For Thought x - TopCatMath · + sin2 x= cos2 x+ sin2 x= 1 15. sin(x) 1=sin(x) + cos(x) 1=cos(x) =...

6.1 BASIC IDENTITIES 347

For Thought

1. False, for example sin 0 = 0 but cos 0 = 1.

2. True, sincesin(x)cos(x)

· cos(x)sin(x)

= 1.

3. True, since f(−x) = (sin(−x))2 =(− sin(x))2 = sin2(x) = f(x).

4. False, it is even sincef(−x) = (cos(−x))3 = (cos(x))3 = f(x).

5. True 6. False, since (sinx+ cosx)2 =1 + 2 sin(x) cos(x) 6= 1 = sin2 x+ cos2 x.

7. False, since tan(x) = ±√

sec2(x)− 1.

8. True, since [sin(−3) csc(−3)] · [cos(−3) sec(3)]·[tan(−3) cot(3)] = [1] · [1] · [−1] = −1.

9. False, sin2(−π/9) + cos2(−π/9) = 1.

10. True, 1− sin2(−π/7) = cos2(−π/7) =cos2(π/7).

6.1 Exercises

1. even

2. odd

3. Pythagorean

4. identity

5. odd

6. even

7.sinxcosx

· cosx = sinx

8. sinx · cosxsinx

= cosx

9.1

cosx· cosx = 1

10. sinx · 1sinx

= 1

11.1/ cos(x)

sin(x)/ cos(x)=

1cos(x)

· cos(x)sin(x)

=

1sin(x)

= csc(x)

12.cos(x)/ sin(x)

1/ sin(x)=

cos(x)sin(x)

· sin(x)1

= cos(x)

13.sinx

1/ sinx+ cos2 x = sin2 x+ cos2 x = 1

14.cosx

1/ cosx+ sin2 x = cos2 x+ sin2 x = 1

15.sin(x)

1/ sin(x)+

cos(x)1/ cos(x)

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

16. csc2 x− cot2 x = 1 17. 1− sin2 α = cos2 α

18. sec2 α− 1 = tan2 α

19. (sinβ + 1)(sinβ − 1) = sin2 β − 1 = − cos2 β

20. (1 + cosβ)(1− cosβ) = 1− cos2 β = sin2 β

21.1 + cosα · sinα

cosα· 1

sinα1/ sinα

=1 + 1

1/ sinα= 2 sinα

22.(sinα+ 1)(sinα− 1)

cos2 α=

sin2 α− 1cos2 α

=

− cos2 α

cos2 α= −1

23. Since cot2 x = csc2 x− 1, we obtaincotx = ±

√csc2 x− 1.

24. Since sec2(x) = tan2(x) + 1, we havesec(x) = ±

√tan2(x) + 1.

25. sin(x) =1

csc(x)=

1

±√

1 + cot2(x)

26. cos(x) =1

sec(x)=

1

±√

tan2(x) + 1

27. Since cot2(x) = csc2(x)− 1, we obtain

tan(x) =1

cot(x)=

1±√

csc2(x)− 1

28. cot(x) =1

tan(x)=

1±√

sec2(x)− 1

29. Since secα =√

1 + (1/2)2 =√

5/2,

cosα = 2/√

5, sinα =√

1− (2/√

5)2 = 1/√

5.So cscα =

√5 and cotα = 2.

Copyright 2013 Pearson Education, Inc.

348 CHAPTER 6. TRIGONOMETRIC IDENTITIES AND CONDITIONAL EQUATION

30. Since α is in Quadrant II,cosα = −

√1− (3/4)2 = −

√7/4,

secα = −4/√

7, tanα =3/4−√

7/4= −3/

√7,

cotα = −√

7/3, and cscα = 4/3.

31. Since sinα = −√

1− (−√

3/5)2 =−√

1− 3/25 = −√

22/5, cscα = −5/√

22,

secα = −5/√

3, tanα =−√

22/5−√

3/5=√

22/√

3,

and cotα =√

3/√

22.

32. cosα =5

−4√

5·√

5√5

= −5√

5−20

= −√

5/4,

sinα =√

1− (−√

5/4)2 =√

11/4

Also, tanα =√

11/4−√

5/4= −√

11/√

5 =

−√

55/5, cotα = −5/√

55, andcscα = 4/

√11.

33. Since α is in Quadrant IV, we getcscα = −

√1 + (−1/3)2 = −

√10/3,

sinα = −3/√

10, cosα =√

1− (−3/√

10)2 =√1− 9/10 = 1/

√10, secα =

√10,

and tanα = −3.

34. Since sinα = 1/√

3, we obtain

cos(α) =√

1− (1/√

3)2 =√

2/√

3,

sec(α) =√

3/√

2, tan(α) =1/√

3√2/√

3= 1/

√2,

and cot(α) =√

2.

35. Let θ = arccosx. Then cos θ = x and θ lies inquadrant 1 or 2. Since sin2 θ = 1 − cos2 θ =1 − x2, we obtain sin(arccosx) = sin θ =±√

1− x2. Since sine is positive in bothquadrants 1 and 2, we have sin(arccosx) =√

1− x2.

36. Let θ = arcsinx. Then sin θ = x and θ lies inquadrant 1 or 4. Since cos2 θ = 1 − sin2 θ =1 − x2, we obtain cos(arcsinx) = cos θ =±√

1− x2. Since cosine is positive in bothquadrants 1 and 4, we have cos(arcsinx) =√

1− x2.

37. Note, by Exercise 26, cos(x) =±1√

tan2(x) + 1.

