C H A P T E R 2 Analytic Trigonometry€¦ · 2 2 2 csc x = 1 + 7 = sin = cos x 2 214 Chapter 2...

2

4

= −

= −

2

5

3

= −

= −

= −

C H A P T E R 2

Analytic Trigonometry

Section 2.1 Using Fundamental Identities 1. tan u

2. csc u

9. sin θ 3

= − , cos θ 4

> 0 θ is in Quadrant IV.

3 9 7

3. cot u cos θ = 1 − − =

− 3

1 − = 16 4

4. csc u tan θ =

sin θ = 4 = − 3 3 7

5. 1 cos θ 7 7 7

4

6. − sin u

sec θ 1 1 4 = = = = 4 7

7. sec x

5

, tan x < 0 x is in Quadrant II.

cos θ 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

csc θ

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

sin θ

= − 1 − 2

= − 1 − 4

= − 5

5

csc x = 1

= 5

= 5 21

9 3

− 5

sin x 21 21 tan θ = sin θ = 2 = −

5

cot x = 1

= − 2

= − 2 21

cos θ 2 2

3

tan x 21 21 1 1 3

8. csc x

7

, tan x > 0 x is in Quadrant III.

sec θ = cos θ

= 2

= 2

3

6 cot θ =

1 =

1 = −

2 2 5

sin x = 1

= 1

= − 6 tan θ

− 5 5 5

csc x −

7 7 6

csc θ

2

= 1

= 1

= − 3 3 5

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2

7 cos x = − 1 −

−

6

− 6

= − 1 − 36

= − 13

49 7

sin θ −

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





2

2

2

csc x = − 1 + 7

= −

= −

cos x

2

214 Chapte r 2 A nalytic Trigo nomet ry

11.

tan x = 2, cos x > 0 x is in Quadrant I. 15. cos x(1 + tan 2 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 tan 2 x

sin 2 x 1

2 17. = = ⋅ = sec2 x

cos x = 1

= 1

= 3

= 3 13

sin 2 x sin 2 x cos2 x sin 2 x

sec x 13 13 13

3

Matches (e).

cos2 (π 2) − x

sin 2 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.

tan θ cot θ

1 tan θ

= tan θ

tan x = 1

= 1

= 4 sec θ 1

cot x 7 7

4

4

16 65

cos θ

= 1 1

sec x = − 1 + = − 7

2

= −

1 + = − 49 7

1 + 49

= − 65

π

cos θ = cos θ

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. tan 2 x − tan 2 x sin 2 x = tan 2 x(1 − sin 2 x) = tan 2 x cos2 x

2

13. sec x cos x = 1

cos x

= sin x

⋅ cos2 x cos2 x

= 1

Matches (c).

22.





= si

n 2

x

sin 2 x sec2 x − sin 2 x = sin 2 x(sec2 x − 1)

14. cot 2 x − csc2 x = (csc2 x − 1) − csc2 x

= −1 = sin 2 x tan 2 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





= −

2

Section 2.1 U s i ng Fundam ental Ident i ties 215

25. 1 − 2 cos

x + cos x = (1 − cos x)

26. sec4 x − tan 4 x = (sec2 x + tan 2 x)(sec2 x − tan 2 x)2 4 2

2 ( 2 2

)( ) = (sin 2 x) = sec x + tan x 1

27.

28.

= sin 4 x

cot3 x + cot 2 x + cot x + 1 = cot 2 x(cot x + 1) + (cot x + 1) = (cot x + 1)(cot 2 x + 1) = (cot x + 1)csc2 x

sec3 x − sec2 x − sec x + 1 = sec2 x(sec x − 1) − (sec x − 1) = (sec2 x − 1)(sec x − 1) = tan 2 x(sec x − 1)

= sec 2 x + tan 2 x

29.

3 sin 2

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 cos2 x + 5 cos x − 6 = (3 cos x − 2)(2 cos x + 3) sin u

= cos u + sin u

cos u

2 2 2

31. cot 2 x + csc x − 1 = (csc2 x − 1) + csc x − 1 39. 1 − sin x

= cos x = cos2 x tan 2 x = (cos2 x)

sin x

2 2 2

= csc2 x + csc x − 2

= (csc x − 1)(csc x + 2)

csc x − 1 cot x

= sin 2 x

cos x

32.

sin 2 x + 3 cos x + 3 = (1 − cos2 x) + 3 cos x + 3

= − cos2 x + 3 cos x + 4

= −(cos2 x − 3 cos x − 4) = −(cos x + 1)(cos x − 4)

40. cos2 y

1 − sin y

1 − sin 2 y =

1 − sin y

= (1 + sin y)(1 − sin y)

1 − sin y

= 1 + sin y

33.

tan θ csc θ

= sin θ

⋅ 1

= 1

= sec θ

41. (sin x + cos x)2 = sin 2 x + 2 sin x cos x + cos2 x

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

cos θ sin θ cos θ = 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 csc2 x − 4

= 4(csc2 x − 1) = 4 cot 2 x

35.

sin φ (csc φ − sin φ ) = (sin φ ) 1

− sin 2 φ

43.

1 +

1 =

1 − cos x + 1 + cos x

sin φ 1 + cos x 1 − cos x (1 + cos x)(1 − cos x)

= 1 − sin 2 φ = cos2 φ = 2 1 − cos2 x

36.

cos x(sec x − cos x) = cos x 1

− cos x

= 2

cos x

sin 2 x

= 1 − cos2 x

= sin 2 x

44.

= 2 csc2 x

1 −

1 =

sec x − 1 − (sec x + 1)

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

sin 2 β = +

cos β

cos2 β

cos β

sec2 x − 1

= −2

sin 2 β + cos2 β =

cos β

tan 2 x

1 = −2 2

= 1 tan x

cos β = sec β

= −2 cot 2 x






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 − sin 2 x

= − 2sin x cos x

cos2 x

= − 2sin x

cos x

= − 2 tan x

46.

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)

= 2sin x 1 − cos2 x

= 2sin x sin 2 x

= 2 sin x

= 2 csc x

47. sec2 x

tan x − = tan 2 x − sec2 x 50.

5 ⋅

tan x − sec x = 5( tan x − sec x)

tan x tan x

= −1

= −cot x tan x

2 2

tan x + sec x tan x − sec x tan 2 x − sec2 x

5( tan x − sec x) =

−1

= 5(sec x − tan x)

48. cos x 1 + sin x cos x + (1 + sin x)

+ =

y = 1

sin x cot x + cos x1 + sin x cos x cos x(1 + sin x)

cos2 x + 1 + 2 sin x + sin 2 x 51. 1 ( )

2

= 1

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

= 2 + 2 sin x

2 sin x

1

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

2(1+ sin x) =

cos x(1 + sin x) = cos x

2

= 2 cos x

= 2 sec x

− 2π 2π

49.

sin 2 y

1 − cos y

− 2

1 − cos2 y =

1 − cos y

(1 + cos y)(1 − cos y) =

1 − cos y





= 1 + cos y





Section 2.1 U s i ng Fundam ental Ident i ties 217

52. y1 = sec x csc x − tan x

1 1 sin x

= −

53. y = tan x + 1

1 sec x + csc x

cos x sin x cos x sin x

+ 1

1 sin 2 x = − =

cos x 1 1

cos x sin x

1 − sin 2 x cos x sin x

cos x +

sin x

= cos x sin x

cos2 x =

cos x sin x

cos x =

sin x + cos x

= cos x sin x + cos x

sin x cos x

sin x + cos x

sin x cos x

sin x =

= cot x 6

= sin x

cos x sin x + cos x

2

− 2π 2π

− 2π 2π

− 6

54.

y1 =

1 1

− 2

− 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

sin 2 x

sin x

= = = = tan x − 5sin x cos x sin x cos x cos x

55. Let x = 3 cos θ . 57. Let x = 2 sec θ .

9 − x2 =

=

=

9 − (3 cos θ )2

9 − 9 cos2 θ

9(1 − cos2 θ )

x2 − 4 =

=

=

(2 sec θ )2

− 4

4(sec2 θ − 1)

4 tan 2 θ

= 9 sin 2 θ

= 3 sin θ = 2 tan θ

56. Let x = 7 sin θ . 58. Let 3x = 5 tan θ .

2 2

49 − x2 = 49 − (7 sin θ )2

9x + 25 = (3x) + 25

= 49 − 49 sin 2 θ

= 49(1 − sin 2 θ )

= 49 cos2 θ





= (5 tan θ )2

+ 25

= 25 tan 2 θ + 25

= 25(tan 2 θ + 1)

= 7 cos θ

= 25 sec2 θ

= 5 sec θ





= ± 1 − 1


59. Let x = 2 sin θ .

4 − x2 = 2

4 − (2 sin θ )2

= 2

4 − 4 sin 2 θ = 2

4(1 − sin 2 θ ) = 2

4 cos2 θ = 2

2 cos θ = 2

cos θ = 2 2

2

sin θ = 1 − cos2 θ = 2 2

1 −

= ±

60. Let x = 2 cos θ .

2 2

61.

x = 6 sin θ

16 − 4x2 = 2 2 3 = 36 − x2

16 − 4(2 cos θ )2 = 2 2 = 36 − (6 sin θ )

2

16 − 16 cos2 θ

16(1 − cos2 θ )

= 2 2

= 2 2

=

=

36(1 − sin 2 θ )

36 cos2 θ

16 sin 2 θ = 2 2 = 6 cos θ

4 sin θ = ±2 2 cos θ = 3

= 1

sin θ = ± 2

2

sin θ

6 2

= ± 1 − cos2 θ

cos θ =

=

1 − sin 2 θ

1 − 1

2

2

2 = ±

3

= 1 4

2

= ± 3

2 2 =

2 62. x = 10 cos θ

5 3 =

5 3 =

5 3 =

5 3 =

100 − x2

100 − (10 cos θ )2

100(1 − cos2 θ )

100 sin 2 θ

5 3 = 10 sin θ

sin θ

= 5 3

= 3

cos θ = 10 2





1 − sin 2 θ = 2

3 1

1 −

= 2 2





Section 2.1 Usi ng Fundamenta l Ident i ties 219

63. sin θ = 1 − cos2 θ 67. μ W cos θ

μ

= W sin θ

W sin θ

=

= tan θ

Let y1 = sin x and y2 = 1 − cos2 x , 0 ≤ x ≤ 2π . W cos θ

y1 = y2 for 0 ≤ x ≤ π .

sec x tan x − sin x = 1

⋅ sin x

− sin x

So, sin θ = 1 − cos2 θ for 0 ≤ θ ≤ π . 68.

cos x

cos x

2

y2

0 2π

y1

sin x = − sin x

cos2 x

sin x − sin x cos2 x =

cos2 x

64.

− 2

cos θ

= − 1 − sin 2 θ

sin x(1 − cos2 x) =

cos2 x

2

sin x sin x =

2Let y1 = cos x and y2 = − 1 − sin 2 x , 0 ≤ x ≤ 2π . cos x

2

y1 = π y2 for

≤ x ≤ 3π

. = sin x tan x

So, cos θ

2 2

= − 1 − sin 2 θ

for π

3π

≤ θ ≤ .

69. True. sin u

2 2

2

0 2π

tan u = cos u

cot u = cos u sin u

sec u = 1 cos u

65.

− 2

sec θ =

1 + tan 2 θ

1

csc u = 1 sin u

Let y1 = and y2 = cos x

π

1 + tan 2 x , 0 ≤ x ≤ 2π .

3π

70. False. A cofunction identity can be used to transform a

tangent function so that it can be represented by a

cotangent function.

y1 = y2 for 0 ≤ x < and 2 2

< x ≤ 2π .

π −

71. As x → , tan x → ∞ and cot x → 0.

So, sec θ = 1 + tan2 θ for 0 ≤ θ < π 2

and 2

+ 1

3π < θ

2 < 2π . 4

72. As x → π , sin x → 0 and csc x = sin x

→ −∞.

y2

0 2π

y1

− 4

73. cos(−θ ) ≠ − cos θ

cos(−θ ) = cos θ

The correct identity is sin θ

= cos(−θ )

sin θ

cos θ

66. csc θ = 1 + cot 2 θ

1

= tan θ

Let y1 = and y2 = si

n x





1 + cot 2 x , 0 ≤ x ≤ 2π .

74. Let u = a tan θ ,

then

y1 = y2 for 0 < x < π .

a 2 + u 2 =

a 2 + (a tan θ )2

So, csc θ = 1 + cot 2 θ for 0 < θ < π .

2 2 2

= a + a 2

tan θ

0 2π

= a 2 (1 + tan 2 θ )

= a 2 sec2 θ

= a sec θ . − 2





= 1 + a

= 1 + b

2


75. Because sin 2 θ + cos2 θ = 1, then cos2 θ = 1 − sin2 θ .

cos θ

= ± 1 − sin θ

tan θ = sin θ =

sin θ

cot θ

cos θ ±

= cos θ

= ±

sin θ

1 − sin 2 θ

1 − sin 2 θ

sin θ

sec θ = 1

= 1

csc θ

cos θ

1 =

sin θ

± 1 − sin 2 θ

76. To derive sin 2 θ + cos2 θ = 1, let sin θ = a

and cos θ = b

a2 + b2 a2 + b2 .

2 2

a b a b2

So, sin 2 θ + cos2 θ = + = +

a2 + b2 a2 + b2 a2 + b2

a2 + b2

= a2 + b2

a2 + b2

= 1.

To derive 1 + ta n 2 θ

= sec2 θ , let tan θ = a

and sec θ b

a 2 + b 2 = .

b

So, 1 + ta n 2 θ

2

b

a2

= 1 + = b2

b2 + a2

b2

2 2 a2 + b2 a2 + b2

= = b2 b

= sec2 θ .

To derive 1 + cot 2 θ

= csc2 θ , let cot θ = b

and csc θ a

a 2 + b 2 = .

a

So, 1 + cot 2 θ 2

a

= 1 + b2

a2

2

a2 + b2 a2 + b2

= = a2 a2

2

a2 + b2

= = csc2 θ .

a

Answers will vary.





1 1 +

sin θ

Section 2.2 Verifying Tri gonom e t r ic Identities 221

77.

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 2.2 Verifying Trigonometric Identities 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. cot 2 u

7. −csc u

18.

sec2 y − cot 2 π

2

− y

= sec2 y − tan 2 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 − sin 2 β = cos2 β − (1 − cos2 β ) = 2 cos2 β − 1

1

= csc x − sin x

13.

14.

cos2 β − sin 2 β

sin 2 α − sin 4 α

= (1 − sin 2 β ) − sin 2 β

= 1 − 2 sin 2 β

= sin 2 α (1 − sin 2 α )

= (1 − cos2 α )(cos2 α )

= cos2 α − cos4 α

21.

1 + sin θ

cos θ

+ cos θ

= 1 + sin θ

=

=

(1 + sin θ )2

+ cos2 θ

cos θ (1 + sin θ )

1 + 2 sin θ + sin 2 θ + cos2 θ


2 + 2 sin θ


2(1+ sin θ )





15. tan π 2 − θ

tan θ = cot θ tan θ =


1 = tan θ

tan θ

= 1

= 2

cos θ

= 2 sec θ





( )

(


cos θ cot θ cos θ cot θ − (1 − sin θ ) 1

22. 1 − sin θ

− 1 =

=

1 − sin θ

cos θ cos θ

sin θ −

1 + sin θ

⋅ 1 − sin θ

cos2 θ − sin θ + sin 2 θ

sin θ

sin θ

25.

26.

27.

sec y cos y = cos y = 1 cos y

cot 2 y(sec2 y − 1) = cot 2 y tan2 y = 1

tan =

( ) = sin θ tan θ

2 θ sin θ cos θ tan θ =

sin θ (1 − sin θ ) sec θ 1 cos θ

= 1 − sin θ sin θ (1 − sin θ )

= 1 sin θ

= csc θ

28.

cot 3 t

csc t

cot t cot 2 t =

csc t

cot t (csc2 t − 1) =

csc t

cos t csc2 t − 1

sin t23.

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

2 cos x

= 1

sin t

= cos t sin t

csc2

sin t

t − 1)

= −sin 2 x = cos t(csc2 t − 1)

= −2 ⋅ 1

⋅ cos x

1 1 + tan 2 β

sin x sin x 29. tan β

+ tan β = tan β

24.

= −2 csc x cot x

cos x − cos x

= cos x(1 − tan x) − cos x

sec2 β =

tan β

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)

= −sin x cos x cos x − sin x

sin x cos x

cos x

= sin x − cos x

30. sec θ − 1

= sec θ − 1

⋅ sec θ sec θ (sec θ − 1)

=

= sec θ

1 − cos θ 1 − (1 sec θ ) sec θ sec θ − 1

cot 2 t cos2 t sin 2 t cos2 t 1 − sin 2 t 1

31. = = = 33. sec x − cos x = − cos x

csc t 1 sin t sin t sin t cos x

1 − cos2 x =





32. sin x

cos x + sin x tan x = cos x + sin x cos x

cos2 x + sin 2 x

cos x

sin 2 x =

cos x

= cos x

= sin x ⋅

sin x

= 1 cos x

= sec x

cos x

= sin x tan x





Section 2.2 Veri fying Tri gono met r ic Id enti ties 223

34.

cot x − tan x = cos x −

sin x

sin x cos x

cos2 x − sin 2 x =

sin x cos x 1 − sin 2 x − sin 2 x

= sin x cos x

1 − 2 sin 2 x =

sin x cos x

1 1 − 2 sin 2 x =

cos x sin x

1 1 2 sin 2 x =

cos x sin x

− sin x

= sec x(csc x − 2 sin x)

cot x cos x sin x cos2 x 1 − sin 2 x 1 sin 2 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 ⋅ cos2 x = cos3 x

sin x

38.

sec6 x(sec x tan x) − sec4 x(sec x tan x) = sec4 x(sec x tan x)(sec2 x − 1) = sec4 x(sec x tan x) tan 2 x = sec5 x tan3 x

39. (1 + sin y)1 + sin(− y) = (1 + sin y)(1 − sin y)

= 1 − sin 2 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

⋅ cot x cot y

(1 + sin θ )2

= 1 − sin 2 θ

(1 + sin θ ) 2

=1 − tan x tan y 1 − 1

⋅ 1 cot x cot y cos2 θ

cot x cot y 1 + sin θ

== cot y + cot x

cot x cot y − 1 cos θ

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 + sin 2 x − sin 2 y

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

(cos2 x + sin 2 x) − (cos2 y + sin 2 y) =

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

43. = 0





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 .





