctan 2 + nπ

Section 7.3

Trigonometric Equations 603

604 Chapter 7 Analytic Trigonometry

tan

5 x = π 4 + nπ x = arctan 5 + nπ 62. sec

0 1

0 tan2

0 (tan

= 0 tan

1 x

arctan

nπ x = arctan(1) + nπ 63. ≈ 1.1071 + nπ 2 sin2 x + 5 cos x = 4 2(1 cos2 x) + 5 cos x 4 = 0 2 cos2 x + 5 cos x 2 = 0 (2 cos x 1)(cos x 2) = 0 = π + nπ 4 2 cos x 1 = 0 or cos x 2 = 0 cos x = 1 2 x = π + 2nπ, 5π + 2nπ cos x = 2 No solution 3 3 64. 2 cos2 x + 7 sin x = 5 2(1 sin2 x) + 7 sin x 5 = 0 2 sin2 x + 7 sin x 3 = 0 (2 sin x 1)(sin x 3) = 0 2 sin x 1 = 0 or sin x 3 = 0 sin x = 1 2 sin x = 3 x = π + 2nπ, 5π + 2nπ No solution 6 6 65. cot2 x 9 = 0 cot2 x = 9 1 = tan2 x 9 ± 1 = tan x 3 x = arctan 1 + nπ, arctan( 1) + nπ

61. sec2 x 6 tan x = 4 1 + tan2 x 6 tan x + 4 = 0 tan2 x 6 tan x + 5 = 0 (tan x 1)(tan x 5) = 0 tan x 1 = 0 tan x 5 = 0

x =

tan x =

2 x + tan x 3 =

+ tan2 x + tan x 3 =

x + tan x 2 =

x + 2)(tan x 1) =

tan x + 2 = 0 tan x 1

x = 2 tan x =

(

) +

Section 7.3 Solving Trigonometric Equations 605

66. cot2 x 6 cot x + 5 = 0 (cot x 5)(cot x 1) = 0 cot x 5 = 0 or cot x 1 = 0 cot x = 5 cot x = 1 1 = tan x 5 1 = tan x x = arctan 1 + nπ x = π + nπ 5 4 67. sec2 x 4 sec x = 0 sec x(sec x 4) = 0 sec x = 0 sec x 4 = 0 No solution sec x = 4 1 = cos x 4 x = arccos 1 + 2nπ, arccos 1 + 2nπ 4 4 68. sec2 x + 2 sec x 8 = 0 (sec x + 4)(sec x 2) = 0 sec x + 4 = 0 or sec x 2 = 0 sec x = 4 sec x = 2 1 = cos x 1 = cos x 4 2 x = arccos  1  + 2nπ, arccos  1  + 2nπ x = π + 2nπ, 5π + 2nπ  4   4  3 3     69. csc2 x + 3 csc x 4 = 0 (csc x + 4)(csc x 1) = 0 csc x + 4 = 0 or csc x 1 = 0 csc x = 4 csc x = 1 1 = sin x 4 1 = sin x x = arcsin 1  + 2nπ, arcsin 1  + 2nπ x = π + 2nπ  4   4  2     70. csc2 x 5 csc x = 0 csc x(csc x 5) = 0 csc x = 0 or csc x 5 = 0 No solution csc x = 5 1 sin x 5 x = arcsin(1) + 2nπ, arcsin( 1) + 2nπ

606 Chapter 7 Analytic Trigonometry

71. 12 sin2 x 13 sin x + 3 = 0 ( 13) ± ( 13)2 4(12)(3) 30 13 ± 5 sin x = = 2(12) 24 0 2π sin x = 1 or sin x = 3 10 3 4 x ≈ 0 3398, 2 8018 x ≈ 0.8481, 2.2935 The x-intercepts occur at x ≈ 0 3398, x ≈ 0.8481, x ≈ 2 2935, and x ≈ 2 8018. 72. 3 tan2 x + 4 tan x 4 = 0 4 ± 42 4(3)( 4) 4 ± 64 2 tan x = = = 2, tan x = 2 tan x = 2(3) 6 3 2 3 50 x = arctan( 2) + nπ ≈ 1.1071 + nπ x = arctan 2  + nπ   ≈ 0.5880 + nπ 0 2 The values of x in [0, 2π) are 0.5880, 3.7296, 2.0344, 5.1760. 10 73. tan2 x + 3 tan x + 1 = 0 3 ± 32 4(1)(1) 3 ± 5 tan x = = 10 2(1) 2 tan x = 3 5 or tan x = 3 + 5 0 2π 2 2 x ≈ 1.9357, 5.0773 x ≈ 2.7767, 5.9183 5 The x-intercepts occur at x ≈ 1.9357, x ≈ 2 7767, x ≈ 5 0773, and x ≈ 5 9183. 74. 4 cos2 x 4 cos x 1 = 0 4 ± ( 4)2 4(4)( 1) 4 ± 32 1 ± 2 cos x = = = cos x = 1 2 2(4) 8 2 cos x = 1 + 2 7 2 2   1 2 x = arccos  No solution  2  0 2π   ≈ 1 7794 1 + 2  3 > 1    2    1 2  Solutions in [0, 2π) are arccos  and 2π arccos 1 2 : 1.7794, 4.5038.

