Chapter 6 Analytic Trigonometry
Copyright © 2014 Pearson Education, Inc. 705
Section 6.1
Check Point Exercises
1. 1 sin
csc tansin cos
xx x
x x
1
cossec
xx
We worked with the left side and arrived at the right side. Thus, the identity is verified.
2. cos
cos cot sin cos sinsin
xx x x x x
x
2
2
2 2
2 2
cossin
sin
cos sinsin
sin sin
cos sin
sin sin
cos sin
sin1
sincsc
xx
x
x xx
x x
x x
x x
x x
x
xx
We worked with the left side and arrived at the right side. Thus, the identity is verified.
3. 2 22
3
sin sin cos sin 1 cos
sin sin
sin
x x x x x
x x
x
We worked with the left side and arrived at the right side. Thus, the identity is verified.
4. 1 cos 1 cos
sin sin sincsc cot
θ θθ θ θ
θ θ
We worked with the left side and arrived at the right side. Thus, the identity is verified.
5. sin 1 cos
1 cos sin
x x
x x
2 2
2 2
sin (sin ) (1 cos )(1 cos )
(1 cos )sin sin (1 cos )
sin 1 2cos cos
(1 cos )sin (1 cos )sin
sin cos 2cos 1
(1 cos )sin
1 1 2cos
(1 cos )sin
2 2cos
(1 cos )sin
2 1 cos
(1 cos )sin
2
sin2cs
x x x x
x x x x
x x x
x x x x
x x x
x x
x
x x
x
x x
x
x x
x
c x
We worked with the left side and arrived at the right side. Thus, the identity is verified.
6.
2
2
cos cos 1 sin
1 sin (1 sin ) 1 sin
cos (1 sin )
1 sincos (1 sin )
cos1 sin
cos
x x x
x x x
x x
xx x
xx
x
We worked with the left side and arrived at the right side. Thus, the identity is verified.
Chapter 6 Analytic Trigonometry
706 Copyright © 2014 Pearson Education, Inc.
7. sec csc( )
sec csc
x x
x x
sec csc
sec csc1 1
cos sin1 1
cos sinsin cos
cos sin cos sin1
cos sinsin cos
cos sin1
cos sinsin cos cos sin
cos sin 1sin cos
x x
x x
x x
x xx x
x x x x
x xx x
x x
x xx x x x
x xx x
We worked with the left side and arrived at the right side. Thus, the identity is verified.
8. Left side:
2
1 1
1 sin 1 sin1(1 sin ) 1(1 sin )
(1 sin )(1 sin ) (1 sin )(1 sin )
1 sin 1 sin
(1 sin )(1 sin )
2
(1 sin )(1 sin )
2
1 sin
θ θθ θ
θ θ θ θθ θθ θ
θ θ
θ
Right side: 2
22
2 2
2 2
2 2
2
2 2
sin2 2 tan 2 2
cos
2cos 2sin
cos cos
2cos 2sin
cos2 2
cos 1 sin
θθθ
θ θθ θθ θ
θ
θ θ
The identity is verified because both sides are equal
to 2
2.
1 sin θ
Concept and Vocabulary Check 6.1
1. complicated; other
2. sines; cosines
3. false
4. (csc 1)(csc 1)x x
5. identical/the same
Exercise Set 6.1
1. 1
sin sec sincos
sin
costan
x x xx
x
xx
2. 1
2cos csc cossin
x x xx
cos
sincot
x
xx
3. tan( ) cos tan cos
sincos
cossin
x x x x
xx
xx
4. cot( ) sin cot sin
cossin
sincos
x x x x
xx
xx
5. sin 1
tan csc cos coscos sin1
xx x x x
x x
6. cos 1
cot sec sin sinsin cos1
xx x x x
x x
7. 2 2sec sec sin sec (1 sin )
1 2coscoscos
x x x x x
xxx
Section 6.1 Verifying Trigonometric Identities
Copyright © 2014 Pearson Education, Inc. 707
8. 2 2csc csc cos csc (1 cos )
1 2sinsinsin
x x x x x
xxx
9. 2 2 2 22 2
2
cos sin 1 sin sin
1 sin sin
1 2sin
x x x x
x x
x
10. 2 2 2 2cos sin cos (1 cos )2 2cos 1 cos
22cos 1
x x x x
x x
x
11. 1
csc sin sinsin
θ θ θθ
2
2
2
1 sin
sin sin
1 sin
sin
cos
sincos
cossincot cos
θθ θ
θθθ
θθ θθθ θ
12. sin cos
tan cotcos sin
θ θθ θθ θ
sin sin cos cos
cos sin sin cos2 2sin cos
cos sin cos sin2 2sin cos
cos sin1
cos sin1 1
cos sinsec csc
θ θ θ θθ θ θ θ
θ θθ θ θ θθ θ
θ θ
θ θ
θ θθ θ
13.
sin costan cot cos sin
1cscsin
θ θθ θ θ θ
θθ
11
sin1
1sin
sin1
1sin
θ
θθ
θ
14.
cos 1cos sec 1 cos
coscotsin
θθ θ θ
θθθ
1cos
sincos
1sin
sin1
costan
θθ
θθ
θθ
θ
15. 2 2 2 2sin (1 cot ) sin (csc )θ θ θ θ
2
2
1sin
sin1
θθ
16. 2 2 2 2cos (1 tan ) cos (sec )θ θ θ θ
2
2
1cos
cos1
θθ
17. 2 21 cos sin
cos cossin
sincos
sin tan
t t
t tt
tt
t t
18.
2 21 sin cos
sin sincos
cossin
cos cot
t t
t tt
tt
t t
Chapter 6 Analytic Trigonometry
708 Copyright © 2014 Pearson Education, Inc.
19. 2 2
2
2
1csc sin
coscotsin1 cos
sinsin1 sin
cossin1 1
sin coscsc sec
t tttt
t
ttt
tt
t tt t
20. 2 2
2
2
1sec cos
sintancos
1 sin
coscos1 cos
sincos1 1
cos sinsec csc
t tttt
t
ttt
tt
t tt t
21. 2 2
2
tan sec 1
sec sec
sec 1
sec secsec cos
t t
t t
t
t tt t
22. 2 2
2
cot csc 1
csc csc
csc 1
csc csccsc sin
t t
t t
t
t tt t
23. 1 cos 1 cos
sin sin sincsc cot
θ θθ θ θ
θ θ
24. 1 sin 1 sin
cos cos cossec tan
θ θθ θ θ
θ θ
25.
2 2
sin cos sin cos1 1csc sec
sin cos1 1
sin cossin cos
sin cossin cos
1 1
sin cos
1
t t t t
t tt t
t tt t
t tt t
t t
26. sin cos sin cossin costan cotcos sin
sin cossin cos
cos sincos sin
sin cossin cos
cos sin
sin cos
t t t tt tt tt t
t tt t
t tt t
t tt t
t t
t t
27. costan1 sin
sin cos
cos 1 sinsin 1 sin cos cos
cos 1 sin 1 sin cos2 2sin sin cos
cos (1 sin ) cos (1 sin )2 2sin sin cos
cos (1 sin )
1 sin
cos (1 sin )
1
cossec
tt
tt t
t tt t t t
t t t t
t t t
t t t t
t t t
t t
t
t t
tt
Section 6.1 Verifying Trigonometric Identities
Copyright © 2014 Pearson Education, Inc. 709
28. sin cos sincot1 cos sin 1 cos
cos 1 cos sin sin
sin 1 cos 1 cos sin2 2cos cos sin
sin (1 cos ) sin (1 cos )2 2cos cos sin
sin (1 cos )
cos 1
sin (1 cos )
1
sincsc
t t tt
t t tt t t t
t t t t
t t t
t t t t
t t t
t t
t
t t
tt
29. 2 2
2
2
2
2
sin sin 1 cos1 1
1 cos 1 cos 1 cos
sin (1 cos )1
1 cos
sin (1 cos )1
sin1 1 cos
cos
x x x
x x x
x x
x
x x
xx
x
30. 2 2
2
2
2
2
cos cos 1 sin1 1
1 sin 1 sin 1 sin
cos (1 sin )1
1 sin
cos (1 sin )1
cos1 1 sin
sin
x x
x x x
x x
x
x x
xx
x
31.
2
2
cos 1 sin
1 sin coscos 1 sin 1 sin
1 sin 1 sin coscos (1 sin ) 1 sin
cos1 sincos (1 sin ) 1 sin
coscos1 sin 1 sin
cos cos2
cos1
2cos
2sec
x x
x xx x x
x x xx x x
xxx x x
xxx x
x x
x
xx
32.
2
2
2
sin cos 1
cos 1 sinsin cos 1 cos 1
cos 1 cos 1 sinsin (cos 1) cos 1
sincos 1sin (cos 1) cos 1
sinsinsin (1 cos ) cos 1
sinsin1 cos cos 1
sin sin0
sin0
x x
x xx x x
x x xx x x
xxx x x
xxx x x
xxx x
x x
x
33. 2 2 2 2
2 2 2
22
2 2
22
2 2
2 2
sec csc (1 tan )csc
csc tan csc
sin 1csc
cos sin1
csccos
csc sec
sec csc
x x x x
x x x
xx
x x
xx
x x
x x
Chapter 6 Analytic Trigonometry
710 Copyright © 2014 Pearson Education, Inc.
34. 2 2
2
2
2
2
csc sec (1 cot )sec
sec cot sec
cos 1sec
cossincos
secsin
1 cossec
sin sinsec csc cot
x x x x
x x x
xx
xxx
xx
xx
x xx x x
35.