( )( )

( 2 2 )


45. (a)

3

− 2π 2π

− 1

Identity

(b)

Identity

(c) 1 + cot 2 x cos2 x = csc2 x cos2 x = 1

⋅ cos2 x = cot 2 x sin 2 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) = sin 2 x(2 + 3 cos2 x) ≠ sin2 x(3 + 2 cos2 x)

48. (a)

− π

5

y1 y2

π

(b)

− 5

Not an identity

sin 4 x

sin 2 x

Not an identity

(c) tan 4 x + tan 2 x − 3 = + − 3

cos4 x

1 cos2 x

sin 4 x

= cos2 x

cos2 x

+ sin 2 x − 3

1 sin 4 x + sin 2 x cos2 x =

cos2 x

cos2 x − 3

1 sin 2 x = sin x + cos x − 3

cos2 x cos2 x

1 sin 2 x =

cos2 x cos2 x

⋅ 1 − 3





= sec2 x tan 2 x − 3

≠ sec2 x(4 tan 2 x − 3)





.

cos2 x

( )

Section 2.2 Veri fying Tri gono met r ic Id entit ies 225

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) =

Not an identity

(c) cot α

is the reciprocal of csc α + 1

sin x sin x(1 − cos x) csc α + 1 cot α

1 − cos2 x =

sin x(1 − cos x) sin 2 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 tan 2 x

5

1 − cos x = tan x

2 4

52. (tan 2 x + tan 4 x) sec2 x = sin x

sin x 1 +

cos4 x cos2 x

1 = sin 2 x + sin 4 x

cos4 x cos2 x

1 sin 2 x cos2 x + sin 4 x =

cos4 x cos2 x

1 sin 2 x cos2 x + sin 2 x =

cos4 x cos2 x

1 sin 2 x =

⋅ 1 = sec4 x ⋅ tan 2 x

cos4 x cos2 x

53. (sin 2 x − sin 4 x) cos x = sin 2 x(1 − sin 2 x) cos x

= sin 2 x cos2 x cos x

= sin 2 x cos3 x

54. sin 4 x + cos4 x = sin 2 x sin 2 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. sin 2 25° + sin 2 65° = sin 2 25° + cos2 (90° − 65°)

= sin 2 25° + cos2 25°

= 1

56. tan 2 63° + cot 2 16° − sec2 74° − csc2 27° = tan 2 63° + cot 2 16° − csc2 (90° − 74°) − sec2 (90° − 27°)

= tan 2 63° + cot 2 16° − csc2 16° − sec2 63°





=

(t

a

n 2

6

3

°

−

s

e

c2

6

3

°

) +

(c

o

t 2

1

6

°

− c

s

c2

1

6

°

) =

−

1

+

(−

1

)

=

−2





θ 15° 30° 45° 60° 75° 90°

s 18.66 8.66 5 2.89 1.34 0

1


57. Let θ

= sin−1 x sin θ = x = x

.

60. Let θ = cos−1 x + 1 cos θ = 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 x1 −

sin 2 x x

θ

1 − x2

= cos x(1 − csc2 x) = −cos x(csc2 x − 1)

= −cos x cot 2 x

From the diagram,

cos(sin −1 x) = cos θ

1 − x 2

= = 1

1 − x2 .

62. (a)

(b)

h sin(90° −θ ) sin θ

= h cos θ

= h cot θ sin θ

59. Let θ = sin −1 x − 1 sin θ = x − 1

.

4 4

4

x − 1

(c) Maximum: 15°

Minimum: 90°

(d) Noon

θ 63. False. tan x2 = tan(x ⋅ x) and

16 − (x − 1)2

From the diagram,

tansin−1 x − 1 = tan θ

x − 1

= .

tan 2 x = (tan x)(tan x), tan x2

64. True. Cosine is an even function,

≠ tan 2 x.

π π 4 16 − (x − 1)

2 cosθ − = 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.





2

b 2

b

2 2

2

Section 2.3 Solving Trigo nomet ric Equ a tion s 227

66. If sin θ = a , sec θ =

c , and 68. tan θ = sec2 θ − 1

c b True identity: tan θ

= ± sec2 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 − cos θ

= sin θ

b2 − 1

= c2

(1 − cos θ )

= (sin θ )

2 2

b2 1 − 2 cos θ + cos θ = sin θ

c2 − b2

= b2

c2

1 − 2 cos θ + cos2 θ

2 cos2 θ − 2 cos θ

= 1 − cos2 θ

= 0

b2

c2 − b2 b2

= ⋅ b2 c2

c2 − b2

2 cos θ (cos θ − 1) = 0

The equation is not an identity because it is only true

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

the equation is not true is − π

.

= c2

a2 70. =

c2

1 + tan θ

(1 + tan θ )2

2

= sec θ

= (sec θ )2

a 2 2=

1 + 2 tan θ + tan θ = sec θ

c

= sin 2 θ . 1 + 2 tan θ + tan 2 θ

2 tan θ

= 1 + tan 2 θ

= 0

67. Because sin 2 θ

= 1 − cos2 θ , then tan θ = 0

sin θ = ± 1 − cos2 θ ; sin θ ≠ 1 − cos2 θ

if θ This equation is not an identity because it is only true

lies in Quadrant III or IV. when tan θ = 0. So, one angle for which the equation

One such angle is θ = 7π

. 4

is not true is π

. 6

Section 2.3 Solving Trigonometric Equations 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 −





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





=

2 cos2 2

2

2 cos2 2

2 = −


7. 3 tan 2

2 x − 1 = 0 10. csc4

x − 4 csc2

x = 0

(a) x =

π 12

2

3 tan 2 π

− 1 = 3 tan 2 π − 1

(a) x =

π 6

csc4 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 tan 2 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 4 x − 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

9. 2 sin 2 x − sin x − 1 = 0 3 3





2

(a)

x = π 2

2 sin 2 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 sin 2 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





Section 2.3 Solving Trigo nomet ric Equ a t ion s 229

15. 3 sec2 x − 4 = 0 20. (2 sin 2 x − 1)(tan 2 x − 3) = 0

sec2 x = 4 3

2 sin 2 x − 1 = 0 or tan 2 x = 3

sec x = ± 2 3

sin 2 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 cot 2 x − 1 = 0

cot 2 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 sin 2 x = 0

sin 2 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 tan 2 x − 1) = 0

4

x = 5π 4

x = 7π

+ 2nπ

+ 2nπ

tan x = 0

x = nπ

or 3 tan 2 x − 1 = 0

tan x = ±

x = π

3

3

+ nπ , 5π

+ nπ





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





= −


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 sin 2 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 + sin 2 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






31. 2 sin 2 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 cos 2 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.

tan 2 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

sin 2 x = 3 cos2 x

csc2 x + 2 csc x cot x + cot 2 x = 1

cot 2 x + 1 + 2 csc x cot x + cot 2 x = 1

2 cot 2 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

sin 2 x − 3 cos2 x = 0 = = − 2 2 sin x

sin x

sin 2 x − 3(1 − sin 2 x) = 0

4 sin 2 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 + tan 2 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(tan 2 x + 1) + tan 2 x − 3 = 0 1 + sin x =

cos x

3 tan 2 x − 1 = 0 (1 + sin x)2

= cos2 x

tan x = ± 3

3 1 + 2 sin x + sin 2 x = cos2 x

1 + 2 sin x + sin 2 x = 1 − sin 2 x

x = π

, 5π

, 7π

, 11π

6 6 6 6

2 sin 2

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 sin 2 x + 1 = 0

sin 2 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





2

2


39. 2 cos 2x − 1 = 0 45. 3 tan

x −

3 = 0

cos 2x = 1 2

2

tan x

= 3

2 x = π

+ 2nπ

or 2 x = 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 2 x + 3 = 0 2

x

sin 2 x = − 3

2

tan 2

x

= − 3

= 2π

+ nπ

x = 4π

+ 2nπ

2 x = 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 4 x − 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 = −cos π x

4 x = π + 2nπ or 4 x =

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

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






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 tan 2 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 sin 2

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. tan 2 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. tan 2 x − tan x − 2 = 0

(tan x + 1)(tan x − 2) = 0

tan x + 1 = 0 or tan x − 2 = 0


x = 3π 4

+ nπ

x = arctan 2 + nπ





3 3


61. sec2 x − 6 tan x = − 4

1 + tan 2 x − 6 tan x + 4 = 0

tan 2 x − 6 tan x + 5 = 0

(tan x − 1)(tan x − 5) = 0

tan x − 1 = 0 tan x − 5 = 0

tan x = 1 tan x = 5

x = π 4

+ nπ

x = arctan 5 + nπ

62. sec2 x + tan x − 3 = 0

1 + tan 2 x + tan x − 3 = 0

tan 2 x + tan x − 2 = 0

(tan x + 2)(tan x − 1) = 0

tan x + 2 = 0 tan x − 1 = 0


x = arctan(−2) + nπ x = arctan(1) + nπ

63.

≈ −1.1071 + nπ

2 sin 2 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 − sin 2 x) + 7 sin x − 5 = 0

− 2 sin 2 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. cot 2 x − 9 = 0

cot 2 x = 9

1 = tan 2 x

9

± 1

= tan x 3

x = arctan 1 + nπ , arctan(− 1 ) + nπ





5 5

=


66. cot 2 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





3


71. 12 sin 2 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 tan 2 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. tan 2 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.





2 2





−

−

−


75. 3 tan 2 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) 2 sin x cos x − sin x = 0

sin x(2 cos x − 1) = 0

0 2π

sin x = 0 or 2 cos x − 1 = 0

x = 0, π

− 2 ≈ 0, 3.1416 cos x =

1 2

Maximum: (1.0472, 1.25)

Maximum: (5.2360, 1.25)

Minimum: (0, 1)

Minimum: (3.1416, −1)

x = π

, 5π

3 3

≈ 1.0472, 5.2360

80. (a) f (x) = cos2 x − sin x 2

(b) −2 sin x cos x − cos x = 0

−cos x(2 sin x + 1) = 0

−cos x = 0 2 sin x + 1 = 00 2π

− 2

cos x = 0 sin x = − 1 2

x = π

, 3π

x = 7π

, 11π

Maximum: (3.6652, 1.25) 2 2 6 6

81. (a)

Maximum: (5.7596, 1.25)

Minimum: (1.5708, −1)

Minimum: (4.7124, 1)

f (x) = sin x + cos x

3

≈ 1.5708, 4.7124

(b) cos x − sin x = 0

cos x = sin x

≈ 3.6652, 5.7596

0 2π

1 = sin x cos x





− 3

Maximum: (0.7854, 1.4142)

Minimum: (3.9270, −1.4142)

tan x = 1

x = π

, 5π

4 4

≈ 0.7854, 3.9270





2

+ =


82. (a) f (x) = 2 sin x + cos 2 x

3

(b) 2 cos x − 4 sin x cos x = 0

2 cos x(1 − 2 sin x) = 0

2 cos x = 0 1 − 2 sin x = 00 2π

x = π

, 3π sin x =

1

2 2 2− 3

Maximum: (0.5236, 1.5)

Maximum: (2.6180, 1.5)

Minimum: (1.5708, 1.0)

Minimum: (4.7124, − 3.0)

≈ 1.5708, 4.7124 x = π

, 5π

6 6 ≈ 0.5236, 2.6180

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) −sin 2 x + cos2 x = 0 cos x cos x cos x

−sin 2 x + 1 − sin 2 x = 0

−2 sin 2 x + 1 = 0

sin x + 1 − 1 = 0

cos2 x

sin x + 1 cos2 x − = 0

sin 2 x = 1 cos2 x cos2 x

2 sin x + 1 − cos2 x

2 = 0

sin x = ± 1

= ± 2 cos x

2 2 sin x + sin 2 x = 0

x = π

, 3π

, 5π

, 7π

4 4 4 4

≈ 0.7854, 2.3562, 3.9270, 5.4978

cos2 x

sin x + sin 2 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

3π is undefined in original function. So, it is not

2

a solution.

85. The graphs of y1 = 2 sin x and y2

equation 2 sin x = 3x + 1.

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

86. The graphs of y1 = 2 sin x and y2 1 x 1 appear to have three points of intersection. This implies there are three solutions 2





+ to the equation 2 sin x = 1 x 1. 2





x

32

32

Month

ly s

ales

(in t

housa

nds

of

do

llar

s)


87. f (x) = sin x 90. Graph the following equations. 4

x y1 = 1.56t −1 2 cos 1.9t 0 10

(a) Domain: all real numbers except x = 0. y2 = 1

(b) The graph has y-axis symmetry.

(c) As x → 0, f (x) → 1.

y3 = −1 − 4

(d) sin x

x

= 0 has four solutions in the interval [−8, 8]. The rightmost point of intersection is at approximately

(1.91, −1).

The displacement does not exceed one foot from

sin x 1

= 0

equilibrium after t ≈ 1.91 seconds.

π t

sin x = 0

x = −2π , −π , π , 2π

91. Graph y1 = 58.3 + 32 cos 6

y2 = 75.

88. f (x) = cos

1

x

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

(b) The graph has y-axis symmetry and a horizontal

asymptote at y = 1.

(c) As x → 0, f (x) oscillates between −1 and 1.

Left point of intersection: (1.95, 75)

Right point of intersection: (10.05, 75)

So, sales exceed 7500 in January, November,

and December. S

(d) There are infinitely many solutions in the interval

2

100

75

[−1, 1]. They occur at x = (2n + 1)π

any integer.

where n is 50

25

x

89.

(e) The greatest solution appears to occur at

x ≈ 0.6366.

y = 1

(cos 8t − 3 sin 8t )

92.

2 4 6 8 10 12

Month (1 ↔ January)

Range = 300 feet

12 v0

= 100 feet per second

1 (cos 8t − 3 sin 8t ) = 0

12

r = 1 v02 sin 2θ

cos 8t = 3 sin 8t

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°





0

16 2

16 2

16 2

6

1 1

6 6


93. (a) and (c) 100

94. h(t ) = 53 + 50 sin π

t − π

(a) h(t ) = 53 when 50 sin

π t −

π = 0.

1 12

0 π

t − π

= 0 or π

t − π

= π

The model fits the data well. 16 2 16 2

π π π 3π

(b) C = a cos(bt − c) + d t = t =

16 2 16 2

a = 1[high − low] =

1[84.1 − 31.0] = 26.55 t = 8 t = 24

2 2

p = 2[high time − low time] = 2[7 − 1] = 12 A person on the Ferris wheel will be 53 feet

above ground at 8 seconds and at 24 seconds

b = 2π p

= 2π

= π

12 6

(b) The person will be at the top of the Ferris wheel

when

The maximum occurs at 7, so the left end point is sin π

t − π

= 1

c = 7 c = 7

π =

7π b 6

π t −

π = π

16 2 2

d = [high + low] = [93.6 + 62.3] = 57.55 2 2

π

16

t = π

C = 26.55 cos π

t − 7π

+ 57.55

(d) The constant term, d, gives the average maximum

t = 16.

The first time this occurs is after 16 seconds.

2π

temperature. The period of this function is π 16

= 32.

95.

The average maximum temperature in Chicago is

57.55°F.

(e) The average maximum temperature is above 72°F

from June through September. The average

maximum temperature is below 70°F from October

through May.

A = 2 x cos x, 0 < x < π 2

During 160 seconds, 5 cycles will take place and

the person will be at the top of the ride 5 times,

spaced 32 seconds apart. The times are: 16 seconds,

48 seconds, 80 seconds, 112 seconds, and

144 seconds.

(a) 2

π 2

− 2

(b)

The maximum area of A ≈ 1.12 occurs when x ≈ 0.86.

A ≥ 1 for 0.6 < x < 1.1

96. f (x) = 3 sin(0.6 x − 2)





(a) Zero : sin(0.6x − 2) = 0 (b) g (x) =

0.6 x − 2 = 0

0.6x = 2

x = 2

= 10

4

0

−0.45x2 + 5.52 x − 13.70

6

0.6 3 f

g

− 4

For 3.5 ≤ x ≤ 6 the approximation appears to be good.





=

= −

=


(c) −0.45x2 + 5.52x − 13.70 = 0

−5.52 ±

(5.52)2

− 4(−0.45)(−13.70)

x =

x ≈ 3.46, 8.81

2(−0.45)

The zero of g on [0, 6] is 3.46. The zero is close to the zero 10

3

≈ 3.33 of f.

97. f (x) = tan π x 4

Because tan π 4 = 1, x = 1 is the smallest nonnegative

fixed point.

100. False.

sin x = 3.4 has no solution because 3.4 is outside the

range of sine.

98. Graph y = cos x and y = x on the same set of axes.

101. cot x cos2 x = 2 cot x

cos2 x = 2

Their point of intersection gives the value of c such that

f (c) = c cos c = c.

2 (0.739, 0.739)

− 3 3

− 2

c ≈ 0.739

cos x = ± 2

No solution

Because you solved this problem by first dividing by

cot x, you do not get the same solution as Example 3.

When solving equations, you do not want to divide each

side by a variable expression that will cancel out because

you may accidentally remove one of the solutions.

102. The equation 2 cos x − 1 = 0 is equivalent to

99. True. The period of 2 sin 4t − 1 is π

and the period of

cos x = 1 . So, the points of intersection of y cos x 2

2 and y 1 2

represent the solutions of the equation

2 sin t − 1 is 2π . 2 cos x − 1 = 0. In the interval (− 2π , 2π ) the solutions

In the interval [0, 2π ) the first equation has four cycles

whereas the second equation has only one cycle, so the

first equation has four times the x-intercepts (solutions)

as the second equation.

of the equation are x 5π

, − π

, π

, and 3 3 3

5π .

3

103. (a)

3

0 2π

− 2

The graphs intersect when x = π 2

and x = π .