Solving Trigonometric Equations

75. 3 tan2 x + 5 tan x 4 = 0,   3 π , π  2 2  77. 4 cos2 x 2 sin x + 1 = 0,   6 π , π  2 2  π π 2 2 p p 2 2 76. 7 x ≈ 1 154, 0 534 cos2 x 2 cos x 1 = 0, [0, π] 3 78. 2 x ≈ 1 110 2 sec2 x + tan x 6 = 0,   π , π  2 2  4 0 π 3 x ≈ 1 998 π π 2 2 6 x ≈ 1 035, 0 870 79. (a) f (x) = sin2 x + cos x 2 (b) 0 2π 2 Maximum: (1.0472, 1.25) Maximum: (5 2360,1 25) Minimum: (0,1) Minimum: (3.1416, 1) 80. (a) f (x) = cos 2 x sin x 2 (b) 0 2π 81. (a) 2 Maximum: (3 6652, 1 25)Maximum: (5 7596,1 25)Minimum: (1 5708, 1)Minimum: (4.7124,1) f (x) = sin x + cos x 3 0 2π 3 Maximum: (0.7854,1.4142) Minimum: (3.9270, 1.4142) (b)

Section 7.3

607

608 Chapter 7 Analytic Trigonometry

3π is undefined in original function So, it is not 2 a solution. 85.

= 3x + 1 appear to have one point of intersection. This implies there is one solution to the 86.

x 1 appear to have three points of intersection. This implies there are three solutions

82. (a) f (x) = 2 sin x + cos 2x 3 (b) 0 2π 3 Maximum: (0.5236,1.5) Maximum: (2 6180,1 5) Minimum: (1 5708,1.0) Minimum: (4.7124, 3.0) 83. (a) f (x) = sin x cos x 2 84. (a) f (x) = sec x + tan x x 6 0 2π 0 2π 2 Maximum: (0.7854, 0 5) Maximum: (3.9270, 0 5) Minimum: (2.3562, 0.5) (b) 8 Maximum: (3.1416, 4.1416) Minimum: (0,1) sec x tan x + sec2 x 1 = 0 Minimum: (5 4978, 0.5) 1 ⋅ sin x + 1 1 = 0 (b) sin2 x + cos2 x = 0 cos x cos x cos x sin2 x + 1 sin2 x = 0 2 sin2 x + 1 = 0 sin x + 1 1 = 0 cos2 x sin x + 1 cos2 x = 0 sin2 x = 1 cos2 x cos2 x 2 sin x + 1 cos2 x 2 = 0 sin x = ± 1 = ± 2 cos x 2 2 sin x + sin2 x = 0 x = π , 3π , 5π , 7π 4 4 4 4 ≈ 0.7854, 2.3562, 3.9270, 5.4978 cos2 x sin x + sin2 x = 0 sin x(1 + sin x) = 0 sin x = 0 or 1 + sin x = 0 x = 0, π ≈ 0, 3.1416 sin x = 1 x = 3π 2

of y1 = 2 sin x and y2 equation 2 sin x = 3x + 1.

of y1 = 2 sin x

The graphs

and

(a) Domain: all real numbers except x = 0.

(b) The graph has y-axis symmetry

(d) sin x x = 0 has four solutions in the interval [ 8, 8].

Section 7.3 Solving Trigonometric Equations 609

90. y1 = 1.56e 0.22t cos 4.9

Right-most point of intersection: (1.96, 1)

The displacement does not exceed one foot from equilibrium after t = 1 96 seconds.

(a) Domain: all real numbers x except x = 0.

(b) The graph has y-axis symmetry and a horizontal asymptote at y = 1.

(d) There are infinitely many solutions in the interval 2

Left point of intersection: (1.95, 75)

Right point of intersection: (10 05, 75)

So, sales exceed 7500 in January, November, and December.

(e) The greatest solution appears to occur at

© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.  x  6 1   32 32 Monthly sales (in thousands of dollars)

) = sin x x

87. f (x

0 10 sin x  1  = 0   sin x = 0 x = 2π, π, π, 2π 88. f (x) = cos 1 4

 πt    y2 = 75. x

91. Graph y = 58.3 + 32 cos

S 100 75 [ 1,1] They occur at x = (2n + 1)π any integer where n is 50 25 89.

x ≈

6366. y = 1 (cos 8t 3 sin 8t) x 2 4 6 8 10 12 Month (1 ↔ January) 12 92. Range = 300 feet 1 (cos 8t 3 sin 8t) = 0 v0 = 100 feet per second 12 cos 8t = 3 sin 8t r = 1 v0 2 sin 2θ 1 = tan 8t 3 8t ≈ 0.32175 + nπ t ≈ 0.04 + nπ 8 1 (100)2 sin 2θ sin 2θ 2θ θ or = 300 = 0.96 = arcsin(0.96) ≈ 73.74° ≈ 36.9° In the interval 0 ≤ t ≤ 1, t ≈ 0.04, 0.43, and 0.83 2θ = 180° arcsin(0.96) ≈ 106.26° θ ≈ 53.1°