1 1sec csc cos sin
1 1sec csccos sin
x x x xx x
x x
1 1sincos sin
1 1 sincos sinsin
1cossin
1costan 1
tan 1
xx xx
x xx
xx
xx
x
36.
1 1csc sec sin cos
1 1csc secsin cos
x x x xx x
x x
1 1cossin cos
1 1 cossin coscos
1sincos
1sincot 1
cot 1
xx xx
x xx
xx
xx
x
37. 2 2sin cos (sin cos )(sin cos )
sin cos sin cossin cos
x x x x x x
x x x xx x
38. 2 2tan cot (tan cot )(tan cot )
tan cot tan cottan cot
x x x x x x
x x x xx x
39. 2 2 2 2
2
tan 2 sin 2 cos 2 tan 2 1
sec 2
x x x x
x
40. 2 2 2 2
2
cot 2 cos 2 sin 2 cot 2 1
csc 2
x x x x
x
41.
2
sin 2 cos 2tan 2 cot 2 cos 2 sin 2
1csc 2sin
sin 2 sin 2 cos 2 cos 2
cos 2 sin 2 sin 2 cos 21
sin
sin 2 cos2
cos 2 sin 21
sin 21 1
cos 2 sin 2 sin 21 sin 2
cos 2 sin 2 11
sec 2cos 2
θ θθ θ θ θ
θθ
θ θ θ θθ θ θ θ
θθ θθ θ
θ
θ θ θθ
θ θ
θθ
Section 6.1 Verifying Trigonometric Identities
Copyright © 2014 Pearson Education, Inc. 711
42.
2 2
2 2
sin 2 cos 2tan 2 cot 2 cos 2 sin 2
1sec 2cos 2
sin 2 sin 2 cos 2 cos 2
cos 2 sin 2 sin 2 cos 21
cos2
sin 2 cos 2
cos 2 sin 2 cos 2 sin 21
cos 2
sin 2 cos 2
cos2 sin 21
cos 21 1
cos 2 sin 2 cos21 cos 2
cos 2 sin 2
θ θθ θ θ θ
θθ
θ θ θ θθ θ θ θ
θθ θ
θ θ θ θ
θθ θ
θ θ
θ
θ θ θ
θ θ
1
1
sin 2csc 2
θ
θθ
43. sin sin
tan tan cos coscos cossin sin1 tan tan cos cos1cos cos
sin cos cos sin
cos cos sin sin
x y
x y x yx yx yx y x yy y
x y x y
x y x y
44. cos cos
cot cot sin sincos cos1 cot cot 1sin sin
cos cos
sin sinsin sincos cos sin sin1sin sin
cos sin sin cos sin sin
sin 1 sin 1cos cos sin sin
sin sinsin sin 1
cos sin sin co
x y
x y x yx yx yx y
x y
x yx yx y x yx y
x x y y x y
x yx y x y
x yx y
x y x
ssin sin cos cos
y
x y x y
Chapter 6 Analytic Trigonometry
712 Copyright © 2014 Pearson Education, Inc.
45. Left side: 2
2
2
2
2
1 sin(sec tan )
cos cos
1 sin
cos
(1 sin )
cos
xx x
x x
x
x
x
x
Right side:
2
2
2
2
1 sin 1 sin 1 sin
1 sin 1 sin 1 sin
(1 sin )
1 sin
(1 sin )
cos
x x x
x x x
x
x
x
x
The identity is verified because both sides are equal to 2
2
(1 sin )
cos
x
x
.
46. Left side: 2 2 2
22
1 cos 1 cos (1 cos )(csc cot )
sin sin sin sin
x x xx x
x x x x
Right side: 2 2
2 2
1 cos 1 cos 1 cos (1 cos ) (1 cos )
1 cos 1 cos 1 cos 1 cos sin
x x x x x
x x x x x
The identity is verified because both sides are equal to2
2
(1 cos )
sin
x
x
.
47.
2
2
tan tan sec 1
sec 1 sec 1 sec 1tan (sec 1)
sec 1tan (sec 1)
tansec 1
tan
t t t
t t tt t
tt t
tt
t
48.
2
2
cot cot csc 1
csc 1 csc 1 csc 1cot (csc 1)
csc 1cot (csc 1)
cotcsc 1
cot
t t t
t t tt t
tt t
tt
t
Section 6.1 Verifying Trigonometric Identities
Copyright © 2014 Pearson Education, Inc. 713
49. Left side:
2
2
2
2
1 cos 1 cos 1 cos
1 cos 1 cos 1 cos
(1 cos )
1 cos
(1 cos )
sin
t t t
t t t
t
t
t
t
Right side: 2
2
2
2
2
1 cos(csc cot )
sin sin
1 cos
sin
(1 cos )
sin
tt t
t t
t
t
t
t
The identity is verified because both sides are equal to 2
2
(1 cos )
sin
t
t
.
50. Left side: 2cos 4cos 4 (cos 2)(cos 2)
cos 2 cos 2cos 2
t t t t
t tt
Right side: 2sec 1 2sec 1
sec sec sec2 cos
cos 2
t t
t t tt
t
The identity is verified because both sides are equal to cos 2t .
51.
4 4 2 2 2 2
2 2
2 2
2
cos sin cos sin cos sin
cos sin 1
1 sin sin
1 2sin
t t t t t t
t t
t t
t
52.
4 4 2 2 2 2
2 2
2 2
2
sin cos sin cos sin cos
sin cos 1
1 cos cos
1 2cos
t t t t t t
t t
t t
t
Chapter 6 Analytic Trigonometry
714 Copyright © 2014 Pearson Education, Inc.
53. sin cos cos sin
sin cos
θ θ θ θθ θ
2 2
2 2
(sin cos )cos (cos sin )sin
cos sin cos sin
sin cos cos sin cos sin
sin cos
2sin cos cos sin
sin cos2sin cos 1
sin cos2sin cos 1
sin cos sin cos1 1
2sin cos
2 csc sec
2 sec csc
θ θ θ θ θ θθ θ θ θ
θ θ θ θ θ θθ θ
θ θ θ θθ θ
θ θθ θ
θ θθ θ θ θ
θ θθ θθ θ
54. sin cos
1 cot tan 1
θ θθ θ
2
sin coscos sin
1 1sin cossin cos
sin cos sin cos
sin sin cos cossin cos
sin cos sin cos
sin cossin cos sin cos
sin cossin cos
sin cossin cos
sin cos sin cos
sin
sin cos
θ θθ θθ θθ θ
θ θ θ θθ θ θ θ
θ θθ θ θ θ
θ θθ θ θ θθ θ
θ θθ θθ θ
θ θ θ θθ
θ
2
2 2
cos
sin cos
sin cos
sin cos(sin cos )(sin cos )
sin cossin cos
θθ θ θ
θ θθ θ
θ θ θ θθ θ
θ θ
Section 6.1 Verifying Trigonometric Identities
Copyright © 2014 Pearson Education, Inc. 715
55. 2 2tan 1 cos 1θ θ 2 2 2 2
22 2 2
2
2 2 2
2 2 2
2
2
tan cos tan cos 1
sincos tan cos 1
cos
sin tan cos 1
sin cos tan 1
1 tan 1
tan 2
θ θ θ θθ θ θ θθ
θ θ θ
θ θ θθ
θ
56. 2 2(cot 1)(sin 1)θ 2 2 2 2
22 2 2
2
2 2 2
2
2
cot sin cot sin 1
cossin cot sin 1
sin
cos sin cot 1
1 cot 1
cot 2
θ θ θ θθ θ θ θθθ θ θ
θθ
57. 2 2
2 2 2 2
2 2 2 2
(cos sin ) (cos sin )
cos 2cos sin sin cos 2cos sin sin
cos sin cos sin
1 1 2
θ θ θ θθ θ θ θ θ θ θ θθ θ θ θ
58. 2 2(3cos 4sin ) (4cos 3sin )θ θ θ θ 2 2
2 2
2 2 2 2
2 2 2 2
9cos 24cos sin 16sin
16cos 24cos sin 9sin
9cos 9sin 16sin 16cos
9(cos sin ) 16(sin cos )
9(1) 16(1)
25
θ θ θ θθ θ θ θθ θ θ θθ θ θ θ
59. 2 2
2
cos sin
1 tan
x x
x
2 2 2 2
2 2 2
2 2
2 2 2 2
2
2 2 22
2 2
cos sin cos sin
sin cos sin1
cos cos
cos sin cos sin
1 cos
cos sin coscos
1 cos sin
x x x x
x x x
x x
x x x x
x
x x xx
x x
Chapter 6 Analytic Trigonometry
716 Copyright © 2014 Pearson Education, Inc.
60. sin cos cos sin
sin cos
x x x x
x x
2 2
2 2
(sin cos )cos (cos sin )sin
sin cos sin cos
sin cos cos cos sin sin
sin cos
cos sin
sin cos1
sin cos1 1
sin coscsc sec
sec csc
x x x x x x
x x x x
x x x x x x
x x
x x
x x
x x
x xx x
x x
61. Conjecture: left side is equal to cos x 2 2(sec tan )(sec tan ) sec tan
sec sec1
seccos
x x x x x x
x x
xx
62. Conjecture: left side is equal to sin x
2 2 22
2 2 2 2
2 2
2 2
1 1sec csc cos sinsincos
1 1sec csc cos sincos sin
sin
sin cossin
1sin
x x x xxxx x x x
x xx
x xx
x
63. Conjecture: left side is equal to 2sin x
cossin
cos cot sin cos cot sin
cot cot cotcos cot sin
cot
cos sinsin
cossin sin
2sin
xx
x x x x x x
x x xx x x
x
x xx
xx x
x
Section 6.1 Verifying Trigonometric Identities
Copyright © 2014 Pearson Education, Inc. 717
64. Conjecture: left side is equal to cos 1x cos tan tan 2cos 2 cos tan 2cos tan 2
tan 2 tan 2cos tan 2cos tan 2
tan 2 tan 2cos (tan 2)
1tan 2
cos 1
x x x x x x x x
x xx x x x
x xx x
xx
65. Conjecture: left side is equal to 2sec x 2 2 2 21 1 sec tan sec tan
sec tan sec tan sec tan sec tan(sec tan )(sec tan ) (sec tan )(sec tan )
sec tan sec tansec tan sec tan
2sec
x x x x
x x x x x x x xx x x x x x x x
x x x xx x x x
x
66. Conjecture: left side is equal to 2csc x
2 2
2 2
1 cos sin 1 cos 1 cos sin sin
sin 1 cos sin 1 cos 1 cos sin
1 2cos cos sin
sin 1 cos sin 1 cos
1 2cos cos sin
sin 1 cos
1 2cos 1
sin 1 cos
2 2cos
sin 1 cos
2(1 cos )
sin 1 cos
2
sin
x x x x x x
x x x x x x
x x x
x x x x
x x x
x x
x
x x
x
x x
x
x x
x
2csc x
Chapter 6 Analytic Trigonometry
718 Copyright © 2014 Pearson Education, Inc.
67.