(b)

3

0 2π

− 2

The x-intercepts are π

, 0

and (π , 0).

2

(c) Both methods produce the same x-values. Answers will vary on which method is preferred.





4 6 4 6


Section 2.4 Sum and Difference Formulas 1. sin u cos v − cos u sin v π π π π π π

7. (a) cos + = cos cos − sin sin

2. cos u cos v − sin u sin v

3. tan u + tan v 1 − tan u tan v

4 3 4 3 4 3

2 1 2 3 = ⋅ − ⋅

2 2 2 2

2 − 6 =

4

4. sin u cos v + cos u sin v (b)

cos

π + cos

π

= 2

+ 1

= 2 + 1

5. cos u cos v + sin u sin v 4 3 2 2 2

7π π

5π π 1

6. tan u − tan v 8. (a) sin

6 −

3 = sin

6 = sin =

6 21 + tan u tan v

(b) sin 7π − sin

π = − 1

− 3 =

−1 − 3

9. (a)

sin(135° − 30°) = sin 135° cos 30° − cos 135° sin 30°

6 3 2 2 2

2 3 2 1 6 + 2

=

− − =

2 2 2 2 4

(b)

sin 135° − cos 30° = 2

− 3

= 2 − 3

10. (a)

2 2 2

cos(120° + 45°) = cos 120° cos 45° − sin 120° sin 45°

1 2 3 2 − 2 − 6

= −

−

=

2 2 2 2 4

(b)

cos 120° + cos 45° = − 1

+ 2

= −1 + 2

11.

sin 11π

= sin 3π

+ π

2 2 2

tan 11π

= tan 3π

+ π

12

4

= sin 3π

cos π + cos

3π sin

π tan 3π + tan

π

4 6 4 6 = 4 6

2 3 2 1 1 − tan 3π

tan π

= ⋅ + −

4 62 2 2 2

−1 + 3

co

s

11π





(

4 6

(

2 =

4

= cos 3π

3 − 1)

+ π

=

3

1

−

(−

1

)

3

3 12 =

−3 + 3 ⋅

3 − 3

= cos 3π

cos π − sin

3π sin

π 3 + 3 3 − 3

4 6 4 6 =

−12 + 6 3 = −2 + 3

2 3 2 1 2 = − ⋅ − ⋅ = − 3 + 1)

6

2 2 2 2 4





4 6

3 4

(

(

3 4

(

4 6

3 4

)

)

)

Section 2.4 Sum a nd Dif f erence Formu l as 243

12. 7π

= π

+ π

13.

sin 17π

= sin 9π 5π

−

12 3 4 12

sin 7π

= sin π

+ π

= sin 9π

cos 5π

− cos 9π

sin 5π

12

= sin π

cos π

+ sin π

cos π

4 6 4 6

2 3 2 1 =

−

−

3 4 4 3

3 2 2 1

2 2 2 2

= ⋅ + ⋅ 2 2 2 2 = −

2 3 + 1

4

= 2

3 + 1 4

cos

17π

12 = cos

9π

4

5π −

6

cos 7π

= cos π

+ π

= cos 9π

cos 5π

+ sin 9π

sin 5π

12

4 6 4 6

= cos π

cos π

− sin π

sin π 2 3 2 1

3 4 3 4 = 2

− 2

+ 2

2

= 1

⋅ 2

− 3

⋅ 2

2

2 2 2 2 = 4

(1 − 3)

2 =

4 1 − 3

tan

17π

= tan 9π

− 5π

tan 7π

= tan π

+ π

12

12 tan(9π 4) − tan(5π 6)

= 1 + tan(9π 4) tan(5π 6)

tan π

+ tan π

= 3 4 1 − (− 3 3)

1 − tan π

tan π

3 4

= 1 + (− 3 3)

3 + 1 = =

3 + 3 ⋅

3 + 3

1 − 3 3 − 3 3 + 3

= −2 − 3 =

12 + 6 3 6

= 2 + 3

14. − π

= π

− π

12 6 4

sin − π

= sin π

− π

cos − π

= cos π

− π

tan − π

= tan π

− π

12

6 4

12

6 4

12

6 4





( ( ) )

= sin π

cos π − sin

π cos

π = cos π

cos π + sin

π sin

π tan π − tan

π

6 4 4 6

1 2 2 3 = ⋅ − ⋅

2 2 2 2

2 = 1 − 3

4

6 4 6 4

3 2 1 2 = ⋅ + ⋅

2 2 2 2

= 2

3 + 1 4

= 6 4

1 + tan π

tan π

6 4

3 − 1

= 3

1 + 3

3

= −2 + 3





(

(

(

(

(

(

(

)

)

)

)

)


15. sin 105° = sin(60° + 45°)

= sin 60° cos 45° + cos 60° sin 45°

3 2 1 2 = ⋅ + ⋅

2 2 2 2

17. sin(−195°) = sin(30° − 225°) = sin 30° cos 225° − cos 30° sin 225°

= sin 30°(− cos 45°) − cos 30°(− sin 45°)

= 1

− 2

− 3

− 2

2 2 2 2

= 2

3 + 1)

4

cos 105° = cos(60° + 45°) = −

2 1 − 3

4

= cos 60° cos 45° − sin 60° sin 45°

1 2 3 2

= 2

4 3 − 1)

= ⋅ − ⋅ 2 2 2 2

2 = 1 − 3

4

cos(−195°) = cos(30° − 225°)

= cos 30° cos 225° + sin 30° sin 225°

= cos 30°(− cos 45°) + sin 30°(− 45°)

tan 105° = tan(60° + 45°) 3 2 1 2 =

−

+

−

tan 60° + tan 45° =

2 2 2 2

1 − tan 60° tan 45° = −

2 3 + 1

4

3 + 1 3 + 1 1 + 3 = = ⋅ tan(−195°) = tan(30° − 225°)

1 − 3 1 − 3 1 + 3

= tan 30° −tan 225°

= 4 + 2 3

−2 = −2 − 3 1 + tan 30° tan 225°

= tan 30° −tan 45°

16. 165° = 135° + 30°

sin 165° = sin (135° + 30°)

= sin 135° cos 30° + sin 30° cos 135°

1 + tan 30° tan 45°

3

3 − 1

3 − 3 3 − 3 = = ⋅

= sin 45° cos 30° − sin 30° cos 45° 3

1 + 3 + 3 3 − 3

3 2 3 1 2

= ⋅ − ⋅ 2 2 2 2

= −12 + 6 3

= − 2 + 3 6

= 2

3 − 1 4

18. 225° = 300° − 45°

cos 165° = cos (135° + 30°)

= cos 135° cos 30° − sin 135° sin 30°

= −cos 45° cos 30° − sin 45° cos 30°

sin 255° = sin(300° − 45°)

= sin 300° cos 45° − sin 45° cos 300°

= − sin 60° cos 45° − sin 45° cos 60°

2 3 2 1 = − ⋅ − ⋅ = −

3 ⋅

2 − 2

⋅ 1 = −

2 ( 3 + 1)

2 2 2 2 2 2 2 2 4

= − 2

3 + 1 4

tan 165° = tan (135° + 30°)

cos 255° = cos(300° − 45°)

= cos 300° cos 45° + sin 300° sin 45°

= cos 60° cos 45° − sin 60° sin 45°





( ) tan 135° + tan 30° =

1 2 3 2 = ⋅ − ⋅

2 = 1 − 3

1 − tan 135° tan 30°

= − tan 45° + tan 30° 1 + tan 45° tan 30°

−1 + 3

= 3

1 + 3

2 2 2 2 4

tan 255° = tan(300° − 45°)

tan 300° −tan 45° =

1 + tan 300° tan 45°

= − tan 60° − tan 45° 1 − tan 60° tan 45°

3 − 3 − 1

= −2 + 3 =

1 − 3 = 2 + 3





4 3 4 3

4 3

(

3 4

(

3 4

(

3 4

)

)

)


19. 13π =

3π + π

12 4 3

sin 13π

= sin 3π

+ π

tan 13π

= tan 3π

+ π

12

= sin 3π

cos π

+ cos 3π

sin π

12

tan 3π

+ tan π

4 3 4 3 =

2 1 2 3 1 − tan 3π

tan π

= ⋅ + −

4

3 2 2

= 2 (1 −

2 2

3)

= −1 + 3

( )4 1 − −1 ( 3)

cos 13π

= cos 3π

+ π 1 − 3 1 − 3

12 4 3

= − ⋅

1 + 3 1 − 3

= cos 3π

cos π

− sin 3π

sin π 4 − 2 3

4 3 4 3 = −

−2

2 1 2 3 = − ⋅ − ⋅

2 = − 1 + 3 = 2 − 3

20. 19π

2 2 2 2 4

π 5π = +

12 3 4

sin 19π

= sin π 5π

+ 12

= sin π

cos 5π

+ sin 5π

cos π

3 4 4 3

3 =

−

2

+

− 2 1

⋅

2 2 2 2

= − 2

3 + 1 4

cos 19π

= cos π 5π

+ 12

= cos π

cos 5π

− sin π

sin 5π

3 4 3 4

1 2 =

−

−

3 2

−

2 2 2 2

2 =

4 −1 + 3

tan 19π

= tan π

+ 5π

12

tan π + tan

5π





= 3 4

1 − tan π

tan 5π

3 4

tan π + tan

π

= 3 4

1 − tan π

tan π

3 4

3 + 1 1 + 3 = ⋅

1 − 3 1 + 3

= 4 + 2 3

− 2

= − 2 − 3





(

(

− =

(

= + − = (1 − )


21. − 5π = −

π − π

12 4 6

sin − π

− π

= sin − π

cos π

− cos − π

sin π

4 6

4

6

4

6

2 3 2 1 2 =

−

−

= −

3 + 1) 2 2 2 2 4

cos − π

− π

= cos − π

cos π

+ sin − π

sin π

4 6

4

6

4

6

2 3 2 1 2 =

+

− =

3 − 1) 2 2 2 2 4

tan − π tan

− π

− tan π

π 4

6

4 6 1 + tan −

π tan

π

4 6

−1 − 3

= 3 =

− 3 − 3

−

1 + (−1) 3

3 3

3

= − 3 − 3

⋅ 3 + 3

3 − 3 3 + 3

22. − 7π

= − π

= −12 − 6 3

6

− π

= − 2 − 3

12 3 4

sin −

7π = sin

− π

− π

= sin − π

cos π

− cos − π

sin π

12

3 4

3

4

3

4

3 2 1 2 2

= − − = − 3 + 1)

2 2 2 2 4

cos −

7π = cos

− π

− π

= cos − π

cos π

+ sin − π

sin π

12

3 4

3

4

3

4

1 2 3 2 2

3 2 2 2 2 4

tan

−

7π = tan

− π





π 3 4 tan

− π

− tan π

− =

= − 3 − 1

= 2 + 3 12 3 4

1 + tan − π

tan π 1 + (− 3)(1)

3 4





− = − (

− − = (

(

(

(

)


23. 285° = 225° + 60°

sin 285° = sin(225° + 60°) = sin 225° cos 60° + cos 225° sin 60°

2 1 2 3 2 = −

3 + 1)

2 2 2 2 4

cos 285° = cos(225° + 60°) = cos 225° cos 60° − sin 225° sin 60°

2 1 2 3 2 = −

3 − 1)

2 2 2 2 4

tan 285° = tan(225° + 60°) = tan 225° + tan 60° 1 − tan 225° tan 60°

1 + 3 1 + 3 = ⋅ =

4 + 2 3 = −2 − 3 = −(2 + 3)

1 − 3 1 + 3 −2

24. 15° = 45° − 30°

sin 15° = sin(45° − 30°) = sin 45° cos 30° − cos 45° sin 30°

2 3 2 1 2 ( 3 − 1) 2 =

−

= =

3 − 1) 2 2 2 2 4 4

cos 15° = cos(45° − 30°) = cos 45° cos 30° + sin 45° sin 30°

2 3 2 1 2 ( 3 + 1) 2 =

+

= =

3 + 1) 2 2 2 2 4 4

tan 15° = tan(45° − 30°) = tan 45° −tan 30° 1 + tan 45° tan 30°

1 − 3 3 − 3

= 3 = 3 = 3 − 3

⋅ 3 − 3

12 − 6 3

=

= 2 − 3

3 3 + 3 3 + 3 3 − 3 6

1 + (1) 3 3

25.

−165° = −(120° + 45°)

sin(−165°) = sin −(120° + 45°) = − sin(120° + 45°) = −[sin 120° cos 45° + cos 120° sin 45°]

3 2 1 2 2 (

3 1)= −

⋅ − ⋅

2 2 2 2 = − − 4

cos(−165°) = cos−(120° + 45°) = cos(120° + 45°) = cos 120° cos 45° − sin 120° sin 45°

1 2 3 2 = − ⋅ − ⋅

2 = − 1 + 3

2 2 2 2 4

tan(−165°) = tan −(120° + 45°) = −tan(120° + tan 45°) = − tan 120° + tan 45°

1 − tan 120° tan 45°

= − − 3 + 1

= − 1 − 3

⋅ 1 − 3

= − 4 − 2 3

= 2 − 3





1 − (− 3)(1) 1 + 3 1 − 3 −2





= − − = − (1 +

(

7 5

8 8

)

)


26. −105 = 30° − 135°

sin(30° − 135°) = sin 30° cos 135° − cos 30° sin 135° = sin 30°(−cos 45°) − cos 30° sin 45°

1 2 3 2 2

3 2 2 2 2 4

cos(30° − 135°) = cos 30° cos 135° + sin 30° sin 135° = cos 30°(−cos 45°) + sin 30° sin 45°

3 2 1 2 2

= − + = 1 − 3

2 2 2 2 4

tan(30° − 135°) = tan 30° −tan 135°

= tan 30° −(− tan 45°)

1 + tan 30° tan 135°

3 − (−1)

= 3 = 2 + 3

1 + (−1)

1 + tan 30°(− tan 45°) 3

3

27.

sin 3 cos 1.2 − cos 3 sin 1.2 = sin(3 − 1.2) = sin 1.8

36.

cos π

cos 3π

− sin π

sin 3π

= cos π

+ 3π

16 16 16 16 16 16

28. cos π

cos π

− sin π

sin π

= cos π

+ π π 2

7 5 7 5

= cos12π

= cos = 4 2

29.

30.

35

sin 60° cos 15° + cos 60° sin 15° = sin(60° + 15°)

= sin 75°

cos 130° cos 40° − sin 130° sin 40° = cos(130° + 40°)

= cos 170°

37.

38.

cos 130° cos 10° + sin 130° sin 10° = cos(130° − 10°)

= cos 120°

1 = −

2

sin 100° cos 40° − cos 100° sin 40° = sin (100° − 40°)

= sin 60°

3

31. tan (π 15) + tan(2π 5) =

= tan(π 15 + 2π 5) 2 1 − tan (π 15) tan (2π 5)

= tan(7π 15) 39.

tan(9π 8) − tan(π 8) 9π

= tan −

π

32.

tan 1.1 − tan 4.6

= tan(1.1 − 4.6) = tan(− 3.5) 1 + tan 1.1 tan 4.6

1 + tan(9π 8) tan(π 8)

= tan π

= 0

33.

34.

cos 3x cos 2 y + sin 3x sin 2 y = cos(3x − 2 y)

sin x cos 2x + cos x sin 2 x = sin(x + 2x) = sin(3x)

40.

tan 25° + tan 110° = tan(25° + 110°)

1 − tan 25° tan 110°

= tan 135°

= −1

35. sin π

cos π

+ cos π

sin π

= sin π

+ π

12 4 12 4 12 4

= sin π 3

3





= 2





(15, 8

17

v

17 17 15

− −

− +

5 85

4

15

−

25 25 24

Section 2.4 Sum a n d Diff erence Formu l as 249

For Exercises 41 – 46, you have:

sin u = – 3 , u in Quadrant IV cos u = 4 , tan u = – 4

5 5 3

cos v = 15 , v in Quadrant I sin v = 8 , tan v = 8

y y

)

u

x x

5

(4, − 3)

41.

Figures for Exercises 41–46

sin(u + v) = sin u cos v + cos u sin v

= −

3 15 +

4 8

44.

csc(u − v) = 1

= 1

( )

5 17

5 17

sin u − v

1 sin u cos v − cos u sin v

13 = 3 15 4 8

= − 85

5 17 5 17 1 85

42.

cos(u − v) = cos u cos v + sin u sin v

= 4 15

+ −

3 8

= = −

− 77 77 85

5 17

5 17 1 1

45. sec(v − u) = cos(v − u)

= cos v cos u + sin v sin u

60 − 24 36 = + = 1 1

85 85 85 = 15 4 8

= 3 60

24

3 8

+ − 17 5 17 1 85

+ −

85

43. tan(u + v) = tan u + tan v

= 4 15 =

36 =

36

1 − tan u tan v 1 −

−

3 8 85

4 15

− 13

60

13 5

13

tan u + tan v

−

3 +

8

= = −

= − 46. tan(u + v) = =

3 8

1 + 32

60

60 7 84 1 − tan u tan v 1 −

−

4 15

13

= 60 7

5

= − 13 84

1 1 84

For Exercises 47– 52, you have:

sin u = – 7 , u in Quadrant III cos u = – 24 , tan u = 7

cot(u + v) = = tan(u + v)

= −

− 13 13 84





u

25

(− 24, − 7)

cos v = – 4 , v in Quadrant III sin v = – 3 , tan v = 3 5 5 4

y y

v

x x

5

(− 4, − 3)






25 5 25 5

25 5 25 5

25 5 25 5

24

4

− 5 25 5

4

24

(

=

=


47. cos(u + v) = cos u cos v − sin u sin v

51. csc(u − v) = 1

= 1

= (− 24 )(− 4 ) − (− 7 )(− 3 ) sin(u − v) sin u cos v − cos u sin v

1

3 = 5 7

− −

4 − −

24 3 −

48. sin(u + v) = sin u cos v + cos u sin v

= (− 7 )(− 4 ) + (− 24 )(− 3 ) 1

= 44

− 125

= 28 + 72 = 100 = 4

49.

125 125 125 5

tan(u − v) = tan u − tan v 1 + tan u tan v

7 3 11

52.

125 = −

44

sec(v − u) = 1 cos(v − u)

− − 24 4 24 44 1

= = = −

1 + 7 3 39

32

117 = cos v cos u + sin v sin u

1 =

50.

tan(v − u) = tan v − tan u

=

3 7 4 24

4 − −

24 + −

3 −

7

25

1 + tan v tan u

11

1 + 3 7

1 =

117

125

125

= 24 44 =

39 117

32

117

1 1 117cot(v − u) =

tan(v − u) =

44 =

44

117

53. sin(arcsin x + arccos x) = sin(arcsin x) cos(arccos x) + sin(arccos x) cos(arcsin x)

= x ⋅ x +

1 − x2 ⋅

1 − x2

= x2 + 1 − x2

= 1

1

x 1

1 − x2

θ θ

1 − x2 x

54.

sin(arctan 2x − arccos x) = sin(arctan 2x − arccos x)

θ = arcsin x

θ = arccos x

= sin(arctan 2x) cos(arccos x) − cos(arctan 2x) sin(arccos x)

= 2 x

(x) − 1

4x2 + 1 4x2 + 1

1 − x2 )

2 x2 − =





1 − x2

4 x2 + 1

4x2 + 1 2x

1 1 − x2

θ

1

θ = arctan 2x

θ

x

θ = arccos x





(


55. cos(arccos x + arcsin x) = cos(arccos x) cos(arcsin x) − sin(arccos x) sin(arcsin x)

= x ⋅

= 0

(Use the triangles in Exercise 53.)

1 − x2 −

1 − x2 ⋅ x

56. cos(arccos x − arctan x) = cos(arccos x − arctan x)

= cos(arccos x) cos(arctan x) + sin(arccos x) sin(arctan x)

= (x)

1 + ( 1 − x2 ) x

= x + x

1 + x2

1 − x2

1 + x2 1

1 − x2

1 + x2

x

1 + x2

θ

x

θ = arccos x

θ

1

θ = arctan x

57.

sin π

− x

= sin π

cos x − cos π

sin x

59.

sin π

+ x

= sin π

cos x + cos π

sin x

2

2 2

6

6 6

= (1)(cos x) − (0)(sin x)

= cos x

=

5π

1

cos x + 2

5π

3 sin x)

5π

58.

sin π

+ x

= sin π

cos x + sin x cos π 60. cos

4 − x = cos cos x + sin

4 sin x

4

2

2 2

θ + π

= (1)(cos x) + (sin x)(0)

= cos x

tan θ + tan π tan θ + 0

= = =

tan θ

= θ

= − 2 (cos x + sin x)

2

61.

tan( ) 1 − tan θ tan π 1 − (tan θ )(0) tan 1

π tan

π − tan θ

4

1 − tan θ

62. tan 4

− θ = 1 + tan

π tan θ

4

= 1 + tan θ

63.

cos(π − θ ) + sin π

+ θ

= cos π cos θ + sin π sin θ + sin π

cos θ + cos π

sin θ

2

2 2

= (−1)(cos θ ) + (0)(sin θ ) + (1)(cos θ ) + (sin θ )(0)

= −cos θ + cos θ

= 0

64. cos(x + y) cos(x − y) = (cos x cos y − sin x sin y)(cos x cos y + sin x sin y)

= cos2 x cos2 y − sin 2 x sin 2 y

= cos2 x(1 − sin 2 y) − sin 2 x sin 2 y





=

c

o

s2

x

− c

o

s2

x

s

i

n 2

y

−

s

i

n 2

x

s

i

n 2

y

=

c

o

s2

x

−

s

i

n 2

y

(c

o

s2

x

+ s

i

n 2

x

)

=

c

o

s2

x

−

s

i

n

2 y






65.

cos

3π − θ

= cos

3π cos θ + sin

3π sin θ

68.

tan(π + θ ) = tan π + tan θ

2

2 2

1 − tan π tan θ

= (0)(cos θ ) + (−1)(sin θ )

= − sin θ

2

0 + tan θ =

1 − (0) tan θ

= tan θ

1 1

− 2π 2π cot(π + θ ) = =

tan(π + θ )

tan θ = cot θ

5

− 2

The graphs appear to coincide, so

− 2π 2π

cos 3π 2

− θ

= − sin θ .

66.

sin (π + θ ) = sin π cos θ + cos π sin θ

= (0) cos θ + (−1) sin θ

= − sin θ

69.

− 5

The graphs appear to coincide, so cot (π + θ ) = cot θ

sin(x + π ) − sin x + 1 = 0

sin x cos π + cos x sin π − sin x + 1 = 0

2

− 2π 2π

(sin x)(−1) + (cos x)(0) − sin x + 1 = 0

−2 sin x + 1 = 0

sin x = 1 2

− 2


sin (π + θ ) = − sin(θ ).

70.

x = π

, 5π

6 6

cos(x + π ) − cos x − 1 = 0

67.

sin 3π

+ θ

= sin 3π

cos θ + cos 3π

sin θ

cos x cos π − sin x sin π − cos x − 1 = 0

2

2 2

= (−1)(cos θ ) + (0)(sin θ )

= − cos θ

(cos x)(−1) − (sin x)(0) − cos x − 1 = 0

−2 cos x − 1 = 0

cos x 1

csc 3π = −

+ θ

= 1

= 1

= − sec θ 2

2

sin(3π

+ θ )

− cos θ

2π 4π 2

5

x = , 3 3

− 2π 2π

− 5


csc 3π 2 + θ

= − sec θ .






71. cos x +

π − cos

x −

π = 1

4

4

cos x cos π

− sin x sin π

− cos x cos

π + sin x sin

π = 1

4 4

4 4

2 −2 sin x

= 1

2

− 2 sin x = 1

sin x = − 1 2

sin x = − 2

2

x = 5π

, 7π

4 4

72.

sin x +

π − sin

x −

7π =

3

6

6

2

sin x cos π

+ cos x sin π

− sin x cos 7π

− cos x sin 7π

= 3

6 6

6 6

2

(sin x) 3

+ (cos x) 1

− (sin x) − 3

+ (cos x) − 1 =

3 2

2 2

2 2

73.

tan(x + π ) + 2 sin(x + π ) = 0

tan x + tan π + 2(sin x cos π + cos x sin π ) = 0

1 − tan x tan π

3 sin x = 3 2

sin x = 1 2

x = π

, 5π

6 6

tan x + 0 +

− + =

1 − tan x(0) 2 sin x( 1) cos x(0) 0

tan x − 2 sin x = 0

1

sin x

cos x

= 2 sin x

sin x = 2 sin x cos x

sin x(1 − 2 cos x) = 0

sin x = 0 or cos x = 1 2

x = 0, π x = π

, 5π

3 3





1 4 4

1

3 4

3


74.

sin x +

π − cos2 x = 0

76.

tan(x + π ) − cos x +

π = 0

2

2

sin x cos π

+ cos x sin π

− cos2 x = 0

x = 0, π

2 2

(sin x)(0) + (cos x)(1) − cos2 x = 0

cos x − cos2 x = 0

cos x(1 − cos x) = 0

cos x = 0 or 1 − cos x = 0

4

0 2π

− 4

x = π

, 3π

2 2

cos x = 1 77.

sin x +

π + cos2 x = 0

x = 0

2

75. cos x +

π + cos

x −

π = 1

4

4

Graph y = cos x +

π + cos x −

π and y2

= 1. 0 2π

x = π

, 7π

4 4 2

− 1

x = π

, π , 3π

2 2

0 2π 78.

cos

x −

π − sin 2 x = 0

2

− 2

1

0 2

79.

y = 1

sin 2t + 1

cos 2t 3 4

− 3

x = 0, π

, π 2

(a) a =

1, b =

1 , B = 2

3 4

C = arctan b

a = arctan

3 ≈ 0.6435

4

y ≈ 1

2 2

+

sin(2t + 0.6435) = 5

sin(2t + 0.6435) 12

(b) Amplitude: 5 12

feet

(c) Frequency: 1

= B

= 2

= 1

cycle per secondperiod 2π 2π π





λ

λ

λ T λ

λ λ T T


t x 80. y1 = A cos 2π −

T

t x y2 = A cos 2π +

T

t x t x y1 + y2 = A cos 2π

T − + A cos 2π +

t x t x t x t x t x

y1 + y2 = Acos 2π

cos 2π λ

+ sin 2π T

sin 2π + Acos 2π

cos 2π λ

− sin 2π T

sin 2π = 2 A cos 2π T

cos 2π λ

81. True.

sin(u + v) = sin u cos v + cos u sin v

sin(u − v) = sin u cos v − cos u sin v

So, sin(u ± v) = sin u cos v ± cos u sin v.

82. False.

cos(u + v) = cos u cos v − sin u sin v

cos(u − v) = cos u cos v + sin u sin v

So, cos(u ± v) = cos u cos v sin u sin v.

83. sin (α + β ) = sin α cos β + sin β cos α = 0

sin α cos β + sin β cos α

sin α cos β

= 0

= − sin β cos α

False. When α and β are supplementary, sin α cos β = − cos α sin β .

84. cos( A + B) = cos(180° − C )

= cos(180°) cos(C ) + sin(180°) sin(C )

= (−1) cos(C ) + (0) sin(C )

= − cos(C )

True. cos( A + B) = − cos C. When A, B and C form Δ ABC , A + B + C

= 180°, so A + B = 180° − C.

85. The denominator should be 1 + tan x tan (π 4).

tan x −

π =

tan x − tan(π 4)

87. cos(nπ + θ ) = cos nπ cos θ − sin nπ sin θ

= (−1)n (cos θ ) − (0)(sin θ )

4

1 + tan x tan(π 4)

= tan x − 1 1 + tan x

88.

sin(nπ

= (−1)n (cos θ ), where n is an integer.

+ θ ) = sin nπ cos θ + sin θ cos nπ

n

86. (a) Using the graph, sin(u + v) ≈ 0 and

sin u + sin v ≈ 0.7 + 0.7 = 1.4. Because

= (0)(cos θ ) + (sin θ )(−1)

= (−1)n (sin θ ), where n is an integer.

0 ≠ 1.4, sin(u + v) ≠ sin u + sin v.

(b) Using the graph, sin(u − v) ≈ −1 and

sin u − sin v ≈ 0.7 − 0.7 = 0. Because

89.

−1 ≠ 0,

C = arctan b

sin(u − v) ≠ sin u − sin v.

sin C = b

, cos C = a

a a2 + b2 a2 + b2





a2 + b2 sin(Bθ + C ) = a b

a2 + b2 sin Bθ ⋅ + ⋅ cos Bθ = a sin Bθ + b cos Bθ

a2 + b2 a2 + b2

90. C = arctan a sin C =

a , cos C =

b

b a2 + b2 a2 + b2

a2 + b2 cos(Bθ − C ) = b a

a2 + b2 cos Bθ ⋅ + sin Bθ ⋅

= b cos Bθ + a sin Bθ

a2 + b2

= a sin Bθ + b cos Bθ

a2 + b2





θ


91. sin θ + cos θ

a = 1, b = 1, B = 1

94. sin 2θ + cos 2θ

a = 1, b = 1, B = 2

(a)

C = arctan b

a = arctan 1 =

π 4

(a)

C = arctan b

a = arctan(1) =

π 4

sin θ + cos θ = a2 + b2 sin(Bθ + C )

sin 2θ + cos 2θ = a2 + b2 sin(Bθ + C )

π = 2 sinθ +

= 2 sin 2θ +

π

(b)

C = arctan

a

b

4

= arctan 1 = π 4

(b)

C = arctan a

4

= arctan(1) = π

sin θ + cos θ =

=

a2 + b2 cos(Bθ − C ) b 4

π ( ) 2 cosθ −

4

sin 2θ + cos 2θ =

=

a2 + b2 cos Bθ − C

2 cos 2θ −

π

4

92. 3 sin 2θ + 4 cos 2θ

a = 3, b = 4, B = 2 b π

(a) C = arctan b = arctan

4

≈ 0.9273 95. C = arctan

a = a = b, a > 0, b > 0

4

a

3 sin 2θ + 4 cos 2θ =

3

a2 + b2 sin(Bθ + C ) a2 + b2

B = 1

= 2 a = b = 2

≈ 5 sin(2θ + 0.9273) 2 sin

+ π

=

2 sin θ +

2 cos θ

(b) C = arctan a

b = arctan

3 4

≈ 0.6435 4

3 sin 2θ + 4 cos 2θ = a2 + b2 cos(Bθ − C )

96. C = arctan b

= π

a 4 a = b, a > 0, b > 0

93. 12 sin 3θ + 5 cos 3θ

a = 12, b = 5, B = 3

≈ 5 cos(2θ − 0.6435) a2 + b2

B = 1

= 5 a = b = 5 2

2

(a) C = arctan

b = arctan 5

≈ 0.3948 π 5 2 5 2 5 cosθ − = sin θ + cos θ

a

12 sin 3θ + 5 cos 3θ

12

= a2 + b2 sin(Bθ + C )

97.

4 2 2

y

(b)

C = arctan a

b

≈ 13 sin(3θ + 0.3948)

= arctan 12

≈ 1.1760 5

≈ 13 cos(3θ − 1.1760)

y1 = m1 x + b1

θ

δ

α β x

y2 = m2 x + b2





m1 = tan α and m2 = tan β

β + δ = 90° δ = 90° − β

α + θ + δ = 90° α + θ + (90° − β )

= 90° θ = β − α

So, θ = arctan m2 − arctan m1. For y = x and

y = 3x you have m1 = 1 and m2 = 3.

θ = arctan 3 − arctan 1 = 60° − 45° = 15°





2 2

B


98. For m2 > m1 > 0, the angle θ between the lines is: 99. y1 = cos(x + 2), y2

2

= cos x + cos 2

m2 − m1

θ = arctan 1 + m1m2

m2 = 1

1

y2

0 2π y1

m1 =

3

1 −

1

− 2

No, y1 ≠ y2 because their graphs are different.

θ = arctan 3 = arctan(2 −

1 + 1

3) = 15° 100. y1 = sin(x + 4), y2

2

= sin x + sin 4

3

y1

0 2π

y2

− 2

No, y1 ≠

y2 because their graphs are different.

101. (a) To prove the identity for sin(u + v) you first need to prove the identity for cos(u − v).

y

C 1 u − v

Assume 0 < v < u < 2π and locate u, v, and u − v on the unit circle. B

The coordinates of the points on the circle are: D

A = (1, 0), B = (cos v, sin v), C = (cos(u − v), sin(u − v)), and D = (cos u, sin u). Because ∠DOB = ∠COA, chords AC and BD are equal. By the Distance Formula:

− 1

u

v A x

O 1

2 2 2 2

cos(u − v) − 1 + sin(u − v) − 0 = (cos u − cos v) + (sin u − sin v) − 1

cos2 (u − v) − 2 cos(u − v) + 1 + sin 2 (u − v) = cos2 u − 2 cos u cos v + cos2 v + sin 2 u − 2 sin u sin v + sin 2 v

cos2 (u − v) + sin 2 (u − v) + 1 − 2 cos(u − v) = (cos2 u + sin 2 u) + (cos2 v + sin 2 v) − 2 cos u cos v − 2 sin u sin v

2 − 2 cos(u − v) = 2 − 2 cos u cos v − 2 sin u sin v

−2 cos(u − v) = −2(cos u cos v + sin u sin v) cos(u − v) = cos u cos v + sin u sin v

Now, to prove the identity for sin(u + v), use cofunction identities.

sin(u + v) = cosπ

− (u + v)

= cos π

− u − v

2

2

= cos π

2 − u

cos v + sin

π

2 − u

sin v

= sin u cos v + cos u sin v

(b) First, prove cos(u − v) = cos u cos v + sin u sin v using the figure containing points

A(1, 0)

B(cos(u − v), sin(u − v)) C(cos v, sin v)

y

1 D u − v

C

u

v A x

D(cos u, sin u)

on the unit circle.

− 1 u − v 1

Because chords AB and CD are each subtended by angle u − v, their lengths are equal. Equating − 1





2 2

2

d ( A, B) = d (C, D) you have (cos(u − v) − 1) + sin 2 (u − v) = (cos u − cos v) + (sin u − sin v) .

Simplifying and solving for cos(u − v), you have cos(u − v) = cos u cos v + sin u sin v.

Using sin θ = cos π

2

− θ ,

sin(u − v) = cos π

2 − (u − v) = cos

π 2

− u − (−v) = cos π

2 − u cos(−v) + sin

π

2 − u sin(−v)

= sin u cos v − cos u sin v





h 0.5 0.2 0.1 0.05 0.02 0.01

f (h) 0.267 0.410 0.456 0.478 0.491 0.496

g (h) 0.267 0.410 0.456 0.478 0.491 0.496

= −

= −


102. (a) The domains of f and g are the same, all real numbers h, except h = 0.

(b) (c) 2 (d) As h → 0*,

f → 0.5 and− 3 3

− 2

Section 2.5 Multiple-Angle and Product-to-Sum Formulas

g → 0.5.

1. 2 sin u cos u

2. cos2 u − sin 2 u = 2 cos2 u − 1 = 1 − 2 sin 2 u

8. sin 2x sin x = cos x

2 sin x cos x sin x − cos x = 0

cos x(2 sin 2 x − 1) = 0

1 3.

2 sin(u + v) + sin(u − v)

cos x = 0 or 2 sin 2

x = π

+ 2nπ

x − 1 = 0

sin 2 x = 1

4. tan 2 u 2 2

sin x = ± 2

25. ±

1 − cos u

2

x = π

+ nπ

6. −2 sin u + v

sin u − v

4 2

9. cos 2 x − cos x = 0

2

2

7. sin 2 x − sin x = 0

2 sin x cos x − sin x = 0

sin x(2 cos x − 1) = 0

sin x = 0 or 2 cos x − 1 = 0

cos 2 x = cos x

cos2 x − sin 2 x = cos x

cos2 x − (1 − cos2 x) − cos x = 0

2 cos2 x − cos x − 1 = 0

(2 cos x + 1)(cos x − 1) = 0

x = nπ cos x = 1 2

2 cos x + 1 = 0 or cos x − 1 = 0

x = π

+ 2nπ , 5π

+ 2nπ cos x 1 cos x = 1

10.