sin costan cot cos sin
1cscsin
x xx x x x
xx
2
2
2 2
sin cos sin
cos sin 1
sin sin cos
cos sin
1 cos cos
cos 1
1 cos cos
cos cos1
cos
x x x
x x
x x x
x x
x x
x
x x
x x
x
68.
1 1sec csc sin coscos sin
sin1 tan sin cos1cos
x x x xx xxx x xx
2
2
sin cos sin cos
cos sinsin cos
sin coscos
sin cos
sin cos sinsin cos
sin cos sin
1
sin
x x x x
x xx x
x xx
x x
x x xx x
x x x
x
69. cos cos cos sin 1 sin
tan1 sin 1 sin cos cos 1 sin
x x x x xx
x x x x x
2 2
2 2
2 2
cos sin sin
1 sin cos 1 sin cos
cos sin sin
1 sin cos
sin cos sin
1 sin cos
sin 1
1 sin cos
1
cos
x x x
x x x x
x x x
x x
x x x
x x
x
x x
x
Section 6.1 Verifying Trigonometric Identities
Copyright © 2014 Pearson Education, Inc. 719
70. 1 1 sin
cotsin cos sin cos cos
xx
x x x x x
2
2
1 cos cos
sin cos sin cos
1 cos
sin cos
sin
sin cossin
costan
1
cot
x x
x x x x
x
x x
x
x xx
xx
x
71. 1 cos 1 1 cos cos 1 cos
1 cos 1 cos 1 cos 1 cos 1 cos 1 cos
x x x x
x x x x x x
2
2 2
2
2
2
2
2
2 2
2 2
2 2
2
1 cos cos cos
1 cos 1 cos
1 cos cos cos
1 cos
1 cos
sin
1 cos
sin sin
csc cot
csc csc 1
2csc 1
x x x
x x
x x x
x
x
x
x
x x
x x
x x
x
72. (sec csc )(sin cos ) 2 cot sec sin sec cos csc sin csc cos 2 cotx x x x x x x x x x x x x x
sec sin sec cos csc sin csc cos 2 cot
tan 1 1 cot 2 cot
tan
x x x x x x x x x
x x x
x
Chapter 6 Analytic Trigonometry
720 Copyright © 2014 Pearson Education, Inc.
73. 1 1
1csc sin sinsin
x x xx
2
2
2
2
11
sinsin
1
1 sin
sin sin1
1 sin
sin1
cos
sinsin
cos1 sin
cos cossec tan
xx
x
x x
x
x
x
xx
xx
x xx x
74. 1 sin 1 sin (1 sin )(1 sin ) (1 sin )(1 sin )
1 sin 1 sin (1 sin )(1 sin ) (1 sin )(1 sin )
x x x x x x
x x x x x x
2 2
2 2
2
2
(1 sin )(1 sin ) (1 sin )(1 sin )
(1 sin )(1 sin ) (1 sin )(1 sin )
1 2sin sin 1 2sin sin
1 sin 1 sin4sin
1 sin4sin
cos4 sin
cos cos4sec tan
x x x x
x x x x
x x x x
x xx
xx
xx
x xx x
75. – 78. Answers may vary.
Section 6.1 Verifying Trigonometric Identities
Copyright © 2014 Pearson Education, Inc. 721
79.
1
sec (sin cos ) 1 (sin cos ) 1cossin cos
1cos costan 1 1
tan
x x x x xxx x
x xx
x
80.
sin( )cos tan( ) cos
cos( )
sincos sin
cos
xx x x
x
xx x
x
81.
The graphs do not coincide. Values for x may vary.
82.
The graphs do not coincide. Values for x may vary.
83.
The graphs do not coincide. Values for x may vary.
84.
The graphs do not coincide. Values for x may vary.
85.
2 2
sin sin 1csc
sin1 cos sin
x xx
xx x
86.
2 2sin sin cos sin (1 cos )
2 3sin sin sin
x x x x x
x x x
87.
The graphs do not coincide. Values for x may vary.
88. makes sense
89. makes sense
90. makes sense
91. does not make sense; Explanations will vary. Sample explanation: The most efficient way to simplify the identity is to multiply out the numerator and then use a Pythagorean identity.
Chapter 6 Analytic Trigonometry
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92. 3 3sin cossin cos
2 2(sin cos )(sin sin cos cos )
sin cos2 2sin cos sin cos
1 sin cos
x x
x x
x x x x x x
x x
x x x x
x x
93. sin cos 1sin cos 1
sin cos 1 sin cos 1
sin cos 1 sin cos 12 2sin 2cos sin cos 1
2 2sin 2sin cos 12 2sin cos 2cos sin 1
2 2sin 2sin (1 sin ) 1
1 2cos sin 12 2sin 2sin sin2cos sin
x x
x xx x x x
x x x x
x x x x
x x x
x x x x
x x x
x x
x x xx x
22sin 2sin
2sin cos
2sin (sin 1)
cos
sin coscos sin 1
sin 1 sin 1cos (sin 1)
2sin 1cos (sin 1)2 2cos 1 cos
cos (sin 1)2cos
sin 1
cos
x xx x
x x
x
x xx x
x xx x
xx x
x xx x
xx
x
94.
-
11ln | cos | ln | cos | ln| cos |
1ln ln | sec |
cos
x xx
xx
95. 2 2tan sec
2 2
2 2
2 2
ln
tan sec
( tan sec )
(sec tan )
1
x xe
x x
x x
x x
96. Answers may vary.
97. Answers may vary.
98. 3
cos302
1sin 30
21
cos602
3sin 60
2
cos90 0
sin 90 1
99. a. No, they are not equal.
cos(30 60 ) cos30 cos60
3 1cos90
2 2
1 30
2
b. Yes, they are equal.
cos(30 60 ) cos30 cos60 sin 30 sin 60
3 1 1 3cos90
2 2 2 2
3 30
4 40 0
100. a. No, they are not equal.
sin(30 60 ) sin 30 sin 60
1 3sin 90
2 2
1 31
2
Section 6.2 Sum and Difference Formulas
Copyright © 2014 Pearson Education, Inc. 723
b. Yes, they are equal.
sin(30 60 ) sin 30 cos60 cos30 sin 60
1 1 3 3sin 90
2 2 2 2
1 31
4 41 1
Section 6.2
Check Point Exercises
1. cos30 cos(90 60 )
cos90 cos60 sin 90 sin 60
1 30 1
2 2
30
2
3
2
2. cos70 cos40 sin 70 sin 40
cos(70 40 )
cos30
3
2
3. cos( ) cos cos sin sincos cos cos cos
cos cos sin sin
cos cos cos cos
1 1 tan tan
1 tan tan
α β α β α βα β α β
α β α βα β α β
α βα β
We worked with the left side and arrived at the right side. Thus, the identity is verified.
4. 5sin sin12 6 4
sin cos cos sin6 4 6 4
1 2 3 2
2 2 2 2
2 6
4 4
2 6
4
π π π
π π π π
5. a. 4sin5
y
rα
Find x: 2 2 2
2 2 24 5
2 16 25
2 9
x y r
x
x
x
Because α is in Quadrant II, x is negative.
9 3x 3 3
cos5 5
x
rα
b. 1sin2
y
rβ
Find x: 2 2 2
2 2 21 2
2 1 4
2 3
x y r
x
x
x
Because β is in Quadrant I, x is positive.
3x
3cos
2
x
rβ
c. cos( ) cos cos sin sin
3 3 4 1
5 2 5 2
3 3 4
10 10
3 3 4
10
α β α β α β
d. sin( ) sin cos cos sin
4 3 3 1
5 2 5 2
4 3 3
10 10
4 3 3
10
α β α β α β
6. a. The graph appears to be the sine curve, sin .y x
It cycles through intercept, maximum, intercept, minimum and back to intercept. Thus, siny x also describes the graph.
Chapter 6 Analytic Trigonometry
724 Copyright © 2014 Pearson Education, Inc.
b. 3 3 3cos cos cos sin sin2 2 2
x x xπ π π
cos 0 sin ( 1)
sin
x x
x
This verifies our observation that
3cos
2y x
π and siny x describe the
same graph.