3 3

cos 2x + sin x = 0

1 − 2 sin 2 x + sin x = 0

2 sin 2 x − sin x − 1 = 0

(2 sin x + 1)(sin x − 1) = 0

2

x = 2nπ 3

x = 0

2 sin x + 1 = 0 or sin x − 1 = 0

sin x 1 2

sin x = 1

x = 7π + 2nπ ,

11π

+ 2nπ x = π

+ 2nπ

6 6 2





( )

Section 2.5 Mult iple Angl e and Product - to-S um Fo rmulas 259

11. sin 4 x = −2 sin 2x

sin 4 x + 2 sin 2x = 0

2 sin 2x cos 2 x + 2 sin 2 x = 0

2 sin 2x(cos 2x + 1) = 0

2 sin 2x = 0 or cos 2 x + 1 = 0

sin 2 x = 0 cos 2x = −1

2x = nπ 2x = π + 2nπ

x = nπ x =

π

+ nπ

2 2

12. (sin 2x + cos 2x)2

= 1

sin 2 2 x + 2 sin 2 x cos 2x + cos2 2 x = 1

2 sin 2 x cos 2 x = 0

sin 4 x = 0

4 x = nπ

x = nπ 4

13. tan 2x − cot x = 0

2tan x = cot x

1 − tan 2 x

2 tan x = cot x(1 − tan 2 x)

2 tan x = cot x − cot x tan 2 x

2 tan x = cot x − tan x

3 tan x = cot x

3 tan x − cot x = 0

3 tan x − 1

= 0 tan x

3 tan 2 x − 1= 0

tan x

1 3 tan 2 x − 1 = 0

tan x

cot x(3 tan 2 x − 1) = 0

cot x = 0 or 3 tan 2 x − 1 = 0

x = π

+ nπ tan 2 x = 1

2 3

tan x = ±

x = π

3

3

+ nπ , 5π

+ nπ

6 6





tan 2x − 2 cos x 0

2tan x

1 − tan 2 x 2 cos x

2 tan x

2 tan x

2 cos x(1 − tan 2 x)

2 cos x − 2 cos x tan 2 x

cos x

2

2

=


14. =

=

=

=

2 tan x = 2 cos x − 2 cos x

sin 2 x

cos2 x

2 tan x = 2 cos x − 2

sin 2 x

cos x

sin x

sin 2 x

tan x = cos x −

sin x

= cos x − cos x

sin 2 x

cos x

sin 2 x

cos x

cos x + − cos x = 0

cos x

sin x + sin 2 x − cos2 x = 0

cos x

1 sin x + sin 2 x − (1 − sin 2 x) = 0

sec x 2 sin 2 x + sin x − 1 = 0

sec x(2 sin x − 1)(sin x + 1) = 0

sec x = 0 or 2 sin x − 1 = 0 or sin x + 1 = 0

No solution

sin x = 1 sin x = −1

2 3π x =

x = π

, 5π 2

6 6

Also, values for which cos x = 0 need to be checked.

π ,

3π are solutions.

2 2

x = π + 2nπ ,

π + nπ , 5π

+ 2nπ

15.

6 2 6

6 sin x cos x = 3(2 sin x cos x)

= 3 sin 2x

19.

4 − 8 sin 2 x = 4(1 − 2 sin 2 x) = 4 cos 2x

16. sin x cos x =

=

1 (2 sin x cos x)

1 sin 2 x

20.

10 sin 2 x − 5 = 5(2 sin 2 x − 1) 2

17.

2

6 cos2 x − 3 = 3(2 cos2 x − 1) = 3 cos 2x

= −5(1 − 2 sin x) = −5 cos 2 x

18. cos2 x − 1 = 1 2(cos2 x − 1 )2 2 2

= 1 (2 cos2 x − 1) 1 cos 2 x





2





5 3

u

− 4

u

4

5 −

34

3

5

, ,

2 2

3

tan 2u = =

2

2

= −


21.

sin u 3 3π

= − 5 2

y

< u < 2π

22. cos u 4 π 5 2

y

< u < π

x x

3

sin 2u = 2 sin u cos u = −

3 4 = −

24 sin 2u = 2 sin u cos u = 3

− 4 24 = −

5 5 25 5 5 25

cos 2u = cos2 u − sin 2 u = 16

− 9

= 7

25 25 25 3

2 −


− 9

= 7

25 25 25

2 −

tan 2u = 2tan u

= 4 3 16 24 2tan u

4 3 16 24= − = − tan 2u = =

9 = − = −

23.

1 − tan 2 u

tan u = 3

, 0 < u < π

5 2

1 − 9 16

2 7 7 1 − tan 2 u 1 −

16

2 7 7

y 3 5 15 sin

2u = 2 sin u cos u = 2 = 34 34 17


− 9

= 8

34 34 17

3 2u

x 2tan u 5 =

6 25 =

15

24.

sec u = − 2, π

< u < 3π 2

1 − tan 2 u 1 − 9 25

5 16

8

y

sin 2u = 2 sin u cos u = −

3 −

1 =

3 2 2 2

cos 2u = cos2 u − sin 2 u = −

1

2

3

− −

= 1

− 3

= − 1

u

2 2 4 4 2

− 1 x 3

2

2 tan u 1 2 3

− 3 2 tan 2u = 1 − tan 2 u

= 2

= = − 3 3 − 2

1 − 1

25.

cos 4x = cos(2x + 2x) = cos 2x cos 2x − sin 2x sin 2x

= cos2 2x − sin 2 2x

= cos2 2x − (1 − cos2 2x) = 2 cos2 2x − 1

=

2

(c

o

s

2

x)2

− 1





1 − tan 2 x

=

2

=

2

6.

tan 3x = tan(2x + x)

=

tan

2x +

tan x 1 − tan 2x tan x

2 tan x

1 − tan 2 x + tan x

1 − 2 tan x

(tan x)

2 tan x + tan x − tan3 x= 2(2 cos2

x − 1) − 1 1 − tan 2 x 2 2

= 2(4 cos4 x − 4 cos x + 1) − 1 1 − tan x − 2 tan x

1 − tan 2 x

= 8 cos4 x − 8 cos x + 1 3 tan x

tan3 x

= −

1 − 3 tan 2 x





2

= 1 cos 2 x 1 cos 2 x

3

2


27.

cos4 x = (cos2 x)(cos2 x) =

1 + cos 2 x 1 + cos 2 x =

1 + 2 cos 2 x + cos 2 x

2

2

4

1 + 2 cos 2x + 1 + cos 4 x

= 2 4

= 2 + 4 cos 2 x + 1 + cos 4 x

8

= 3 + 4 cos 2 x + cos 4 x

8

= 1(3 + 4 cos 2x + cos 4x)

8

8 4 4 2 2

2 2

28. sin x = (sin x)(sin x) = (sin x) (sin x)

2 2

− − 2 2

1 − 2 cos 2 x + cos2 2 x 1 − 2 cos 2x + cos2 2x =

4 4

1 − 2 cos 2 x + cos2 2 x − 2 cos 2 x + 4 cos2 2 x − 2 cos3 2 x + cos2 2x − 2 cos3 2 x + cos4 2 x =

16

1 − 4 cos 2 x + 6 cos2 2x − 4 cos3 2 x + cos4 2x =

16 2 3 2

2

1 − 4 cos 2 x + 6 cos 2 x − 4 cos 2 x + (cos 2 x) =

16

1 + cos 4 x 1 + cos 4 x 1 − 4 cos 2 x + 6 − 4 cos 2x +

= 2 2 16

1 + 2 cos 4x + cos2 4x 1 − 4 cos 2 x + 3 + 3 cos 4x − 4 cos3 2 x +

= 4

16

4 − 16 cos 2x + 12 + 12 cos 4x − 16 cos3 2 x + 1 + 2 cos 4x + cos2 4x =

64

17 − 16 cos 2 x + 14 cos 4x − 16 cos3 2x + 1 + cos 8x

= 2 64

34 − 32 cos 2x + 28 cos 4x − 32 cos3 2x + 1 + cos 8x =

128

35 − 32 cos 2x + 28 cos 4x − 32 cos3 2x + cos 8x =

128

35 − 32 cos 2x + 28 cos 4 x − 32 cos2 2x cos 2x + cos 8x =

128

1 + cos 4 x 35 − 32 cos 2x + 28 cos 4x − 32

2 cos 2x + cos 8x

= 128

= 35 − 32 cos 2 x + 28 cos 4 x − 16 cos 2 x − 16 cos 4 x cos 2 x + cos 8x

128

= 35 − 48 cos 2 x + 28 cos 4 x − 16 cos 4 x cos 2 x + cos 8x

128

= 1

(35 − 48 cos 2 x + 28 cos 4 x + cos 8x − 16 cos 2x cos 4x) 128





= 1 cos 4 x

( 2 )

= 1 cos 4 x

( 2 )

( 2 )

2 2

2


29. sin 4 2 x = (sin 2 2 x)

31. tan 4 2 x = (tan 2 2x)

2

−

2

1 − cos 4 x =

2 1 + cos 4 x

1 = 1 − 2 cos 4x + cos 4x

4

= 1 1 − 2 cos 4 x +

1 + cos 8x

1 − 2 cos 4 x + cos2 4 x =

1 + 2 cos 4x + cos2 4x

1 + cos 8x

4 2

1 1 1 1 = − cos 4 x + + cos 8x

4 2 8 8

3 1 1 = − cos 4x + cos 8x

8 2 8

1 = (3 − 4 cos 4x + cos 8x)

8

1 − 2 cos 4 x +

= 2

1 + 2 cos 4x + 1 + cos 8x

2

1 (2 − 4 cos 4x + 1 + cos 8x)

= 2 1 (2 + 4 cos 4x + 1 + cos 8x)

2

= 3 − 4 cos 4 x + cos 8 x

30. cos4 2x = (cos2 2 x) 3 + 4 cos 4x + cos 8x

2

+

32.

tan 2 2 x cos4 2 x = 1 − cos 4x

(cos2 2x)2

2

= 1 (1 + 2 cos 4 x + cos2 4x)

1 + cos 4 x

1 − cos 4 x 1 + cos 4 x 2

4 = 1 + cos 4 x

2

1 1 + cos 8x

= 1 + 2 cos 4 x +

(1 − cos 4 x)(1 + cos 4 x)(1 + cos 4 x)

4 2

1 1 1 1 = + cos 4x + + cos 8x

4 2 8 8

3 1 1 = + cos 4 x + cos 8x

8 2 8

1 = (3 + 4 cos 4 x + cos 8x)

8

= 4(1 + cos 4 x)

(1 − cos 4 x)(1 + cos 4 x) =

4

1 = 1 − cos 4 x

4

= 1

− 1 + cos 8x

14 2

33.

sin 2 2 x cos2 2 x = 1 − cos 4 x 1 + cos 4 x

1 1 1 = − − cos 8x

4 8 8

1 1 = − cos 8x

8 8

1 = (1 − cos 8x)

8

2

2

1 = 1 − cos 4 x

4

1 1 + cos 8 x

= 1 −

4 2

= 1

− 1

− 1

cos 8x 4 8 8

= 1

− 1

cos 8x





8 8

= 1(1 − cos 8x)

8





( 2 )

( 2 3 )

[ ]

[ ]

2 +

3


34. sin 4 x cos2 x = sin 2 x sin 2 x cos2 x

= 1 − cos 2 x 1 − cos 2 x 1 + cos 2 x

2

2

2

1 = (1 − cos 2 x) 1 − cos 2x

8

1 = 1 − cos 2x − cos 2 x + cos 2 x

8

+

+ 1 1 cos 4 x 1 cos 4 x = 1 − cos 2x − + cos 2x

8 2 2

1 = 2 − 2 cos 2x − 1 − cos 4 x + cos 2 x + cos 2 x cos 4x

16

1 = 1 − cos 2x − cos 4 x + cos 2 x cos 4 x

16

1

1 − cos 150° 1 + ( 3 2)

35. sin 75° = sin ⋅ 150° = =

2 2 2

= 1

2 + 3 2

1

1 + cos 150° 1 − ( 3 2)

cos 75° = cos ⋅ 150° = =

2 2 2

= 1

2 − 3 2

tan 75° = tan 1

⋅ 150°

= sin 150°

= 1 2

2

1 + cos 150°

1 ( 3 2) −

1 2 + 3 = ⋅

= 2 + 3

= 2 + 3

2 − 3 2 +

1

3 4 − 3

1 − cos 330°

1 − ( 3 2)36. sin 165° = sin ⋅ 330° = =

2 2 2

= 1

2 − 3 2

1

1 + cos 330° 1 + ( 3 2)

cos 165° = cos ⋅ 330° = − = −

2 2 2

1 = −

2

tan 165° = tan 1

⋅ 330°

= sin 330°

= −1 2

2

1 + cos 330°

1 ( 3 2) +

−1 2 − 3

− 2 + 3

= ⋅ = = − 2 + 32 + 3 2 − 3 4 − 3





1


1

1 − cos 225° 1 − (− 2 2) 1

37. sin 112° 30′ = sin ⋅ 225° = = = 2 + 2

2 2 2 2

1

1 + cos 225° 1 + (− 2 2) 1

cos 112° 30′ = cos ⋅ 225° = − = − = − 2 − 2

2 2 2 2

tan 112° 30′ = tan 1

⋅ 225°

= sin 225°

= − 2 2

= − 2

⋅ 2 + 2

= − 2 2 − 2

= −1 − 2

2

1 + cos 225° 1 (

2 2)

2 − 2 2 + 2 2

+ −

1

1 − cos 135° 1 + ( 2 2) 1

38. sin 67° 30′ = sin ⋅ 135° = = = 2 + 2

2 2 2 2

1

1 + cos 135° 1 − ( 2 2) 1

cos 67° 30′ = cos ⋅ 135° = = = 2 − 2

2 2 2 2

tan 67° 30′ = tan 1

⋅ 135°

= sin 135°

= 2 2

= 1 + 2 2

1 + cos 135°

1 ( 2 2) −

π 1 π 1 − cos π 4 1

39. sin = sin = = 2 − 2

8 2 4 2 2

ππ 1 π 1 + cos

4

cos = cos = = 2 + 2

8 2 4 2 2

sin π 2

tan π

= tan 1 π

= 4 = 2 =

2 − 1

8 2 4 1 + cos

π 1 +

2

4 2

40.

sin 7π

= sin 1 7π

= 1 − cos

7π 6 =

1 + 3

2 = 1

2 + 3

12 2

6

2 2 2

1 + cos 7π

1 − 3

cos 7π

= cos1 7π

= − 6 = −

2 = − 1

2 − 3

12 2

6

2 2 2

sin 7π

− 1

tan 7π

= tan 1 7π

= 6 = 2 = −2 − 3

12 2 6 1 + cos 7π

1 − 3

6 2





2

2

,

2

2

= −


41. cos u = 7

, 0 < u < π

42. sin u = 5

, π

< u < π cos u 12

25 2

(a) Because u is in Quadrant I, u

is also in Quadrant I. 2

1 − 7

13 2 13

(a) Because u is in Quadrant II, u

is in Quadrant I. 2

1 + 12

(b)

sin u

= 1 − cos u

= 25 =

9 3

= (b)

sin

u = 1 − cos u

= 13 5 26

=

2 2 2 25 5

1 + 7

2 2 26

1 − 12

cos u

= 1 + cos u

= 25 =

16 4 =

cos u

= 1 + cos u

= 13 26

=

2 2 2 25 5

1 − 7

2 2 26

5

tan u

= 1 − cos u

= 25 = 3

tan u

= sin u

= 13 = 5

2 sin u

24 4

2

1 + cos u

1 12

43.

tan u

5 3π = −

12 2

25

< u < 2π

− 13

(a) Because u is in Quadrant IV, u

is in Quadrant II. 2

1 − 12

(b) sin u

= 1 − cos u = 13 = 1 26

=

2 2 2 26 26

1 + 12

cos u 1 + cos u

= − = − 13 25 5 26 = − = −

2 2 2 26 26

1 − 12

tan u =

1 − cos u = 13 = − 1

2 sin u −

5 5

13

44.

cot u = 3, π

< u < 3π 2

(a) Because u is in Quadrant III, u


1 + 3

(b)

sin u

= 1 − cos u

= 10

= 10 + 3 10

= 1 10 + 3 10

2 2 20 2 5

1 − 3

cos u

= − 1 + cos u

= − 10 10 − 3 10

= −

1 10 − 3 10

= −

2 2 20 2 5

1 + 3

tan u

= 1 − cos u

= 10

= −

10 − 3





2

sin u

1 −

10






45. sin x

+ cos x = 0 2

47. cos x

− sin x = 0 2

± 1 − cos x

2

1 − cos x

2

= −cos x

= cos2 x

± 1 + cos x

2

1 + cos x

2

= sin x

= sin 2 x

cos x = 1 2

0 = 2 cos2 x + cos x − 1

= (2 cos x − 1)(cos x + 1)

or cos x = −1

1 + cos x = 2 sin 2 x

1 + cos x = 2 − 2 cos2 x

2 cos2 x + cos x − 1 = 0

(2 cos x − 1)(cos x + 1) = 0

x = π

, 5π

3 3 x = π

2 cos x − 1 = 0 or cos x + 1 = 0

2 1 cos x =

2

cos x = −1

0 2π

x = π

, 5π

3 3

x = π

− 2

By checking these values in the original equation,

x = π 3 and x = 5π 3 are extraneous, and x = π

is the only solution.

x = π

, π , 5π

3 3

π 3, π , and 5π 3 are all solutions to the equation.

2

46. h(x) = sin x

+ cos x − 1 2

sin x

+ cos x − 1 = 0 2

0 2π

− 2

± 1 − cos x

2

1 − cos x

= 1 − cos x

= 1 − 2 cos x + cos2 x

48.

g (x) = tan x

− sin x 2

x

2

1 − cos x = 2 − 4 cos x + 2 cos2 x

tan − sin x = 0 2

1 − cos x

2 cos2 x − 3 cos x + 1 = 0

sin x = sin x

(2 cos x − 1)(cos x − 1) = 0

2 cos x − 1 = 0 or cos x − 1 = 0

1 − cos x = sin 2 x

1 − cos x = 1 − cos2 x

2

cos x = 1 cos x = 1 cos x − cos x = 0

2

x = π

, 5π

3 3

x = 0

cos x(cos x − 1) = 0

cos x = 0 or cos x − 1 = 0

0, π

, and 5π

are all solutions to the equation. x =

π ,

3π 2 2 cos x = 1

3 3 x = 0

1

0, π

, and 3π

are all solutions to the equation. 0 2π 2 2





3

− 2

0 2π

− 3





2

π π

π


49. sin 5θ sin 3θ = 1 cos(5θ − 3θ ) − cos(5θ + 3θ ) = 1 (cos 2θ − cos 8θ )2 2

50. 7 cos(−5β ) sin 3β = 7 ⋅ 1 sin(−5β + 3β ) − sin(−5β − 3β ) = 7 (sin(−2β ) − sin(−8β ))2 2

51. cos 2θ cos 4θ = 1 cos(2θ − 4θ ) + cos(2θ + 4θ ) = 1 cos(−2θ ) + cos 6θ

2 2

52. sin(x + y) cos(x − y) = 1 (sin 2 x + sin 2 y)

54.

sin 3θ + sin θ = 2 sin 3θ + θ

cos 3θ − θ

2 2

53.

sin 5θ − sin 3θ = 2 cos

5θ + 3θ sin

5θ − 3θ = 2 sin 2θ cos θ

2

2

= 2 cos 4θ sin θ 55.

cos 6 x + cos 2 x = 2 cos 6 x + 2 x

cos 6 x − 2 x

2

2

56.

cos x + cos 4 x = 2 cos x + 4 x

cos x − 4 x

= 2 cos 4x cos 2x

2

2

= 2 cos 5x

cos − 3x

2

2

° + °

° − °

57.

75 15 75 15 2 3 6 sin 75° + sin 15° = 2 sin cos = 2 sin 45° cos 30° = 2 =

2 2 2 2 2

° + °

° − ° 58.

120 60 120 60 3 cos 120° + cos 60° = 2 cos cos = 2 cos 90° cos 30° = 2(0) = 0

2 2 2

3π

+ π 3π

− π

59. cos 3π − cos

π = −2 sin 4 4

sin 4 4

= −2 sin sin

4 4 2

2

2 4

cos 3π − cos

π 2 2 = − − = − 2

4 4 2 2

5π +

3π 5π −

3π

60. sin 5π − sin

3π = 2 cos 4 4

sin 4 4

= 2 cos π sin

4 4 2

2

4

sin 5π − sin

3π 2 2 = − − = − 2

4 4 2 2

61. sin 6x + sin 2x = 0

2 sin 6 x + 2 x

cos 6 x − 2 x

= 0 2

2

2(sin 4 x) cos 2x = 0

sin 4x = 0 or cos 2x = 0





− 2

2

4x = nπ 2x = π 2

+ nπ 0 2π

x = nπ x =

π + nπ

4 4 2

In the interval [0, 2π )

x = 0, π

, π

, 3π

, π , 5π

, 3π

, 7π

. 4 2 4 4 2 4






62. h(x) = cos 2x − cos 6x

cos 2x − cos 6 x = 0

−2 sin 4x sin(−2x) = 0

2 sin 4 x sin 2 x = 0

sin 4x = 0 or sin 2x = 0 2

4x = nπ 2x = nπ 0 2π

x = nπ x =

nπ

4 2 − 2

x = 0, π

, π

, 3π

, π , 5π

, 3π

, 7π x = 0,

π , π ,

3π

4 2 4 4 2 4 2 2

63. cos 2 x

− 1 = 0 sin 3x − sin x

cos 2 x = 1

sin 3x − sin x 2

cos 2 x = 1

65.

csc 2θ 1

= sin 2θ

1 =

2 sin θ cos θ

= 1

⋅ 1

2 cos 2 x sin x 0 2π

2 sin x = 1

sin θ

csc θ =

2 cos θ

sin x = 1

− 2

2

2 cos θ

x = π

, 5π 66. ( )( )

cos4 x − sin 4 x =

6 6 = (

cos2 x − sin 2 x

cos 2x)(1)

cos2 x + sin 2 x

64.

f (x) = sin 2 3x − sin 2 x

sin 2 3x − sin 2 x = 0

(sin 3x + sin x)(sin 3x − sin x) = 0

(2 sin 2x cos x)(2 cos 2x sin x) = 0

67. (sin x + cos x)2

= cos 2x

= sin 2 x + 2 sin x cos x + cos2 x

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

= 1 + sin 2x

sin 2x = 0 x = 0, π

, π , 3π

or 2 2

68.

tan u

1 − cos u

=

cos x = 0 x = π

, 3π

or 2 2

2 sin u

1 cos u = −

cos 2x = 0 x = π

, 3π

, 5π

, 7π

or sin u sin u

4 4 4 4

sin x = 0 x = 0, π

1

= csc u − cot u

2 sin x ± y

cos x y

0 2π

69. sin x ± sin y

cos x + cos y = 2 2

2 cos x + y

cos x − y

2

2

− 1 = tan x ± y

2

π + x +

π

− x

π + x −

π

− x





3

70. π

cosπ

+ x + cos − x = 2 cos 3 3

cos 3 3

3 3 2

2

= 2 cos π

cos(x)

1 = 2 cos x = cos x

2





2

= 1

2


71. (a) sin θ

= ±

1 − cos θ

2

1 =

M

(c) When M = 4.5, cos θ (4.5)

2 − 2

= (4.5)

2

2

1 − cos θ

±

cos θ ≈ 0.901235.

2 M So, θ ≈ 0.4482 radian.

1 − cos θ

2

1 =

M 2 (d) When M = 2, speed of object

= M speed of sound

M 2 (1 − cos θ ) = 2 speed of object = 2

1 − cos θ

− cos θ

2 =

M 2

2 = − 1

760 mph

speed of object = 1520 mph.

speed of object

cos θ

M 2

= 1 − 2 M 2

M 2 − 2

When M = 4.5,

speed of sound

speed of object

760 mph

= M

= 4.5

cos θ =

M 2

22 − 2 1 π

speed of object = 3420 mph.

(b) When M = 2, cos θ = = 22 2

. So, θ = . 3

72.

1 (75)

2 sin 2θ

32

sin 2θ

= 130

130(32) =

752

75. True. Using the double angle formula and that sine is an

odd function and cosine is an even function,

sin(− 2 x) = sin 2(− x)

= 2 sin(− x) cos(− x)

θ = 1

sin −1130(32)

= 2(− sin x) cos x

2

752

θ ≈ 23.85° = − 2 sin x cos x.

76. False. If 90° < u < 180°,

73. x

= 2r sin 2 θ = 2r

1 − cos θ

2 2

2

u is in the first quadrant and

= r(1 − cos θ ) 2

So, x = 2r(1 − cos θ ). sin u

= 1 − cos u .

74. (a) Using the graph, sin 2u ≈ 1 and

2 2

77. Because φ and θ are complementary angles,

2 sin u cos u ≈ 2(0.7)(0.7) ≈ 1.

Because 1 = 1, sin 2u = 2 sin u cos u. sin φ = cos θ and cos φ = sin θ .

(b) Using the graph, cos 2u ≈ 0 and

cos2 u − sin 2 u ≈ (0.7)2

− (0.7)2

= 0.

(a) sin(φ − θ ) = sin φ cos θ − sin θ cos φ

= (cos θ )(cos θ ) − (sin θ )(sin θ )

= cos2 θ − sin 2 θ

Because 0 = 0, cos 2u = cos2 u − sin 2 u.

Rev

iew Exercises for Chapter 2





(b)

= c

o

s

2

θ

c

o

s

(φ

−

θ

)

=

c

o

s

φ

c

o

s

θ

+

s

i

n

φ

s

i

n

θ

=

(s

i

n

θ

)

(c

o

s

θ ) + (cos θ )(sin θ )

= 2 sin θ cos θ

= sin 2θ





1. cot x 3. cos x

2. sec x 4. cot 2 x + 1 = csc2 x =

csc x





= −

= −

tan 2 x

2

Review Exer cis e s for Chapter 2 271

5. cos θ 2

, tan θ 5

> 0, θ is in Quadrant III.

sec θ

= 1

= − 5

cos θ 2

sin θ

= − 1 − cos2 θ = −

1 − 4

= −

21 = −

21

csc θ

25 25 5

= 1

= − 5

= − 5 21

sin θ 21 21

− 21

tan θ = sin

= 5 = 21

cos θ −

2 2 5

cot θ = 1

= 2

= 2 21

tan θ 21 21

6. cot x 2

, cos x < 0, x is in Quadrant II. 3

tan x = 1

= − 3

cot x 2

csc x =

1 + cot 2 x =

1 + 4

=

13 13 =

9 9 3

sin x = 1

= 3

= 3 13

csc x 13 13

cos x = − 1 − sin 2 x = −

1 − 9

4 2 2 13 = − = − = −

sec x = 1

= − 13

13 13 13 13

cos x 2

1 1

7. = = sin 2 x cot 2 x + 1 csc2 x

13.

cos2 x + cos2 x cot 2 x = cos2 x(1 + cot 2 x)

= cos2 x(csc2 x)sin θ

1

8. tan θ

= cos θ

= 1

= cos2 x sin 2 x

1 − cos2 θ sin 2 θ sin θ cos θ

cos2 x

= csc θ sec θ = sin 2 x

2

9. tan 2 x(csc2 x − 1) = tan 2 x(cot 2 x) = cot x

= tan 2 x 1

14. (tan x + 1)2

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

= 1 = (sec

x + 2 tan x) cos x

10.





( ) cos x

=

2

cot 2

x sin 2 x =

sin 2 x =

cos2

x sin 2

x

sin x = sec2 x cos x + 2 cos x

cos x

= sec x + 2 sin x

cot π

− u

11. 2

cos u

tan u

cos u

= tan u sec u

12.

sec (−θ )

sec θ

1 cos θ

sin θ 2

2 2 2 2

= = = = tan θcsc2 θ csc2 θ 1 sin 2 θ cos2 θ





2


15.

1 1 (csc θ − 1) − (csc θ + 1) − =

16.

tan x

sec x − 1

2 2

=csc θ + 1 csc θ − 1 (csc θ + 1)(csc θ − 1)

= −2 csc2 θ − 1

= −2 cot 2 θ

= −2 tan 2 θ

1 + sec x 1 + sec x

(sec x + 1)(sec x − 1) =

sec x + 1

= sec x − 1

17. Let x = 5 sin θ , then

25 − x2 =

25 − (5 sin θ )2

=

25 − 25 sin 2 θ =

25(1 − sin 2 θ ) =

25 cos2 θ

= 5 cos θ .

18. Let x = 4 sec θ , then

x2 − 16 =

(4 sec θ )2

− 16 =

16 sec2 θ − 16 =

16(sec2 θ − 1) =

16 tan 2 θ

= 4 tan θ .

19. cos x(tan 2 x + 1) = cos x sec2 x 25. sin5 x cos2 x = sin 4 x cos2 x sin x

1 = sec2 x

sec x = (1 − cos2

x) cos2

x sin x

= sec x = (1 − 2 cos

x + cos

x) cos

x sin x

20.

sec2 x cot x − cot x = cot x(sec2 x − 1)

= cot x tan 2 x

26.

2 4 2

= (cos2 x − 2 cos4 x + cos6 x) sin x

cos3 x sin 2 x = cos x cos2 x sin 2 x

1 = tan 2 x = tan x

tan x

= cos x(1 − sin 2 x) sin 2 x

4

= cos x(sin 2

x − sin x)

21. sin π 2

− θ tan θ

= cos θ tan θ

= (sin 2 x − sin 4 x) cos x

sin θ = cos θ

cos θ

= sin θ

27. sin x =

sin x =

3 − sin x

3

2

x = π + 2π n,

2π + 2π n

22. cot π 2 − θ

csc θ

3 3 = tan θ csc θ

sin θ 1 =

cos θ sin θ

28. 4 cos θ

2 cos θ

= 1 + 2 cos θ

= 1

1

1 =

cos θ

cos θ = 2

π 5π





= sec θ θ = + 2nπ or

3 + 2nπ

3

23. 1

= 1

= cos θ 29. 3 3 tan u = 3

tan θ csc θ sin θ

cos θ ⋅

1 sin θ

tan u = 1

3

π24.

1 1 1 = =

u = + nπ 6

tan x csc x sin x (tan x)

1 (sin x)

sin x tan x

= cot x






30. 1

sec x − 1 = 0 2

1 sec x = 1

2

35. cos2 x + sin x = 1

1 − sin 2 x + sin x − 1 = 0

−sin x(sin x − 1) = 0

sec x = 2 sin x = 0 sin x − 1 = 0

cos x = 1 2

x = 0, π sin x = 1

x = π

x = π + 2nπ or

5π 2 + 2nπ

31.

3 3

3 csc2 x = 4

36. sin 2 x + 2 cos x = 2

1 − cos2 x + 2 cos x = 2

2

csc2 x = 4 0 = cos x − 2 cos x + 1

3

sin x = ± 3

2

0 = (cos x − 1)2

cos x − 1 = 0

cos x = 1

x = π + 2π n,

2π + 2π n, 4π + 2π n,

5π + 2π n x = 0

3 3 3 3

These can be combined as:

37. 2 sin 2x −

2 = 0

x = π

+ nπ or x = 2π

+ nπ

sin 2x = 2

3 3

2 x = π

+ 2π n, 3π

2

+ 2π n

32. 4 tan 2 u − 1 = tan 2 u 4 4

3 tan 2 u − 1 = 0 x =

π + π n,

3π

+ π n

tan 2 u = 1 3

tan u = ± 1 3

= ± 3 3

8 8

x = π

, 3π

, 9π

, 11π

8 8 8 8

x

u = π

+ nπ or 5π

+ nπ 38. 2 cos + 1 = 0

2

6 6 x 1 cos = −

33. sin3 x = sin x 2 2

sin3 x − sin x = 0

sin x(sin 2 x − 1) = 0

sin x = 0 x = 0, π

sin 2 x = 1

x =

2π 2 3

x = 4π 3

x

sin x = ±1 x = π

, 3π

2 2

39. 3 tan 2 − 1 = 0 3

tan 2 x =

1

34.





3 2

c

o

s2

x

+

3

c

o

s

x

=

0

c

o

s

x

(2

c

o

s

x

+

3

)

=

0

cos

x =

0

or

2

cos

x +

3 =

0

3

tan x

=

± 1

3 3

tan x

=

± 3





= −

3 3

x = π

, 3π

2 2 2 cos x = −3 x

= π

, 3 6

5π ,

7π 6 6

cos x 3 2 x =

π ,

2

5π ,

7π 2 2

No solution 5π and

7π 2 2

are greater than 2π , so they are not

solutions. The solution is x = π

. 2





2

(

(

= −


40. 3 tan 3x = 0

tan 3x = 0

3x = 0, π , 2π , 3π , 4π , 5π

43. tan 2 x − 2 tan x = 0

tan x(tan x − 2) = 0

tan x = 0 or tan x − 2 = 0

x = 0, π

, 2π

, π , 4π

, 5π

3 3 3 3 x = nπ tan x = 2

x = arctan 2 + nπ

41. cos 4x(cos x − 1) = 0

44.

2 tan 2 x − 3 tan x = −1

cos 4x = 0 cos x − 1 = 0 2 tan 2 x − 3 tan x + 1 = 0

4x = π + 2π n,

3π + 2π n cos x = 1 (2 tan x − 1)(tan x − 1) = 0

2 2 2 tan x − 1 = 0 or tan x − 1 = 0

x = π +

π n,

3π + π

n x = 0 2 tan x = 1 tan x = 1

8 2 8 2

x = 0, π

, 3π

, 5π

, 7π

, 9π

, 11π

, 13π

, 15π

8 8 8 8 8 8 8 8

tan x = 1 2

x = π

+ nπ 4

42.