7. tan tan
tan( )1 tan tan
tan 0
1 tan 0tan
1tan
xx
xx
xx
x
πππ
Concept and Vocabulary Check 6.2
1. cos cos sin sinx y x y
2. cos cos sin sinx y x y
3. sin cos cos sinC D C D
4. sin cos cos sinC D C D
5. tan tan
1 tan tan
θ φθ φ
6. tan tan
1 tan tan
θ φθ φ
7. false
8. false
Exercise Set 6.2
1. cos(45 30 ) cos45 cos30 sin 45 sin 30
2 3 2 1
2 2 2 2
6 2
4 4
6 2
4
2. o ocos(120 45 )o o o ocos120 cos45 sin120 sin 45
1 2 3 2
2 2 2 2
2 6
4 4
2 6
4
3. 3 3 3
cos cos cos sin sin4 6 4 6 4 6
2 3 2 1
2 2 2 2
6 2
4 4
2 6
4
π π π π π π
4. 2 2 2
cos cos cos sin sin3 6 3 6 3 6
1 3 3 1
2 2 2 2
3 3
4 40
π π π π π π
5. a. cos50 cos20 sin50 sin 20
cos cos sin sin
Thus, 50 and 20 .
α β α βα β
b. cos50 cos20 sin50 sin 20
cos(50 20 )
cos30
c. 3
cos302
6. a. o o o ocos50 cos5 sin50 sin5
cos cos sin sinα β α β
Thus, o o50 and 5α β
b. o o o ocos50 cos5 sin50 sin5 o ocos(50 5 )
ocos 45
Section 6.2 Sum and Difference Formulas
Copyright © 2014 Pearson Education, Inc. 725
c. 2ocos45
2
7. a. 5 5
cos cos sin sin12 12 12 12
cos cos sin sin
5Thus, and .
12 12
π π π π
α β α βπ πα β
b. 5 5
cos cos sin sin12 12 12 12
5cos
12 12
4cos
12
cos3
π π π π
π π
π
π
c. 1
cos3 2
π
8. a. 5 5
cos cos sin sin18 9 18 9
π π π π
cos cos sin sin
5 and
18
α β α βπ πα β
b. 5 5
cos cos sin sin18 9 18 9
π π π π
5cos
18 9
3cos
18
cos6
π π
π
π
c. 3
cos6 2
π
9. cos( ) cos cos sin sin
cos sin cos sin
cos cos sin sin
cos sin cos sin
1 cot tan 1
tan cot
α β α β α βα β α β
α β α βα β α β
β αα β
10. cos( ) cos cos sin sin
sin sin sin sin
cos cos sin sin
sin sin sin sin
cos cos1
sin sin
cot cot 1
α β α β α βα β α β
α β α βα β α βα βα βα β
11. cos cos cos sin sin4 4 4
2 2cos sin
2 2
2(cos sin )
2
x x x
x x
x x
π π π
12. 5 5 5
cos cos cos sin sin4 4 4
2 2cos sin
2 2
2(cos sin )
2
x x x
x x
x x
π π π
13. sin(45 30 ) sin 45 cos30 cos 45 sin 30
2 3 2 1
2 2 2 2
6 2
4 4
6 2
4
14. o osin(60 45 )o o o osin 60 cos 45 cos60 sin 45
3 2 1 2
2 2 2 2
6 2
4 4
6 2
4
Chapter 6 Analytic Trigonometry
726 Copyright © 2014 Pearson Education, Inc.
15. sin(105 ) sin(60 45 )
sin 60 cos 45 cos60 sin 45
3 2 1 2
2 2 2 2
6 2
4 4
6 2
4
16. o o osin 75 sin(30 45 )o o o osin 30 cos 45 cos30 sin 45
1 2 3 2
2 2 2 2
2 6
4 4
2 6
4
17. cos(135 30 ) cos135 cos30 sin135 sin 30
cos 90 45 cos30 sin 90 45 sin 30
cos90 cos 45 sin 90 sin 45 cos30 sin 90 cos45 cos90 sin 45 sin 30
2 2 3 2 2 10 1 1 0
2 2 2 2 2 2
2 3 2 1
2 2 2 2
6 2
4 4
6 2
4
18. cos(240 45 ) cos240 cos 45 sin 240 sin 45
cos 180 60 cos 45 sin 180 60 sin 45
cos180 cos60 sin180 sin 60 cos45 sin180 cos60 cos180 sin 60 sin 45
1 3 2 1 3 21 0 0 ( 1)
2 2 2 2 2 2
1 2 3 2
2 2 2 2
2 6
4 4
6 2
4
Section 6.2 Sum and Difference Formulas
Copyright © 2014 Pearson Education, Inc. 727
19. cos75 cos(45 30 )
cos45 cos30 sin 45 sin 30
2 3 2 1
2 2 2 2
6 2
4 4
6 2
4
20. cos105 cos(45 60 )
cos 45 cos60 sin 45 sin 60
2 1 2 3
2 2 2 2
2 6
4 4
2 6
4
21. tan tan
6 4tan6 4 1 tan tan
6 4
π ππ π
π π
31
33
1 13
3 3
3 33 3
3 3
3 3
3 3
3 3 3 3
3 3 3 3
9 6 3 3
9 3
12 6 3
6
2 3
22. tan tan
3 4tan3 4 1 tan tan
3 4
π ππ π
π π
3 1
1 3 1
1 3
1 3
1 3 1 3
1 3 1 3
1 2 3 3
1 3
4 2 3
2
2 3
23.
4tan tan4 3 4tan
43 4 1 tan tan3 4
π ππ π
π π
3 1
1 3 1
1 3
1 3
1 3 1 3
1 3 1 3
1 2 3 3
1 3
4 2 3
2
2 3
Chapter 6 Analytic Trigonometry
728 Copyright © 2014 Pearson Education, Inc.
24.
5tan tan5 3 4tan
53 4 1 tan tan3 4
π ππ π
π π
5tan tan
3 45
1 tan tan3 4
3 1
1 3 1
1 3
1 3
1 3 1 3
1 3 1 3
1 2 3 3
1 3
4 2 3
2
2 3
π π
π π
25. sin 25 cos5 cos25 sin5 sin(25 5 )
sin 30
1
2
26. o o o osin 40 cos 20 cos 40 sin 20 o osin(40 20 )osin 60
3
2
27. tan10 tan 35
tan(10 35 )1 tan10 tan 35
tan 45
1
28. o otan50 tan 20 o otan(50 20 )o o1 tan50 tan 20
otan 30
3
3
29. 5 5 5
sin cos cos sin sin12 4 12 4 12 4
2sin
12
sin6
1
2
π π π π π π
π
π
30. 7 7 7
sin cos cos sin sin12 12 12 12 12 12
6sin
12
sin2
1
π π π π π π
π
π
31. 5 30
5 30
tan tantan
5 301 tan tan
5tan tan
30 6
3
3
π π
π ππ π
π π
32.
4tan tan 45 5 tan
4 5 51 tan tan5 5
5tan
5tan
0
π ππ π
π π
π
π
33. sin sin cos cos sin2 2 2
sin 0 cos 1
cos
x x x
x x
x
π π π
34. 3 3 3
sin sin cos cos sin2 2 2
sin 0 cos ( 1)
cos
x x x
x x
x
π π π
35. cos cos cos sin sin2 2 2
cos 0 sin 1
sin
x x x
x x
x
π π π
Section 6.2 Sum and Difference Formulas
Copyright © 2014 Pearson Education, Inc. 729
36. cos( ) cos cos sin sin
1 cos 0 sin
cos
x x x
x x
x
π π π
37. tan 2 tan
tan(2 )1 tan 2 tan
0 tan
1 0 tantan
xx
xx
xx
πππ
38. tan tan
tan( )1 tan tan
0 tan
1 0 tantan
xx
xx
xx
πππ
39. sin( ) sin( )
sin cos cos sin
sin cos cos sin
2sin cos
α β α βα β α βα β α β
α β
40. cos( ) cos( )
cos cos sin sin
cos cos sin sin
2cos cos
α β α βα β α βα β α β
α β
41. sin( ) sin cos cos sin
cos cos cos cos
sin cos cos sin
cos cos cos cos
tan 1 1 tan
tan tan
α β α β α βα β α β
α β α βα β α βα βα β
42. sin( ) sin cos cos sincos cos cos cos
sin cos cos sin
cos cos cos cos
tan tan
α β α β α βα β α β
α β α βα β α βα β
43. 4
4
sin coscos coscos sincos cos
sin coscos
cos sincos
tan tantan
4 1 tan tan
tan 1
1 tan
sin cos cos
cos cos sinsin cos
cos sincos sin
cos sin
π
π
θ θθ θθ θθ θθ θ
θθ θ
θ
θπθθ
θθ
θ θ θθ θ θ
θ θθ θθ θθ θ
44. tan tan4tan
4 1 tan tan4
1 tan
1 1 tan1 tan
1 tancos sin
cos coscos sin
cos coscos sin
coscos sin
coscos sin cos sin
cos coscos sin cos
cos cos sincos sin
cos sin
π θπ θ π θ
θθ
θθ
θ θθ θθ θθ θθ θ
θθ θ
θθ θ θ θ
θ θθ θ θ
θ θ θθ θθ θ
Chapter 6 Analytic Trigonometry
730 Copyright © 2014 Pearson Education, Inc.
45.
2 2 2 2
2 2 2 2
2 2 2
2 2 2
2 2
cos( ) cos( )
(cos cos sin sin )
(cos cos sin sin )
cos cos sin sin
1 sin cos sin 1 cos
cos sin cos
sin sin cos
cos sin
α β α βα β α β
α β α βα β α β
α β α β
β α βα α ββ α
46. sin sin(sin cos cos sin )
(sin cos cos sin )
2 2 2 2sin cos cos sin
2 2(1 cos )cos
2 2cos (1 cos )
2 2 2cos cos cos
2 2 2cos cos cos
2 2cos cos
a β α βα β α β
α β α β
α β α β
α β
α β
β α β
α α β
β α
47.
1cos cos
1cos cos
sin cos cos sincos cos
sin cos cos sincos cos
sin cos cos sincos cos cos cos
sin cosco
sin( )
sin( )
sin cos cos sin
sin cos cos sin
sin cos cos sin
sin cos cos sinα β
α βα β α β
α βα β α β
α βα β α βα β α βα β
α βα β
α β α βα β α β
α β α βα β α β
cos sins cos cos cos
tan 1 1 tan
tan 1 1 tan
tan tan
tan tan
α βα β α β
α βα βα βα β
48.
cos( )
cos
cos cos sin sin
cos cos sin sin
1
cos cos sin sin cos cos1cos cos sin sin
cos cos
cos cos sin sin
cos cos cos coscos cos sin sin
cos cos cos cos
1 tan tan
1 tan tan
α βα βα β α βα β α β
α β α β α βα β α β
α βα β α βα β α βα β α βα β α β
α βα β
49. cos( ) cos
cos cos sin sin cos
cos cos cos sin sin
cos (cos 1) sin sin
cos 1 sincos sin
x h x
hx h x h x
hx h x x h
hx h x h
hh h
x xh h
50. sin( ) sin
sin cosh cos sinh sin
sin cosh sin cos sinh
sin (cosh 1) cos sinh
cosh 1 sinhsin cos
x h x
hx x x
hx x x
hx x
h
x xh h
51. sin 2 sin( )
sin cos cos sin
2sin cos
α α αα α α α
α α
52. cos2 cos( )α α α
cos cos sin sin
2 2cos sin
α α α α
α α
Section 6.2 Sum and Difference Formulas
Copyright © 2014 Pearson Education, Inc. 731
53.
2
tan 2 tan( )
tan tan
1 tan tan2 tan
1 tan
α α αα α
α αα
α
54. tan tan4 4
tan tan tan tan4 4
1 tan tan 1 tan tan4 4
1 tan 1 tan
1 tan 1 tan(1 tan )(1 tan ) (1 tan )(1 tan )
(1 tan )(1 tan ) (1 tan )(1 tan )2 21 2 tan tan (1 2 tan tan )
21 tan
π πα α
π πα α
π πα α
α αα αα α α αα α α α
α α α αα
4 tan21 tan
2(2 tan )21 tan
tan tan2
1 tan tan2 tan( )
2 tan 2
αα
αα
α αα α
α αα
55.
1cos cos
1cos cos
sin cos cos sincos cos
cos cos sin sincos cos
sin cos cos sincos cos cos cos
c
sin( )tan( )
cos( )
sin cos cos sin
cos cos sin sin
sin cos cos sin
cos cos sin sinα β
α βα β α β
α βα β α β
α βα β α βα β α β
α βα βα β
α β α βα β α β
α β α βα β α β
os cos sin sincos cos cos cos
tan tan
1 tan tan
α β α βα β α β
α βα β
56. tan( ) tan( ( ))
tan tan( )
1 tan tan( )
tan tan
1 tan tan
α β α βα β
α βα β
α β
57. 3
sin5
y
rα
2 2 2
2 2 2
2
2
3 5
9 25
16
x y r
x
x
x
Because α lies in quadrant I, x is positive. x = 4
3545
4Thus, cos , and
5
sin 3tan .
cos 4
x
rα
ααα
5sin
13
y
rβ
2 2 2
2 2 2
2
2
5 13
25 169
144
x y r
x
x
x
Because β lies in quadrant II, x is negative. x = –12
5131213
12 12Thus, cos , and
13 13
sin 5tan .
cos 12
x
rβ
βββ
a. cos( ) cos cos sin sin
4 12 3 5 63
5 13 5 13 65
α β α β α β
b. sin( ) sin cos cos sin
3 12 4 5 16
5 13 5 13 65
α β α β α β
c.
3 5 44 12 12
633 5484 12
tan tantan( )
1 tan tan
16
631
α βα βα β
Chapter 6 Analytic Trigonometry
732 Copyright © 2014 Pearson Education, Inc.
58. 4
sin5
y
rα
2 2 2
2 2 24 5
2 16 25
2 9
x y r
x
x
x
Because α lies in quadrant I, x is positive. x = 3
3cos
54
sin 45tan3cos 35
x
rα
ααα
7sin
25
y
rβ
2 2 2
2 2 27 25
2 49 625
2 576
x y r
x
x
x
Because β lies in quadrant II, x is negative. x = –24
24cos
257
sin 725tan24cos 24
25
x
rβ
βββ
a. cos( ) cos cos sin sinα β α β α β
3 24 4 7 4
5 25 5 25 5
b. sin sin cos cos sinα β α β α β
4 24 3 4 3
5 25 5 25 5
c. tan tan
tan( )1 tan tan
α βα βα β
4 7 2533 24 24
254 7 41
183 24
59. 3 3
tan4 4
y
xα
2 2 2
2 2 2
2
2
( 4) 3
16 9
25
x y r
r
r
r
Because r is a distance, it is positive. r = 5
4 4Thus, cos , and
5 53
sin .5
x
ry
r
α
α
1cos
3
x
rβ
2 2 2
2 2 2
2
2
1 3
1 9
8
x y r
y
y
y
Because β lies in quadrant I, y is positive. 8 2 2y
2 2313
2 2Thus, sin , and
3
sintan 2 2.
cos
y
rβ
βββ
a. cos( ) cos cos sin sin
4 1 3 2 2
5 3 5 3
4 6 2
15 15
4 6 2
15
4 6 2
15
α β α β α β
b. sin( ) sin cos cos sin
3 1 4 2 2
5 3 5 3
3 8 2
15 15
3 8 2
15
α β α β α β
Section 6.2 Sum and Difference Formulas
Copyright © 2014 Pearson Education, Inc. 733
c.
34
34
3 8 24
4 6 24
tan tantan( )
1 tan tan
2 2
1 2 2
4 6 23 8 2
4 6 2 4 6 2
108 50 2
56
54 25 2
28
α βα βα β
60. 4 4
tan3 3
y
xα
2 2 2
2 2 2( 3) 4
29 16
225
x y r
r
r
r
Because r is a distance, it is positive. r = 5
3cos
54
sin5
x
ry
r
α
α
2cos
3
x
rβ
2 2 2
2 2 22 3
24 9
2 5
x y r
y
y
y
Because β lies in quadrant I, y is positive. 5
sin3
5sin 53tan
2cos 23
y
rβ
βββ
5y
a. cos( ) cos cos sin sinα β α β α β
3 2 4 5 6 4 5
5 3 5 3 15
b. sin sin cos cos sinα β α β α β 4 2 3 5 8 3 5
5 3 5 3 15
c. tan tantan( )
1 tan tan
4 5
3 24 5
13 2
8 3 5
66 4 5
6
8 3 5
6 4 5
8 3 5 6 4 5
6 4 5 6 4 5
108 50 5
44
54 25 5
22
α βα βα β
61. 8
cos17
x
rα
2 2 2
2 2 2
2
2
8 17
64 289
225
x y r
y
y
y
Because α lies in quadrant IV, y is negative. y = –15
15178
17
15 15Thus, sin , and
17 17
sin 15tan .
cos 8
y
rα
ααα
1 1sin
2 2
y
rβ
Chapter 6 Analytic Trigonometry
734 Copyright © 2014 Pearson Education, Inc.
2 2 2
2 2 2
2
2
( 1) 2
1 4
3
x y r
x
x
x
Because β lies in quadrant III, x is negative. 3x
12
32
3 3Thus, cos , and
2 2
sin 1 3tan .
cos 33
x
rβ
βββ
a. cos( ) cos cos sin sin
8 3 15 1
17 2 17 2
8 3 15
34
8 3 15
34
α β α β α β
b. sin( ) sin cos cos sin
15 3 8 1
17 2 17 2
15 3 8
34
α β α β α β
c. tan tantan1 tan tan
15 3
8 3
15 31
8 3
45 8 3
2424 15 3
24
45 8 3 24 15 3
24 15 3 24 15 3
1440 867 3
99
489 289 3
33
α βα βα β
62. 1
cos2
x
rα
2 2 2
2 2 21 2
21 4
2 3
x y r
y
y
y
Because α lies in quadrant IV, y is negative. 3y
3sin
2
3sin 2tan 3
1cos2
y
rα
ααα
1 1
sin3 3
y
rβ
2 2 2
22 21 3
2 1 9
2 8
x y r
x
x
x
Because β lies in quadrant III, x is negative. 8 2 2x
2 2cos
31
sin 1 23tancos 42 2 2 2
3
x
rβ
βββ
a. cos( ) cos cos sin sinα β α β α β
1 2 2 3 1
2 3 2 3
2 6 3
6
b. sin sin cos cos sinα β α β α β
3 2 2 1 1
2 3 2 3
2 6 1
6
Section 6.2 Sum and Difference Formulas
Copyright © 2014 Pearson Education, Inc. 735
c.
tan tantan( )
1 tan tan
23
42
1 34
4 3 2
44 6
4
4 3 2
4 6
4 3 2 4 6
4 6 4 6
16 3 4 18 4 2 12
10
16 3 12 2 4 2 2 3
10
18 3 16 2
10
8 2 9 3
5
α βα βα β
63. 3
tan4
y
xα
Because α lies in quadrant III, x and y are negative.
2 2 2
2 22
2
4 3
25
5
r x y
r
r
r
3 3sin
5 54 4
cos5 5
y
rx
r
α
α
1
cos4
x
rβ
Because β lies in quadrant IV, y is negative.