3 csc2 5x = −4

x = arctan 1

+ nπ

csc2 5x 4 3

csc 5x = ± − 4 3

No real solution

45. tan 2 θ + tan θ − 6 = 0

(tan θ + 3)(tan θ − 2) = 0

tan θ + 3 = 0 or tan θ − 2 = 0

tan θ

θ

= −3 tan θ

= arctan(−3) + nπ θ

= 2

= arctan 2 + nπ

46.

sec2 x + 6 tan x + 4 = 0

1 + tan 2 x + 6 tan x + 4 = 0

tan 2 x + 6 tan x + 5 = 0

47. sin 75° = sin(120° − 45°) = sin 120° cos 45° − cos 120° sin 45°

3 2 1 2

= − −

(tan x + 5)(tan x + 1) = 0 2 2 2 2

tan x + 5 = 0 or tan x + 1 = 0 = 2

4 3 + 1)

tan x = −5 tan x = −1 cos 75° = cos (120° − 45°)

x = arctan(−5) + nπ x = 3π + nπ = cos 120° cos 45° + sin 120° sin 45°

4

= − 1 2 3 2

+

2 2 2 2

= 2

4





3 − 1)

tan 75° = tan(120° − 45°) = tan 120° −tan 45°

1 + tan 120° tan 45°

= − 3 − 1

= − 3 − 1

1 + (− 3)(1) 1 − 3

= − 3 − 1

⋅ 1 + 3

1 − 3 1 + 3

= − 4 − 2 3

− 2

= 2 + 3





(

(

= − + = (

(

)

)


48. sin(375°) = sin(135° + 240°)

= sin 135° cos 240° + cos 135° sin 240°

2 1 2 3

=

− + −

−

2 2 2 2

= 2

3 − 1 4

cos(375°) = cos(135° + 240°)

= cos 135° cos 240° − sin 135° sin 240°

2 1 2 3

= − − −

−

2 2 2 2

2 = 1 + 3

4

tan(375°) = tan(135° + 240°)

= tan 135° + tan 240° 1 − tan 135° tan 240°

= −1 + 3

1 − (−1)( 3)

= −1 + 3

⋅ 1 − 3

= − 4 + 2 3

= 2 − 31 + 3 1 − 3 1 − 3

49.

sin 25π

= sin 11π

+ π

= sin 11π

cos π

+ cos 11π

sin π

12

6 4

6 4 6 4

1 2 3 2 2

3 − 1) 2 2 2 2 4

cos 25π

= cos11π

+ π

= cos 11π

cos π

− sin 11π

sin π

12

6 4

6 4 6 4

3 2 1 2 2

= − − =

3 + 1)

2 2 2 2 4

tan 11π

+ tan π

tan 25π

= tan11π

+ π

= 6 4

12

6 4

1 tan 11π

tan π

− 6 4

3 − + 1

3

= = 2 − 3 3

1 − − (1) 3





5 3

4

− 4

v

− 3 5

(

=

4 4


50.

sin19π

= sin11π

− π

cos19π

= cos11π

− π

12

6 4

12

6 4

= sin 11π

cos π

− cos 11π

sin π

= cos 11π

cos π

+ sin 11π

sin π

6 4 6 4

1 2 3 2

6 4 6 4

3 2 1 2

= − ⋅ − ⋅ = ⋅ + −

2 2 2 2 2 2 2 2

= − 2

(1 + 3) = − 2

( 3 + 1) = 2

3 − 1)4 4 4

tan19π

= tan11π

− π

12

6 4

tan 11π − tan

π

= 6 4

1 + tan 11π

tan π

6 4

− 3

− 1

= 3 =

− 3 − 3

⋅ 3 + 3

3 1 + − (1)

3 − 3 3 + 3

3

−(12 + 6 3) = = −2 − 3

6

51. sin 60° cos 45° − cos 60° sin 45° = sin(60° − 45°)

= sin 15°

52. tan 68° −tan 115°

= tan(68° − 115°) 1 + tan 68° tan 115°

= tan(−47°)

y y

u x x

53. ( )


3 ( 4 )

4 ( 3 ) 24

sin u + v = sin u cos v + cos u sin v = 5

− 5

3

+ 3 3

+ 5

− 5

= − 25

54. tan(u + v) = tan u + tan v

= 4 4 = 2 = 3 16 24

1 − tan u tan v 1 −

3 3

7

16

2 7 7

55. ( )

4 ( 4 )

3 ( 3 )cos u − v = cos u cos v + sin u sin v =





5 −

5 +

5 −

5 = −1

56. ( )

3 ( 4 )

4 ( 3 )sin u − v = sin u cos v − cos u sin v = 5

− 5

− 5

− 5

= 0

57.

cos x +

π = cos x cos

π − sin x sin

π = cos x(0) − sin x(1) = −sin x

2

2 2





, π

2

4

1 − 4

= −

= 2

= −

Review Exer cis es for Chapter 2 277

58. tan x −

π = −tan

π − x

= −cot x

61. sin

x +

π − sin

x −

π = 1

2

2

4

4

59.

tan(π − x) = tan π − tan x

= −tan x 1 − tan π tan x

2 cos x sin π

= 1 4

60.

sin(x − π ) = sin x cos π − cos x sin π

= sin x(−1) − cos x(0)

= − sin x

cos x = 2

2

x = π

, 7π

4 4

62. cos x +

π − cos

x −

π = 1

6

6

cos x cos

π − sin x sin

π −

cos x cos

π + sin x sin

π = 1

6 6

6 6

−2 sin x sin π

= 1 6

63.

sin u 4

< u < 3π

−2 sin x 1

= 1

sin x = −1

x = 3π 2

65.

sin 4x = 2 sin 2x cos 2 x

5 2 = 22 sin x cos x(cos2 x − sin 2 x)cos u = − 1 − sin 2 u =

−3 5 = 4 sin x cos x(2 cos2

x − 1)

tan u = sin u

= 4

cos u 3

= 8 cos3 x sin x − 4 cos x sin x

2

4 3 24 sin

2u = 2 sin u cos u = 2 − − = 5 5 25 − 2π 2π


3 2 2

− −

7

= −

5

5

25 − 2

4 2

1 − cos 2 x

1 − (1 − 2 sin 2 x)

tan 2u = 2 tan u

1 − tan 2 u

3 = − 24

7

3

66. 1 + cos 2 x

= 1 + (2 cos x2 − 1) 2 sin 2 x

= 2 cos2 x

64. cos u = − 2

, π

5 2 < u < π sin u =

1 and 5

= tan 2 x

tan u 1 4

2

1 2 4 sin 2u = 2 sin u cos u = 2 − = −

5 5 5 − 2π 2π





2

2 4


2 2 2

− 1

= 3 − 1

5

5

5

2 sin 2 3x 1 − cos 6x

2

1 − cos 6 x

1 2 − 67. tan 3x =

2 = =

tan 2u = 2tan u

= 2

= −1

= − 4 cos 3x 1 + cos 6 x 1 + cos 6 x

1 − tan 2 u 1 3 3 2

1 − −





2

12

2

3

3

=


68.

sin 2 x cos2 x = 1 − cos 2 x 1 + cos 2 x

2

2

1 − cos2 2 x =

4

1 − 1 + cos 4 x

= 4

= 1 − cos 4 x

8

1 − cos 150°

3 1 − − 2

2 + 3 1

69. sin(−75°) = − = − = − = − 2 + 3 2 2 2 2

3 1 + −

1 + cos 150° 2 2 − 3 1

cos(−75°) = − = = = 2 − 3 2 2 2 2

3

1 − −

− °

tan(−75°) 1 cos 150

2 = −(2 + 3) = −2 − 3

= −

sin 150°

= − 1

2

5π 1 − −

−

5π 1 cos 6 2 2 + 3 1

70. sin = = = = 2 + 3

12 2 2 2 2

5π 1 + − +

5π 1 cos 6 2 2 − 3 1

cos = = = = 2 − 3

12 2 2 2 2

tan 5π

=

1 − cos 5π

6

sin 5π

3 1

− −

= 1

= 2 + 3

71.

tan u =

4

, π

6 2

< u < 3π

(b)

sin u

=

1 − cos u =

1 − −

3

5 4

3 2 2 2 2 5 y

2 5 =

5

1 + −

3





u cos u 1 + cos u

= − = − 5

= − 1

− 3 x 2 2 2 5

− 4 5

5 = −

5

1 − −

3

(a) Because u is in Quadrant III, u

is in Quadrant II. tan u 1 − cos u

= = 5 = −2

2 2 sin u −

4

5





5 3

4

1 −

1 +

1 −

,

= −

=


72. sin u = 3

, 0 < u < π

5 2 y

u x

(a) Because u is in Quadrant I, u

is in Quadrant I. 2

4

(b) sin u

= 1 − cos u = 5 = 1

= 10

2 2 2 10 10

4

cos u

= 1 + cos u = 5 = 9

= 3 10

2 2 2 10 10

4

tan u =

1 − cos u = 5 = 1

2 sin u 3 3

5

73. cos u 2 π 7 2

y

< u < π

7

3 5 u

x − 2

(a) Because u is in Quadrant II, u

is in Quadrant I. 2

1 − −

2

(b) sin u

= 1 − cos u = 7 = 9

2 2 2 14

3 14=

cos u

=

14

1 + cos u =

1 + −

2

7 5

2 2 2 14

70 =

14

1 − −

2

tan u 1 − cos u

= = 7 3 5

=

2 sin u 3 5 5

7





1 −

1 +

1 −

= −

2

2

2

cos

θ


74.

tan u = − 21

, 3π

2 2 y

< u < 2π

u

2 x

5 − 21

(a) Because u is in Quadrant IV, u


2

(b) sin u

= 1 − cos u = 5 = 3

= 30

2 2 2 10 10

2

cos u = −

1 + cos u = − 5 = −

7 = −

70

2 2 2 10 10

2

tan u =

1 − cos u = 5 3 3 21 = − = −

21

2 sin u 21

− 21 21 7

5

75. cos 4θ sin 6θ = 1 sin(4θ + 6θ ) − sin(4θ − 6θ ) = 1 sin 10θ − sin(−2θ )

76.

77.

2 sin 7θ cos 3θ

cos 6θ + cos 5θ

= 2 ⋅ 1 sin(7θ + 3θ ) + sin(7θ − 3θ ) = sin 10θ + sin 4θ

= 2 cos 6θ + 5θ

cos 6θ − 5θ

= 2 cos 11θ

cos θ

2

2

2 2

78. sin 3x − sin x = 2 3x + x

sin 3x − x

79. r = 1

v02

sin 2θ

2 2 32

= 2 cos 2 x sin x range = 100 feet

v0 = 80 feet per second

r = 1

(80)2 sin 2θ

32

= 100

sin 2θ

2θ

θ

= 0.5

= 30°

= 15° or π 12

80. Volume V of the trough will be the area A of the isosceles triangle times the length l of the trough.

V =

(a)

A ⋅ l

A = 1 bh

2

cos θ =

h h = 0.5 cos

θ 4 m

2 0.5 2 b b





sin θ = 2

b = 0.5 sin θ h

2 0.5 2 2 0.5 m 0.5 m

A = 0.5 sin θ

0.5 cos θ = (0.5)

2 sin

θ cos

θ = 0.25 sin θ

cos θ

square meters

2 2 2 2 2 2

V = (0.25)(4) sin θ

cos θ

cubic meters = sin θ

cos θ

cubic meters 2 2 2 2

Not drawn to scale





=

Proble m Solv i ng fo r Cha p t e r 2 281

(b) V

= sin θ

cos θ

= 1 2 sin

θ cos

θ = 1

sin θ cubic meters

2 2 2

2 2

2

Volume is maximum when θ = π

. 2

81. False. If π < θ < π , then

π < θ

< π

, and θ

is in 84. True. It can be verified using a product-to-sum formula.

2

Quadrant I. cos θ

> 0

4 2 2 2 4 sin 45° cos 15° = 4 ⋅

1[sin 60° + sin 30°]

2

2 3 1

82. True. cot x sin 2 x = cos x

sin 2 x = cos x sin x.

sin x

= 2

+ = 2 2

3 + 1

85. Yes. Sample Answer. When the domain is all real

83. True. 4 sin(− x)cos(− x) = 4(−sin x) cos x

= −4 sin x cos x

numbers, the solutions of sin x =

5π

1 are x =

π 2 6

+ 2nπ

= −2(2 sin x cos x)

= −2 sin 2x

Problem Solving for Chapter 2

and x = + 2nπ , so there are infinitely many 6

solutions.

1. sin θ = ± 1 − cos2 θ You also have the following relationships:

tan θ

= sin θ

= ± cos θ

1 − cos2 θ

cos θ sin θ = cos

π 2

− θ

csc θ

= 1

= ± sin θ

1

1 − cos2 θ

tan θ

csc θ

cos (π 2) − θ

cos θ

1 =

sec θ

cot θ

= 1 cos θ

= 1

= ± cos θ

sec θ

cos(π 2) − θ

1 =

cos θ

tan θ 1 − cos2 θ

cot θ cos θ

= cos(π 2) − θ

(2n + 1)π 2nπ + π

2. cos = cos (12n + 1)π 1

3. sin = sin (12nπ + π )

2

2

6

6

= cos nπ +

π

= sin 2nπ +

π

2

6

= cos nπ cos π − sin nπ sin

π = sin π

= 1

2 2

= (±1)(0) − (0)(1) 6 2

(12n + 1)π 1





= 0 So, sin 6

= for all integers n. 2

(2n + 1)π

So, cos = 0 for all integers n.

2





4π 1 2 3 5 6

1

2


4. p(t) = 1

p (t ) + 30 p (t) + p (t ) + p (t) + 30 p (t) 1.4

1.4

p1(t) p2(t)

(a) p1(t ) = sin(524π t ) 1

− 0.006 0.006

− 0.006 0.006

p2 (t) =

p3 (t ) =

p5 (t) =

sin(1048π t) 2

1 sin(1572π t )

3

1 sin(2620π t )

5

1.4

p3(t)

− 1.4

1.4

p5(t)

− 1.4

1.4

p

6(t)

p6 (t) = 1

sin(3144π t) 6

− 0.006 0.006 − 0.006 0.006 − 0.006 0.006

The graph of − 1.4

− 1.4

− 1.4

p(t) = 1

sin(524π t ) + 15 sin(1048π t) + 1

sin(1572π t ) + 1

sin(2620π t ) + 5 sin(3144π t )4π 3 5

yields the graph shown in the text below.

y

1.4

y = p(t)

t

0.006

(b)

− 1.4

Function Period

p (t ) 2π

524π

p (t) 2π

1

= ≈ 0.0038 262

= 1

≈ 0.0019

(c)

1.4 Max

0 0.00382

p3 (t )

1048π

2π

524

= 1

≈ 0.0013

− 1.4 Min

1

1572π

2π 786

1

Over one cycle, 0 ≤ t < , you have five t-intercepts: 262

p5 (t)

p6 (t )

2620π

2π

3144π

= ≈ 0.0008 1310

1 = ≈ 0.0006

1572

t = 0, t ≈ 0.00096, t ≈ 0.00191, t ≈ 0.00285,

t ≈ 0.00382

(d) The absolute maximum value of p over one cycle is

p ≈ 1.1952, and the absolute minimum valueThe graph of p appears to be periodic with a

period of 1

≈ 0.0038. 262

of p over one cycle is p ≈ −1.1952.





2

2

= −

b


5. From the figure, it appears that u + v = w. Assume that u, v, and w are all in Quadrant I.

From the figure:

tan u = s

= 1

3s 3

tan v = s

= 1

2s 2

tan w = s

= 1 s

tan u + v tan u + tan v 1 3 + 1 2 5 6

= = = = 1 = tan w.

( ) 1 − tan u tan v

1 − (1 3)(1 2) 1 − (1 6)

So, tan(u + v) = tan w. Because u, v, and w are all in Quadrant I, you have

arctan tan(u + v) = arctan[tan w]u + v = w.

6. y = − 16 x2 + (tan θ )x + h0

Let h0

v0 2 cos2 θ

= 0 and take half of the horizontal distance:

1 1 v

sin 2θ

= 1

v

2 sin θ cos θ

1

= v

sin θ cos θ

2

2 32

0 0 ( ) 0

2

64 32

Substitute this expression for x in the model.

2

y = − 16 1

v 2 sin θ cos θ sin θ 1

+ v sin θ cos θ

v 2 cos2 θ 32

0

cos θ

32

0

0

1 v 2 sin 2 θ +

1 v 2 sin 2 θ

64 0

1 = v0

2 sin 2 θ 64

32 0

7. (a) 10 θ 10

h

sin θ

b

1 b

= 2

1 2

and cos θ

= h

2 10 2 10

b = 20 sin θ h = 10 cos

θ

2 2

A = 1 bh

2

1 θ θ

= 20 sin 10 cos

2

2

2





2

(b)

= 100 sin θ

cos θ

2 2

θ θ A = 50 2 sin cos

2 2

θ

= 50 sin 2

= 50 sin θ

Because sin π

= 1 is a maximum, θ = π

. So, the area is a maximum at A = 50 sin π

= 50 square meters.

2 2 2





2


8. The hypotenuse of the larger right triangle is:

θ

2 1 2(1 + cos θ )

1

sin 2 θ + (1 + cos θ )2

=

=

sin 2 θ + 1 + 2 cos θ + cos2 θ

2 + 2 cos θ

cos θ θ = 2(1 + cos θ )

sin θ

sin θ

= sin θ

= sin θ

⋅ 1 − cos θ

= sin θ 1 − cos θ

= sin θ 1 − cos θ

=

1 − cos θ

2

θ

2(1 + cos θ )

1 + cos θ

2(1 + cos θ )

(1 + cos θ )2

1 − cos θ

1 + cos θ

2(1 − cos2 θ ) 2 sin θ 2

cos = = = 2 2(1 + cos θ ) 2(1 + cos θ ) 2

tan θ

= sin θ

1 + cos θ

9. F

0.6W sin(θ + 90°) =

sin 12°

(a) F =

0.6W (sin θ cos 90° +cos θ sin 90°) sin 12°

= 0.6W (sin θ )(0) + (cos θ )(1)

sin 12°

(b) Let y1 =

550

0.6(185) cos x.

sin 12°

= 0.6W cos θ

sin 12°

0 90

0

(c) The force is maximum (533.88 pounds) when θ = 0°.

The force is minimum (0 pounds) when θ = 90°.

π (t + 0.2)10. Seward: D = 12.2 − 6.4 cos

11.

d = 35 − 28 cos π

t when t = 0 corresponds to

182.6 6.2

π (t + 0.2) New Orleans: D = 12.2 − 1.9 cos

12:00 A.M.

(a) The high tides occur when cos π

t = −1. Solving

182.6 6.2

(a)

20

0 365 0

yields t = 6.2 or t = 18.6.

These t-values correspond to 6:12 A.M. and 6:36 P.M.

The low tide occurs when cos π

t = 1. Solving 6.2

yields t = 0 and t = 12.4 which corresponds to

12:00 A.M. and 12:24 P.M.

(b) The graphs intersect when t ≈ 91 and when

t ≈ 274. These values correspond to April 1 and

(b) The water depth is never 3.5 feet. At low tide, the

depth is d = 35 − 28 = 7 feet.

October 1, the spring equinox and the fall equinox.

(c) Seward has the greater variation in the number of

daylight hours. This is determined by the

a

m

p

litudes, 6.4 and

1.9.





(d) Period: 2π

= 365.2 days π 182.6

(c)

70

0 24 0





4π

2

2

2

2

2

2

2

2

2

2

.


12. h1 = 3.75 sin 733t + 7.5 θ α

sin +

h = 3.75 sin 733 t + + 7.5

13. (a) n =

2 2

2 3

θ

(a) 15

sin 2

sin θ

cos α

+ cos θ

sin α

=

sin θ

0 1

0

(b) The period for h1 and h2 is

2π

733

≈ 0.0086.

= cos α

+ cot θ

sin α

θ

12 For α = 60°, n = cos 30° + cot sin 30° 2

n = 3

+ 1

cot θ

0 2π

733 3

2 2

(b) For glass, n = 1.50.