2 2 2
2 2 2
2
1 4
15
15
x y r
y
y
y
15 15sin
4 4
15tan 15
1
y
r
y
x
β
β
a. cos( ) cos cos sin sinα β α β α β
4 1 3 15
5 4 5 4
4 3 15
20 20
4 3 15
20
b. sin sin cos cos sinα β α β α β
3 1 4 15
5 4 5 4
3 4 15
20 20
3 4 15
20
c. tan tan
tan( )1 tan tan
α βα βα β
315
43
1 154
3 4 15
4 44 3 15
4 4
3 4 15
4 3 15
3 4 15 4 3 15
4 3 15 4 3 15
12 9 15 16 15 180
16 135
192 25 15
119
192 25 15
119
Chapter 6 Analytic Trigonometry
736 Copyright © 2014 Pearson Education, Inc.
64. 5
sin6
y
rα
Because α lies in quadrant II, x is negative. 2 2 2
2 2 2
2
5 6
11
11
x y r
x
x
x
11 11cos
6 6
5 5 11tan
1111
x
r
y
x
α
α
3
tan7
y
xβ
Because β lies in quadrant III, x and y are both negative.
2 2 2
2 2 2
2
7 3
58
58
r x y
r
r
r
3 3 58sin
5858
7 7 58cos
5858
y
r
x
r
β
β
a. cos( ) cos cos sin sinα β α β α β
11 7 58 5 3 58
6 58 6 58
7 638 15 58
348 348
7 638 15 58
348
b. sin sin cos cos sinα β α β α β
sin cos cos sin
5 7 58 11 3 58
6 58 6 58
35 58 3 638
348 348
3 638 35 58
348
α β α β
c. tan tan
tan( )1 tan tan
α βα βα β
5 11 3
11 7
5 11 31
11 7
35 11 33
77 7777 15 11
77 77
35 11 33
77 15 11
33 35 11 77 15 11
77 15 11 77 15 11
2541 495 11 2695 11 5775
5929 2475
8316 3190 11
3454
22 378 145 11
22 157
378 145 11
157
65. a. The graph appears to be the sine curve,
y = sin x. It cycles through intercept, maximum, minimum and back to intercept. Thus, y = sin x also describes the graph.
b. sin( ) sin cos cos sin
0 cos ( 1) sin
sin
x x x
x x
x
π π π
This verifies our observation that y = sin (π - x) and y = sin x describe the same graph.
66. a. The graph appears to be the cosine curve, y = cos x. It cycles through maximum, intercept, minimum, intercept and back to maximum. Thus, y = cos x also describes the graph.
b. cos( 2 ) cos cos2 sin sin 2x x xπ π π
cos 1 sin 0
cos
x x
x
This verifies our observation that cos( 2 )y x π and y = cos x describe the
same graph.
Section 6.2 Sum and Difference Formulas
Copyright © 2014 Pearson Education, Inc. 737
67. a. The graph appears to be 2 times the cosine curve, y = 2 cos x. It cycles through maximum, intercept, minimum, intercept and back to maximum. Thus y = 2 cos x also describes the graph.
b. sin sin2 2
sin cos cos sin sin cos2 2 2
cos sin2
sin 0 cos 1 1 cos 0 sin
cos cos
2cos
x x
x x x
x
x x x x
x x
x
π π
π π π
π
This verifies our observation that
sin sin2 2
y x xπ π and y = 2cos x
describe the same graph.
68. a. The graph appears to be 2 times the sine curve, y = 2 sin x. It cycles through intercept, maximum, intercept, minimum and back to intercept. Thus, y = 2 sin x also describes the graph.
b. cos cos2 2
x xπ π
cos cos sin sin2 2
cos cos sin sin2 2
2sin sin2
2sin 1
2sin
x x
x x x
x
x
x
π π
π π
π
This verifies our observation that
cos cos2 2
x xπ π and y = 2 sin x
describe the same graph.
69.
cos cos sin sin
cos
cos
α β β α β β
α β βα
70.
sin cos cos sin
sin
sin
α β β α β β
α β βα
71.
sin sin
cos cos
α β α βα β α β
sin cos cos sin sin cos cos sin
cos cos sin sin cos cos sin sin
sin cos cos sin sin cos cos sin
cos cos sin sin cos cos sin sin
cos sin cos sin
cos cos cos cos
2cos sin
2cos cos
sin
α β α β α β α βα β α β α β α β
α β α β α β α βα β α β α β α βα β α βα β α β
α βα β
cos
tan
βββ
72.
cos cos
sin sin
α β α βα β α β
cos cos sin sin cos cos sin sin
sin cos cos sin sin cos cos sin
cos cos sin sin cos cos sin sin
sin cos cos sin sin cos cos sin
cos cos cos cos
cos sin cos sin
2cos cos
2cos sin
c
α β α β α β α βα β α β α β α β
α β α β α β α βα β α β α β α β
α β α βα β α βα βα β
ossin
cot
βββ
73. cos cos sin sin6 6 6 6
π π π πα α α α
cos6 6
cos6 6
cos3
1
2
π πα α
π πα α
π
Chapter 6 Analytic Trigonometry
738 Copyright © 2014 Pearson Education, Inc.
74. sin cos cos sin3 3 3 3
π π π πα α α α
sin3 3
sin3 3
2sin
3
3
2
π πα α
π πα α
π
75. Conjecture: the left side is equal to cos3x . cos2 cos5 sin 2 sin5
cos(2 5 )
cos( 3 )
cos3
x x x x
x x
x
x
76. Conjecture: the left side is equal to sin 3x . sin5 cos 2 cos5 sin 2
sin(5 2 )
sin 3
x x x x
x x
x
77. Conjecture: the left side is equal to sin2
x.
5 5sin cos2 cos sin 2
2 25
sin 22
5 4sin
2 2
sin2
x xx x
xx
x x
x
78. Conjecture: the left side is equal to cos2
x.
5 5cos cos 2 sin sin 2
2 25
cos 22
5 4cos
2 2
cos2
x xx x
xx
x x
x
79. 3
tan2
y
xθ
2 2 2
2 3 2
2
2
2 3
4 9
13
x y r
r
r
r
Because r is a distance, it is positive.
13r 3
Thus, sin132
and cos13
y
r
x
r
θ
θ
13 cos( ) 13(cos cos sin sin )
2 313 cos sin
13 13
cos 2 sin 3
2cos 3sin
t t t
t t
t t
t t
θ θ θ
For the equation 13 cos( ),y t θ the amplitude is
13 13 , and the period is 2 2 .1
π π
80. a. 3sin 2 2sin(2 )
3sin 2 2(sin 2 cos cos 2 cos )
3sin 2 2(sin 2 ( 1) cos 2 0)
3sin 2 2sin 2
sin 2
t t
t t t
t t t
t t
t
ρ ππ π
b. No. The amplitude of p is 1.
81. – 87. Answers may vary.
88.
–2
2
3–
232
3 3 3cos cos cos sin sin
2 2 2
0 cos 1 sin
sin
x x x
x x
x
π π π
Section 6.2 Sum and Difference Formulas
Copyright © 2014 Pearson Education, Inc. 739
89.
–2
2
3–
232
tan tan
tan( )1 tan tan
0 tan
1 0 tantan
1tan
xx
xx
xx
x
πππ
90.
–2
2
3–
232
The graphs do not coincide. Values for x may vary.
91.
–2
2
3–
232
The graphs do not coincide. Values for x may vary.
92.
–2
2
3–
232
cos1.2 cos0.8 sin1.2 sin 0.8
cos(1.2 0.8) cos 2
x x x x
x
93.
–2
2
3–
232
sin1.2 cos0.8 cos1.2 sin 0.8
sin(1.2 0.8 )
sin 2
x x x x
x x
x
Chapter 6 Analytic Trigonometry
740 Copyright © 2014 Pearson Education, Inc.
94. makes sense
95. makes sense
96. does not make sense; Explanations will vary. Sample explanation: The sum and difference formulas allow you to find exact values only for certain angles.
97. makes sense
98. sin( ) sin( ) sin( )cos cos cos cos cos cos
sin cos cos sin sin cos cos sin sin cos cos sin
cos cos cos cos cos cos
sin cos cos sin sin cos cos sin sin
cos cos cos cos cos cos cos cos
x y y z z x
x y y z z x
x y x y y z y z z x z x
x y y z z x
x y x y y z y z
x y y y y z y z
cos cos sincos cos cos cos
sin sin sin sin sin sin
cos cos cos cos cos cos
0
z x z x
z x z x
x y y z z x
x y y z z x
99. 11cos2
1
3
2
31sin5
4
3
5
x
y
r
x
y
r
1 31 1sin cos sin2 5
1 31 1sin cos cossin2 5
1 31 1 coscos sin sin2 5
3 4 1 3
2 5 2 5
4 3 3
10
Section 6.2 Sum and Difference Formulas
Copyright © 2014 Pearson Education, Inc. 741
100. 31sin5
3, 5, 4y r x
41cos5
4, 3, 5x y r
3 41 1sin sin cos5 5
3 4 3 41 1 1 1sin sin coscos cossin sin cos5 5 5 5
3 4 4 3
5 5 5 5
12 12
25 2524
25
101. 41tan3
3
4
5
51cos13
5
12
13
x
y
r
x
y
r
4 51 1cos tan cos3 13
4 51 1cos tan coscos3 13
4 51 1 sin tan sin cos3 13
3 5 4 12
5 13 5 1333
65
Chapter 6 Analytic Trigonometry
742 Copyright © 2014 Pearson Education, Inc.
102. 3 5 11 -1cos sin
2 6 2 6
π π
3 11 1cos cos sin2 2
5cos
6 6
cos
1
π π
π
103. Let 1sin , where - . sin2 2
x xπ πα α α
Because x is positive, sin α is positive. Thus α is in quadrant I. Using a right triangle in quadrant I with sin α = x, the third side can be found using the Pythagorean Theorem.
2 2 21
2 21
21
21 2Thuscos 11
a x
a x
a x
xxα
Because y is positive, cos β is positive. Thus β is in quadrant I. Using a right triangle in quadrant I with cos β = y, the third side can be found using the Pythagorean Theorem.
2 2 21
2 21
21
21 2Thuscos 11
b y
b x
a y
yyα
1 1cos(sin cos ) cos( )
cos cos sin sin
2 21 1
2 21 1
x y
x y x y
y x x y
α βα β α β
104. 1 1tan sinx x
1 1
2 2 1 1
y x y y
x r
r x x y
1 1sin(tan sin )
1 1sin tan cossin
1 1cos tan sin sin
21 1
12 21 1
21
2 1
x y
x y
x y
yxy
x x
x y y
x
105. 1sin 21
1
1cos
21
1
x x
y x
r
y
x y
y y
r
1 1tan(sin cos )
1 1tansin tan cos1 11 tansin tan cos
21
2121
121
2 21 1
2 21 1
x y
x y
x y
yx
yx
yx
yx
xy y x
y x x y
106. Answers may vary.
Section 6.3 Double-Angle Power Reducing, and Half-Angle Formulas
Copyright © 2014 Pearson Education, Inc. 743
107. 1
sin 302
3cos30
2
3sin 60
21
cos602
108. a. No, they are not equal.
sin(2 30 ) 2sin 30
1sin 60 2
2
31
2
b. Yes, they are equal.
sin(2 30 ) 2sin 30 cos30
1 3sin 60 2
2 2
3 3
2 2
109. a. No, they are not equal.
cos(2 30 ) 2cos30
3cos60 2
21
32
b. Yes, they are equal. 2 2
2 2
cos(2 30 ) cos 30 sin 30
3 1cos60
2 2
1 3 1
2 4 41 1
2 2
Section 6.3
Check Point Exercises
1. 4
sin5
y
rθ
Because θ lies in quadrant II, x is negative. 2 2 2
2 2 2
2 2 2
4 5
5 4 9
9 3
x y r
x
x
x
Now we use values for x, y, and r to find the required values.
a. sin 2 2sin cos
4 3 242
5 5 25
θ θ θ
b. 2 2
2 2
cos2 cos sin
3 4 9 16
5 5 25 25
7
25
θ θ θ
c.
2
4 8 83 3 3
2 16 74 9 93
2 tantan 2
1 tan
2
11
8 9 24
3 7 7
θθθ
2. The given expression is the right side of the formula for cos 2θ with 15 .θ
2 2 3cos 15 sin 15 cos(2 15 ) cos302
3.
2
2 2
2
2
3
3
sin 3 sin(2 )
sin 2 cos cos 2 sin
2sin cos cos (2cos 1)sin
2sin cos 2sin cos sin
4sin cos sin
4sin (1 sin ) sin
4sin 4sin sin
3sin 4sin
θ θ θθ θ θ θθ θ θ θ θ
θ θ θ θ θθ θ θθ θ θθ θ θθ θ
By working with the left side and expressing it in a form identical to the right side, we have verified the identity.
Chapter 6 Analytic Trigonometry
744 Copyright © 2014 Pearson Education, Inc.
4. 24 22
2
2
sin sin
1 cos 2
2
1 2cos2 cos 2
41 1 1
cos 2 cos 24 2 41 1 1 1 cos 2(2 )
cos 24 2 4 2
1 1 1 1cos 2 cos4
4 2 8 83 1 1
cos 2 cos 48 2 8
x x
x
x x
x x
xx
x x
x x
5. Because 105° lies in quadrant II, cos105 0.
32
210cos105 cos
2
1 cos 210
2
1
2
2 3
4
2 3
2
6.
2
2
2
2
sin 2 2sin cos
1 cos 2 1 1 2sin
2sin cos
2 2sin2sin cos
2 1 sin
2sin cos
2cossin
tancos
θ θ θθ θ
θ θθ
θ θθ
θ θθ
θ θθ
The right side simplifies to tan ,θ the expression on the left side. Thus, the identity is verified.
7. 1
cos1 1 1
cos sin sin
1cos
cos1cos sin cos sin
1cos
1 coscos sin
sec
sec csc csc
1 cos sin
cos 1 cossin
1 cos
tan2
α
α α α
αα
α α α α
αα
α α
αα α α
α αα α
αα
α
We worked with the right side and arrived at the left side. Thus, the identity is verified.
Concept and Vocabulary Check 6.3
1. 2sin cosx x
2. 2sin A ; 22cos A ; 22sin A
3. 2
2 tan
1 tan
B
B
4. 1 cos 2α
5. 1 cos2α
6. 1 cos 2 y
7. 1 cos x
8. 1 cos y
9. 1 cosα ; 1 cosα ; 1 cosα
10. false
11. false
12. false
13. +
14.
15. +
Section 6.3 Double-Angle Power Reducing, and Half-Angle Formulas
Copyright © 2014 Pearson Education, Inc. 745
Exercise Set 6.3
1. 3 4 24
sin 2 2sin cos 25 5 25
θ θ θ
2. 2 2cos2 cos sinθ θ θ 2 2
4 3 16 9 7
5 5 25 25 25
3.
2
3 34 2
2 93 164
327
16
2 tantan 2
1 tan
2
11
3 16 24
2 7 7
θθθ
Use this information to solve problems 4, 5, and 6.
7tan
24
y
xα
Because r is a distance it is positive. 2 2 2
2 2 2
2
2
24 7
576 49
625
25
7sin
2524
cos25
x y r
r
r
r
r
y
rx
r
α
α
4. 7 24 336
sin 2 2sin cos 225 25 625
α α α
5. 2 2
2 2
cos2 cos sin
24 7 576 49
25 25 625 625
527
625
α α α
6. 2 tantan 221 tan
7 14 14224 24 24
2 49 5277 11 576 57624
14 576 336
24 527 527
ααα
7. 15
sin17
y
rθ
Because θ lies in quadrant II, x is negative. 2 2 2
2 2 2
2 2 2
15 17
17 15 64
64 8
x y r
x
x
x
Now we use values for x, y, and r to find the required values.
a. sin 2 2sin cos
15 8 2402
17 17 289
θ θ θ
b. 2 2
2 2
cos2 cos sin
8 15 64 225
17 17 289 289
161
289
θ θ θ
c.
2
15 15 158 4 4
2 225 16115 64 648
2 tantan 2
1 tan
2
11
15 64 240
4 161 161
θθθ
8. 12
sin13
y
rθ
Because θ lies in quadrant II, x is negative.
2
2 2 2
2 2
2
12 13
25
25 5
x y r
x
x
x
Now we use values for x, y, and r to find the required values.
Chapter 6 Analytic Trigonometry
746 Copyright © 2014 Pearson Education, Inc.
a. sin 2 2sin cos
12 52
13 13
120
169
θ θ θ
b. 2 2cos2 cos sinθ θ θ
2 25 12
13 13
25 144
169 169119
169
c. 2 tantan 221 tan12 2425 5
2 14412 11 255
241205
119 1195
θθθ
9. 24
cos25
x
rθ
Because θ lies in quadrant IV, y is negative. 2 2 2
2 2 2
2 2 2
24 25
25 24 49
49 7
x y r
y
y
y
Now we use values for x, y, and r to find the required values.
a. sin 2 2sin cos
7 24 3362
25 25 625
θ θ θ
b. 2 2
2 2
cos2 cos sin
24 7
25 25
576 49 527
625 625 625
θ θ θ
c.
2
7 7 724 12 12
2 49 5277 576 57624
2 tantan 2
1 tan
2
11
7 576 336
12 527 527
θθθ
10. 40
cos41
x
rθ
Because θ lies in quadrant IV, y is negative. 2 2 2
2 2 2
2
40 41
81
81 9
x y r
y
y
y
Now we use values for x, y, and r to find the required values.
a. sin 2 2sin cosθ θ θ
9 40 720
241 41 1681
b. 2 2cos2 cos sinθ θ θ
240 9 1600 81
41 41 1681 1681
1519
1681
c. 2 tan
tan 221 tan
θθθ
9 9240 20
2 819 11 160040
99 1600 72020
1519 20 1519 15191600
Section 6.3 Double-Angle Power Reducing, and Half-Angle Formulas
Copyright © 2014 Pearson Education, Inc. 747
11. 2
cot 21
x
yθ
Because r is a distance, it is positive. 2 2 2
2 2 2
2
( 2) ( 1)
5
5
r x y
r
r
r
Now we use values for x, y, and r to find the required values.
a. sin 2 2sin cos
1 2 42
55 5
θ θ θ
b. 2 2
2 2
cos2 cos sin
2 1
5 5
4 1 3
5 5 5
θ θ θ
c.
2
12
2 1 31 4 42
2 tantan 2
1 tan
2 1 1
11
4 4(1)
3 3
θθθ
12. 3
cot 31
x
yθ
Because r is a distance, it is positive.
2 2 2
2 22
2
3 1
10
10
r x y
r
r
r
Now we use values for x, y, and r to find the required values.
a. sin 2 2sin cosθ θ θ
1 3 6 3
210 510 10
b. 2 2cos2 cos sinθ θ θ
2 23 1
10 10
9 1 8 4
10 10 10 5
c. 2 tan
tan 221 tan
θθθ
1 223 3
2 11 11 93
22 9 33
8 3 8 49
13. 9 9
sin41 41
y
rθ
Because θ lies in quadrant III, x is negative.