3 1

θ

The graphs intersect twice per cycle. 1.50 = + 2 2

cot 2

There are 1

≈ 116.66 cycles in the interval

3 θ2π 733

21.50 −

= cot

[0, 1], so the graphs intersect approximately 2 2

233.3 times. 1 θ

= tan

3 − 3

θ

= 2 tan −1 1

3 − 3

θ ≈ 76.5°

14. (a)

(b)

sin(u + v + w) = sin (u + v) + w

= sin(u + v) cos w + cos(u + v) sin w

= [sin u cos v + cos u sin v] cos w + [cos u cos v − sin u sin v] sin w

= sin u cos v cos w + cos u sin v cos w + cos u cos v sin w − sin u sin v sin w

tan(u + v + w) = tan (u + v) + w

tan(u + v) + tan w =

1 − tan(u + v) tan w

tan u + tan v + tan w

= 1 − tan u tan v ⋅

(1 − tan u tan v) tan u + tan v

1 − tan w 1 − tan u tan v

(1 − tan u tan v)

tan u + tan v + (1 − tan u tan v) tan w =

(1 − tan u tan v) − (tan u + tan v) tan w

= tan u + tan v + tan w − tan u tan v tan w 1 − tan u tan v − tan u tan w − tan v tan w





6 6 3 3

= 1 cos 2x

+ 1 cos 2 x

4

2

4

.

2

2

2


15. (a) Let y1 = sin x and y2 = 0.5. (b) Let y1 = cos x and y2 = −0.5.

2

0 2π

2

0 2π

− 2

sin x ≥ 0.5 on the interval π

, 5π

− 2

cos x ≤ −0.5 on the interval

2π

, 4π

.

(c) Let y1 = tan x and y2

= sin x.

(d) Let y1 = cos x and y2

= sin x.

2

0 2π

2

0 2π

− 2 − 2

π

5π

tan x < sin x on the intervals π

, π and

3π , 2π .

cos x ≥ sin x on the intervals 0,

and

, 2π .

2

2

4 4

16. (a) f (x) = sin 4 x + cos4 x

= (sin 2 x) + (cos2 x)

2

−

2

+

2 2

= 1 (1 − 2 cos 2x + cos2 2 x) + (1 + 2 cos 2 x + cos2 2x)

= 1 (2 + 2 cos2 2 x)

= 1 (1 + cos2 2x)

1 cos 4x

= 1 +

2 2

= 1 (3 + cos 4x)

4

(b) f (x) = sin 4 x + cos4 x

= (sin 2 x) + cos4 x

2 2

4

= (1 − cos x) + cos x

(c)

= 1 − 2 cos2 2x + cos4 x + cos4 x

= 2 cos4 x − 2 cos2 x + 1

f (x) = sin 4 x + cos4 x

= sin 4 x + 2 sin 2 x cos2 x + cos4 x − 2 sin 2 x cos2 x





2

= (sin 2 x + cos2 x)

− 2 sin 2 x cos2 x

= 1 − 2 sin 2 x cos2 x





2

( 2 )


(d) f (x) = 1 − 2 sin 2 x cos2 x

1 − cos 2x 1 + cos 2x

= 1 − 2

2

2

= 1 − 1 (1 − cos2 2x)

1 1 = + cos2 2x

2 2

1 1 = + 1 − sin 2x

2 2

= 1 − 1

sin 2 2x 2

(e) No; there is often more than one way to rewrite a trigonometric expression, so your result and your friend’s result

could both be correct.





= <


Practice Test for Chapter 2

1. Find the value of the other five trigonometric functions, given tan x 4 , sec x 0. 11

2. Simplify sec2 x + csc2 x

. csc2 x(1 + tan 2 x)

3. Rewrite as a single logarithm and simplify ln tan θ − ln cot θ .

4. True or false:

cos π

− x

= 1

2

csc x

5. Factor and simplify: sin 4 x + (sin 2 x) cos2 x

6. Multiply and simplify: (csc x + 1)(csc x − 1)

7. Rationalize the denominator and simplify:

cos2 x

1 − sin x

8. Verify:

1 + cos θ

sin θ

sin θ

+ = 2 csc θ 1 + cos θ

9. Verify:

tan 4 x + 2 tan 2 x + 1 = sec4 x

10. Use the sum or difference formulas to determine:

(a) sin 105°

(b) tan 15°

11. Simplify: (sin 42°) cos 38° − (cos 42°) sin 38°

12. Verify tan

+ π

= 1 + tan θ

. θ

4 1 − tan θ

13. Write sin(arcsin x − arccos x) as an algebraic expression in x.

14. Use the double-angle formulas to determine:

(a) cos 120°

(b) tan 300°

15. Use the half-angle formulas to determine:

(a) sin 22.5°

(b) tan π 12

16. Given sin θ = 4 5, θ lies in Quadrant II, find cos(θ 2).





1

Practice Te st for C hapte r 2 289

17. Use the power-reducing identities to write (sin 2 x) cos2 x in terms of the first power of cosine.

18. Rewrite as a sum: 6(sin 5θ ) cos 2θ .

19. Rewrite as a product: sin(x + π ) + sin(x − π ).

20. Verify sin 9 x + sin 5x cos 9 x − cos 5x

= −cot 2 x.

21. Verify:

(cos u) sin v =

2 sin(u + v) − sin(u − v).

22. Find all solutions in the interval [0, 2π ):

4 sin 2 x = 1


tan 2 θ + ( 3 − 1) tan θ −

3 = 0


sin 2 x = cos x

25. Use the quadratic formula to find all solutions in the interval [0, 2π ):

tan 2 x − 6 tan x + 4 = 0





2.1 Using Fundamental Identities






Objectives

Recognize and write the fundamental

trigonometric identities.

Use the fundamental trigonometric identities to

evaluate trigonometric functions, simplify

trigonometric expressions, and rewrite

trigonometric expressions.

3





Introduction

4





Introduction

You will learn how to use the fundamental identities to do

the following.

1. Evaluate trigonometric functions.

2. Simplify trigonometric expressions.

3. Develop additional trigonometric identities.

4. Solve trigonometric equations.

5





Introduction

6





Introduction cont’d

7





Introduction

Pythagorean identities are sometimes used in radical form

such as

or

where the sign depends on the choice of u.

8





Using the Fundamental Identities

9





10


One common application of trigonometric identities is to

use given values of trigonometric functions to evaluate

other trigonometric functions.





11

Example 1 – Using Identities to Evaluate a Function

Use the values and tan u > 0 to find the values of all six trigonometric functions.

Solution:

Using a reciprocal identity, you have

.

Using a Pythagorean identity, you have

Pythagorean identity





12

Example 1 – Solution

cont’d

Substitute for cos u.

. Simplify.

Because sec u < 0 and tan u > 0, it follows that u lies in

Quadrant III.

Moreover, because sin u is negative when u is in

Quadrant III, choose the negative root and obtain





13


Knowing the values of the sine and cosine enables you to

find the values of all six trigonometric functions.

cont’d





Example 1 – Solution cont’d

14





15

Example 2 – Simplifying a Trigonometric Expression

Simplify

sin x cos2 x – sin x.

Solution:

First factor out a common monomial factor and then use a fundamental identity.

sin x cos2 x – sin x = sin x(cos2 x – 1)

= –sin x(1 – cos2 x)

= –sin x(sin2 x)

= –sin3 x

Factor out common

monomial factor.

Factor out –1.


Multiply.





16


When factoring trigonometric expressions, it is helpful to

find a special polynomial factoring form that fits the

expression.

On occasion, factoring or simplifying can best be done by

first rewriting the expression in terms of just one

trigonometric function or in terms of sine and cosine only.





17

Example 7 – Rewriting a Trigonometric Expression

Rewrite so that it is not in fractional form.

Solution:

From the Pythagorean identity

cos2 x = 1 – sin2 x = (1 – sin x)(1 + sin x)

multiplying both the numerator and the denominator by

(1 – sin x) will produce a monomial denominator.

Multiply numerator and

denominator by (1 – sin x).

Multiply.





18


cont’d


Write as separate fractions.

Product of fractions

Reciprocal and quotient

identities





2.2 Verifying Trigonometric

Identities






Objective

Verify trigonometric identities.

3





Introduction

4





5

Introduction

In this section, you will study techniques for verifying


Remember that a conditional equation is an equation that is

true for only some of the values in its domain.

For example, the conditional equation

sin x = 0

is true only for

x = nπ

Conditional equation

where n is an integer. When you find these values, you are

solving the equation.





6

Introduction

On the other hand, an equation that is true for all real values in the domain of the variable is an identity. For example, the familiar equation

sin2x = 1 – cos2x

Identity

is true for all real numbers x. So, it is an identity.





Verifying Trigonometric Identities

7





8


Although there are similarities, verifying that a trigonometric

equation is an identity is quite different from solving an

equation. There is no well-defined set of rules to follow in

verifying trigonometric identities, and it is best to learn the

process by practicing.





9


Verifying trigonometric identities is a useful process when

you need to convert a trigonometric expression into a form

that is more useful algebraically.

When you verify an identity, you cannot assume that the

two sides of the equation are equal because you are trying

to verify that they are equal.

As a result, when verifying identities, you cannot use

operations such as adding the same quantity to each side

of the equation or cross multiplication.





10

Example 1 – Verifying a Trigonometric Identity

Verify the identity

Solution:

Start with the left side because it is more complicated.


Simplify.

Reciprocal identity

Quotient identity





11


cont’d

Simplify.

Notice that you verify the identity by starting with the left

side of the equation (the more complicated side) and using

the fundamental trigonometric identities to simplify it until

you obtain the right side.





2.3 Solving Trigonometric

Equations






Objectives

Use standard algebraic techniques to solve

trigonometric equations.

Solve trigonometric equations of quadratic type.

Solve trigonometric equations involving multiple

angles.

Use inverse trigonometric functions to solve

trigonometric equations.

3





Introduction

4





5

Introduction

To solve a trigonometric equation, use standard algebraic

techniques (when possible) such as collecting like terms

and factoring.

Your preliminary goal in solving a trigonometric equation is

to isolate the trigonometric function on one side of the

equation.

For example, to solve the equation 2 sin x = 1, divide each

side by 2 to obtain





6

Introduction

To solve for x, note in the figure below that the equation

has solutions x = π /6 and x = 5π /6 in the interval

[0, 2π).





7

Introduction

Moreover, because sin x has a period of 2π, there are

infinitely many other solutions, which can be written as

and General solution

where n is an integer, as shown above.





8

Introduction

The figure below illustrates another way to show that the equation has infinitely many solutions.

Any angles that are coterminal with π /6 or 5π /6 will also be

solutions of the equation.

When solving trigonometric

equations, you should write

your answer(s) using exact

values rather than decimal

approximations.





9

Example 1 – Collecting Like Terms

Solve

Solution:

Begin by isolating sin x on one side of the equation.

Write original equation.

Add sin x to each side.

Subtract from each side.





10


cont’d

Combine like terms.

Divide each side by 2.

Because sin x has a period of 2π, first find all solutions in the interval [0, 2π).

These solutions are x = 5π /4 and x = 7π /4. Finally, add multiples of 2π to each of these solutions to obtain the general form

and

General solution

where n is an integer.





Equations of Quadratic Type

11





12

Equations of Quadratic Type

Many trigonometric equations are of quadratic type

ax2 + bx + c = 0, as shown below.

Quadratic in sin x Quadratic in sec x

2 sin2 x – sin x – 1 = 0 sec2 x – 3 sec x – 2 = 0

2(sin x)2 – sin x – 1 = 0 (sec x)2 – 3(sec x) – 2 = 0

To solve equations of this type, factor the quadratic or,

when this is not possible, use the Quadratic Formula.





13

Example 4 – Factoring an Equation of Quadratic Type

Find all solutions of 2 sin2 x – sin x – 1 = 0 in the interval

[0, 2π).

Solution:

Treat the equation as a quadratic in sin x and factor.

2 sin2 x – sin x – 1 = 0

(2 sin x + 1)(sin x – 1) = 0


Factor.





14


Setting each factor equal to zero, you obtain the following

solutions in the interval [0, 2π).

2 sin x + 1 = 0 and sin x – 1 = 0

cont’d





Functions Involving Multiple Angles

15





16

Functions Involving Multiple Angles

The next example involve trigonometric functions of

multiple angles of the forms cos ku and tan ku.

To solve equations of these forms, first solve the equation

for ku, and then divide your result by k.





17

Example 7 – Solving a Multiple-Angle Equation

Solve 2 cos 3t – 1 = 0.

Solution:

2 cos 3t – 1 = 0

2 cos 3t = 1


Add 1 to each side.

Divide each side by 2.





18


cont’d

In the interval [0, 2π), you know that 3t = π /3 and 3t = 5π /3

are the only solutions, so, in general, you have

and

Dividing these results by 3, you obtain the general solution

and General solution






Using Inverse Functions

19





20

Example 9 – Using Inverse Functions

sec2 x – 2 tan x = 4

1 + tan2 x – 2 tan x – 4 = 0

tan2 x – 2 tan x – 3 = 0

(tan x – 3)(tan x + 1) = 0

Original equation


Combine like terms.

Factor.

Setting each factor equal to zero, you obtain two solutions

in the interval (–π /2, π /2). [We know that the range of the

inverse tangent function is (–π /2, π /2).]

x = arctan 3 and





21

Example 9 – Using Inverse Functionscont’d

Finally, because tan x has a period of π, you add multiples

of π to obtain

x = arctan 3 + nπ and


General solution

You can use a calculator to approximate the value of

arctan 3.





2.4 Sum and Difference Formulas






3

Objective

Use sum and difference formulas to evaluate

trigonometric functions, verify identities, and

solve trigonometric equations.





4

Using Sum and Difference

Formulas





5

Using Sum and Difference Formulas

In this and the following section, you will study the uses of

several trigonometric identities and formulas.

Example 1 shows how sum and difference formulas can

enable you to find exact values of trigonometric functions

involving sums or differences of special angles.





6

Example 1 – Evaluating a Trigonometric Function

Find the exact value of .

Solution:

To find the exact value of sin π /12, use the fact that

Consequently, the formula for sin(u – v) yields





7


cont’d

Try checking this result on your calculator. You will find that

sin π /12 ≈ 0.259.





8

Example 5 – Proving a Cofunction identity

Use a difference formula to prove the cofunction identity

= sin x.

Solution:

Using the formula for cos(u – v), you have





9

Using Sum and Difference Formulas

Sum and difference formulas can be used to rewrite

expressions such as

and , where n is an integer

as expressions involving only sin θ or cos θ.

The resulting formulas are called reduction formulas.





10

Example 7 – Solving a Trigonometric Equation

Find all solutions of

in the interval .

Solution:

Algebraic Solution

Using sum and difference formulas, rewrite the equation as





11


So, the only solutions in the interval [0, 2π) are x = 5π /4

and x = 7π /4.

cont’d





2.5 Multiple-Angle and

Product-to-Sum Formulas






3

Objectives

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.





4

Objectives

Use product-to-sum and sum-to-product

formulas to rewrite and evaluate trigonometric

functions.

Use trigonometric formulas to rewrite real-life

models.





Multiple-Angle Formulas

5





6


In this section, you will study four other categories of


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 sin(u / 2).

4. The fourth category involves products of trigonometric

functions such as sin u cos v.





7


You should learn the double-angle formulas because

they are used often in trigonometry and calculus.





8

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 x (rather than 2x). Then factor and solve.

2 cos x + sin 2x = 0

2 cos x + 2 sin x cos x = 0

2 cos x(1 + sin x) = 0

2 cos x = 0 and 1 + sin x = 0


Double-angle formula.

Factor.

Set factors equal to zero.





9


cont’d

So, the general solution is

and

Solutions in [0, 2π)

where n is an integer. Try verifying these solutions

graphically.





Power-Reducing Formulas

10





11

Power-Reducing Formulas

The double-angle formulas can be used to obtain the

following power-reducing formulas.





12

Example 4 – Reducing a Power

Rewrite sin4 x in terms of first powers of the cosines of

multiple angles.

Solution:

Note the repeated use of power-reducing formulas.

Property of exponents

Power-reducing formula

Expand.





13


cont’d

Power-reducing formula

Distributive Property

Simplify.

Factor out common factor.





14


You can use a graphing utility to check this result, as

shown below. Notice that the graphs coincide.

cont’d





Half-Angle Formulas

15





16

Half-Angle Formulas

You can derive some useful alternative forms of the

power-reducing formulas by replacing u with u / 2. The

results are called half-angle formulas.





17

Example 5 – Using a Half-Angle Formula

Find the exact value of sin 105°.

Solution:

Begin by noting that 105° is half of 210°. Then, using the

half-angle formula for sin(u / 2) and the fact that 105° lies in

Quadrant II, you have





18


.

cont’d

The positive square root is chosen because sin θ is positive

in Quadrant II.






19





20


Each of the following product-to-sum formulas can be

verified using the sum and difference formulas.

Product-to-sum formulas are used in calculus to solve

problems involving the products of sines and cosines of two

different angles.





21

Example 7 – 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 = [sin(5x + 4x) – sin(5x – 4x)]

= sin 9x – sin x.





22


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.





Application

23





24

Example 10 – Projectile Motion

Ignoring air resistance, the range of a projectile fired at an

angle θ with the horizontal and with an initial velocity of

v0 feet per second is given by

where r is the horizontal distance (in feet) that the projectile

travels.





25

Example 10 – Projectile Motion

A football player can kick a football

from ground level with an initial

velocity of 80 feet per second

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?

cont’d





26

2


a. You can use a double-angle formula to rewrite the projectile motion model as

r = v02 (2 sin θ cos θ)

= v02 sin 2θ.

Rewrite original projectile motion model.

Rewrite model using a double-angle formula.

b. r = v0 sin 2θ

Write projectile motion model.

200 = (80)2 sin 2θ

Substitute 200 for r and 80 for v0.





C H A P T E R 2 Analytic Trigonometry€¦ · 2 2 2 csc x = 1 + 7 = sin = cos x 2 214 Chapter 2...

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Transcript of C H A P T E R 2 Analytic Trigonometry€¦ · 2 2 2 csc x = 1 + 7 = sin = cos x 2 214 Chapter 2...