2 2 2
22 2
2
9 41
1600
1600
40
x y r
x
x
x
x
Now we use values for x, y, and r to find the required values.
a. sin 2 2sin cos
9 40 7202
41 41 1681
θ θ θ
Chapter 6 Analytic Trigonometry
748 Copyright © 2014 Pearson Education, Inc.
b. 2 2
2 2
cos2 cos sin
40 9
41 41
1600 81
1681 16811519
1681
θ θ θ
c.
2
9 9 940 20 20
2 81 15199 1600 160040
2 tantan 2
1 tan
2
11
9 1600 720
20 1519 1519
θθθ
14. 2 2
sin3 3
y
rθ
Because θ lies in quadrant III, x is negative.
2 2 2
22 2
2
2 3
5
5
x y r
x
x
x
Now we use values for x, y, and r to find the required values.
a. sin 2 2sin cosθ θ θ
2 5 4 5
23 3 9
b. 2 2cos2 cos sinθ θ θ
2 25 2
3 3
5 4 1
9 6 9
c. 2 tan
tan 221 tan
θθθ
2 42
5 52 4
2 11 5
5
4
4 5 55 4 51 4 15
15. The given expression is the right side of the formula for sin 2 with 15 .θ θ 2sin15 cos15 sin(2 15 )
1sin 30
2
16. The given expression is the right side of the formula for sin 2θ with θ = 22.5o.
o o o2sin 22.5 cos 22.5 sin(2 22.5 )
2osin 452
17. The given expression is the right side of the formula for cos2 with 75 .θ θ
2 2cos 75 sin 75 cos(2 75 )
3cos150
2
18. The given expression is the right side of the formula for cos 2θ with θ = 105o
2 o 2 o ocos 105 sin 105 cos(2 105 )
3ocos 2102
19. The given expression is the right side of the formula
for cos2 with .8
πθ θ
22cos 1 cos 28 8
2cos
4 2
π π
π
20. The given expression is the right side of the formula
for cos 2θ with θ = .12
π
321 2sin cos 2 cos12 12 6 2
π π π
21. The given expression is the right side of the formula
for tan 2 with .12
πθ θ
122
12
2 tan 3tan 2 tan
12 6 31 tan
π
ππ π
Section 6.3 Double-Angle Power Reducing, and Half-Angle Formulas
Copyright © 2014 Pearson Education, Inc. 749
22. The given expression is the right side of the formula
for tan 2θ with θ = .8
π
2 tan8 tan 2 tan 1
8 421 tan8
ππ π
π
23. 2 2
2 2
2 2
2
2
sincos
2 cos sincos cos
2sincos
cos sincos
2sincos
1cos
2
22 tan
1 tan
2sin cos
cos 12sin cos
sin 2
θθ
θ θθ θ
θθ
θ θθ
θθ
θ
θθ
θ θθθ θθ
24. cos22cot sin2 2 21 cot sin cos
2 2sin sin2cos
sin2 2sin cos
2sin2cos
sin12sin
22cos sin
sin 12cos sin
sin 2
θθ θ
θ θ θθ θ
θθ
θ θθ
θθ
θθ θ
θθ θ
θ
25. 2 2 2
2 2
(sin cos ) sin 2sin cos cos
sin cos 2sin cos
1 2sin cos
1 sin 2
θ θ θ θ θ θθ θ θ θ
θ θθ
26. 2(sin cos )2 2sin 2sin cos cos
2 2sin cos 2sin cos
1 2sin cos
1 sin 2
θ θ
θ θ θ θ
θ θ θ θθ θθ
27. 2 2 2 2
2
sin cos 2 sin cos sin
cos
x x x x x
x
28. 2cos 2 1 2sin
2 2cos cos2 21 sin sin
2cos2 2cos sin
2cos2 2cos sin2 2cos cos
21 tan
x x
x x
x x
x
x x
x
x x
x x
x
29. 2 2
2 2
2 2
2
sin 2 2sin cos
1 cos 2 1 cos sin
2sin cos
1 cos sin2sin cos
sin sin2sin cos
2sincos
sincot
x x x
x x x
x x
x xx x
xx x
xx
xx
30. 2 21 cos 2 1 cos sin
sin 2 2sin cos2 21 sin cos
2sin cos2 2cos cos
2sin cos22cos
2sin coscos
sincot
x x x
x x x
x x
x x
x x
x x
x
x xx
xx
Chapter 6 Analytic Trigonometry
750 Copyright © 2014 Pearson Education, Inc.
31. 22
sintan cos2 2cos 1
cos
2sin cos sin
cos cos2sin cos tan
sin 2 tan
tt t t
t
t t t
t tt t t
t t
32. cos 2cos cos 2 1 2sinsin2cos 2cos sin
sin sincot 2cos sin
2cos sin cot
sin 2 cot
tt t t
t
t t t
t tt t t
t t t
t t
33.
2 23 3
sin 4 sin(2 2 )
sin 2 cos 2 cos 2 sin 2
cos2 (sin 2 sin 2 )
cos2 2sin 2
cos sin 2 2sin cos
4sin cos 4sin cos
t t t
t t t t
t t t
t t
t t t t
t t t t
34. cos4 cos 2(2 )22cos 2 1
2 22(2cos 1) 1
4 22(4cos 4cos 1) 1
4 28cos 8cos 2 1
4 28cos 8cos 1
t t
t
t
t t
t t
t t
35. 46sin x 2
1 cos26
2
21 2cos2 cos 26
4
26 12cos 2 6cos 2
43 3 23cos2 cos 24 23 3 1 cos 4
3cos24 2 2
3 3 1 cos 43cos2
4 2 2 2
3 3 33cos2 cos 4
4 4 49 3
3cos2 cos 44 4
x
x x
x x
x x
xx
xx
x x
x x
36.
1 cos 2 1 cos22 210cos cos 102 2
210(1 2cos 2 cos 2 )
4210 20cos 2 10cos 2
4210 20cos2 10cos 2
4 4 45 5
5cos 2 1 cos 42 45 5 5
5cos 2 cos 42 4 415 5
5cos 2 cos44 4
x xx x
x x
x x
x x
x x
x x
x x
Section 6.3 Double-Angle Power Reducing, and Half-Angle Formulas
Copyright © 2014 Pearson Education, Inc. 751
37. 2 2
2
2
1 cos2 1 cos 2sin cos
2 2
1 cos 2
41 1
cos 24 41 1 1 cos(2 2 )
4 4 2
1 1(1 cos 4 )
4 81 1 1
cos 44 8 81 1
cos 48 8
x xx x
x
x
x
x
x
x
38. 1 cos 2 1 cos 22 28sin cos 8
2 228(1 cos 2 )
428 8(cos 2 )
4 41 cos 2 2
2 22
2 1 cos 4
1 cos 4
x xx x
x
x
x
x
x
39. Because 15° lies in quadrant I, sin15 0.
32
30ºsin15 sin
2
11 cos30
2 2
2 3 2 3
4 2
40. Because 22.5° lies in quadrant I, cos 22.5o >0. o o45 1 cos 45ocos22.5 cos
2 2
21
22
2 2
4
2 2
2
41. Because 157.5° lies in quadrant II, cos157.5 0.
22
315 1 cos315cos157.5 cos
2 2
1 2 2
2 4
2 2
2
42. Because 105° lies in quadrant II, sin 105o > 0. o o210 1 cos210osin105 sin
2 2
31
2
2
2 3
4
2 3
2
43. Because 75° lies in quadrant I, tan 75 0.
3212
150 1 cos150tan 75 tan
2 sin150
12 3
44. Because 112.5° lies in quadrant II, tan 112.50 < 0. o225otan125 tan
2o1 cos225
osin 225
21
2
2
2
2 2
22
12
2 1
Chapter 6 Analytic Trigonometry
752 Copyright © 2014 Pearson Education, Inc.
45. Because 7
8
π lies in quadrant II,
7tan 0.
8
π
7 74 4
74
22
1 cos7tan tan
8 2 sin
1 21
2 2
2
2 1
π π
ππ
46. Because 3
8
πlies in quadrant I, tan
3
8
π > 0.
33 4tan tan8 2
31 cos
43
sin4
21
2
2
22
12
2 1
ππ
π
π
47.
45
1 cossin
2 2
1 1
2 10
1 10
1010
θ θ
48. 1 cos
cos2 2
41
52
9
52
9
103
10
3 10
10
θ θ
49.
45
35
1 costan
2 sin1
1
3
θ θθ
Use this information to solve problems 50, 51, 52 and 54.
7tan
24
y
xα
Because r is a distance, it is positive. 2 2 2
2 2 2
2
24 7
625
25
r x y
r
r
r
7sin
2524
cos25
y
rx
r
α
α
Section 6.3 Double-Angle Power Reducing, and Half-Angle Formulas
Copyright © 2014 Pearson Education, Inc. 753
50. 1 cos
sin2 2
241
252
1
252
1
501
5 2
2
10
α α
51. 242511 cos 49cos
2 2 2 50
7 7 2
105 2
α α
52. 1 cos
tan2 sin
241
257
251
257
251
7
α αα
53. 1 cos 1 cos
2sin cos 22 2 2 2
θ θ θ θ
4 45 51 12
2 2
1 92
10 101 3
210 10
6 3
10 5
54. 1 cos 1 cos
2sin cos 22 2 2 2
α α α α
24 241 1
25 2522 2
1 492
50 501 7
250 50
7
25
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