Since arctanx is an angle in quadrant 1 or 4,and cosine is positive in both quadrants 1and 4, we get

cos(arctanx) =1√

tan2(arctanx) + 1=

1√x2 + 1

.

38. Note, tanx = ±√

sec2 x− 1. Then

tan(arccosx) = ±√

sec2(arccosx)− 1

= ±√

1/x2 − 1

= ±√

1− x2

x.

Observe that tan(arccosx) is positive exactlywhen x > 0, and tan(arccosx) is negativeexactly when x < 0.

Thus, tan(arccosx) =√

1− x2

x.

39. Note, tanx = ±√

sec2 x− 1 andcos(arcsinx) =

√1− x2. Then

tan(arcsinx) = ±√

sec2(arcsinx)− 1

= ±

√(1√

1− x2

)2

− 1

= ±√

11− x2

− 1

= ±

√x2

1− x2

= ±√x2

√1− x2

= ± ±x√1− x2

= ± x√1− x2

.

Note, tan(arcsinx) is positive exactly whenx > 0, and tan(arcsinx) is negative exactlywhen x < 0. Thus, tan(arcsinx) =

x√1− x2

.



40. sec(arcsinx) =1

cos(arcsinx)=

1√1− x2

41. Note, arctanx is an angle in quadrant 1 or 4,and secant is positive in both quadrants 1and 4. Since sec(θ) = ±

√tan2(θ) + 1, we

have sec(arctanx) =√

tan2(arctanx) + 1 =√x2 + 1.

42. csc(arcsinx) =1

sin(arcsinx)=

1x

43. (− sinx) · (− cotx) = sin(x) · cosxsinx

= cos(x)

44. secx−secx = 0 45. sin(y)+(− sin(y)) = 0

46. cos y + cos y = 2 cos y

47.sin(x)cos(x)

+− sin(x)cos(x)

= 0

48.cos(x)− sin(x)

− cos(x)sin(x)

= −2 cot(x)

49. (1 + sinα)(1− sinα) = 1− sin2 α = cos2 α

50. (1− cosα)(1 + cosα) = 1− cos2 α = sin2 α

51. (− sinβ)(cosβ)(1/ sinβ) = − cosβ

52. (− tanβ)(− 1

sinβ

)cosβ = tanβ cotβ = 1

53. Odd, since sin(−y) = − sin(y) for any y,even if y = 2x.

54. Even, since cos(−y) = cos(y) for any y,even if y = 2x.

55. Neither, since f(−π/6) 6= f(π/6) andf(−π/6) 6= −f(π/6).

56. Odd, f(−x) = 2 sin(−x) cos(−x) =2(− sin(x)) cos(x) = −f(x).

57. Even, since sec2(−t)− 1 = sec2(t)− 1.


59. Even, f(−α) = 1 + sec(−α) = 1 + sec(α) =f(α)


61. Even, f(−x) =sin(−x)−x

=− sin(x)−x

= f(x)

62. Odd, f(−x) = (−x) cos(−x) =−x cos(x) = −f(x)

63. Odd, f(−x) = −x+ sin(−x) =−x− sin(x) = −f(x)

64. Even, f(−x) = csc((−x)2) = csc(x2) = f(x)

65. h, since1

csc(x)=

11/ sin(x)

=

1 · sin(x)1

= sin(x)

66. g, for1

sec(x)=

11/ cos(x)

= cos(x)

67. n 68. o 69. m 70. i

71. k, since sin2(x) + cos2(x) = 1

72. j, since sin2(x) + cos2(x) = 1

73. l, since sec2(x) = 1 + tan2(x)

74. a, since sin(x) is an odd function

75. g, since cos(−x) = cos(x) and cos(x) =1

sec(x)

76. c, since tan(−x) =sin(−x)cos(−x)

=− sin(x)cos(x)

=

− tan(x)

77. b, since cot(−x) =cos(−x)sin(−x)

=cos(x)− sin(x)

=

− cot(x)

78. m, since sec(−x) =1

cos(−x)=

1cos(x)

79. f, since csc(−x) =1

sin(−x)=

1− sin(x)

= − csc(x)

80. p, since csc2(x) = 1 + cot2(x)

81. d, since sec2(x) = 1 + tan2(x)



82. e, since csc2(x) = 1 + cot2(x)

83. It is not an identity. If γ = π/3 then(sin(π/3) + cos(π/3))2 = (

√3/2 + 1/2)2 =

(√

3 + 1)2

4=

4 + 2√

34

6= 1 =

sin2(π/3) + cos2(π/3).

84. It is not an identity. If x = π/4then tan2(π/4)− 1 = 1− 1 = 0 andsec2(π/4) = (

√2)2 = 2.

85. It is not an identity. If β = π/6 then(1 + sin(π/6))2 = (1 + 1/2)2 = (3/2)2 = 9/4and 1 + sin2(π/6) = 1 + (1/2)2 = 5/4.

86. It is not an identity. If α = π/6then sin(2π/6) = sin(π/3) =

√3/2 and

sin(π/6) cos(π/6) = (1/2) · (√

3/2) =√

3/4.

87. It is not an identity. If α = 7π/6 thensin(7π/6) = −1/2 while

√1− cos2(7π/6) is a

positive number.

88. It is not an identity. If α = 3π/4 thentan(3π/4) = −1 while

√sec2(3π/4)− 1 is a

positive number.

89. It is not an identity. If y = π/6 thensin(π/6) = 1/2 and sin(−π/6) = −1/2.

90. It is not an identity. If y = π/3 thencos(−π/3) = 1/2 and − cos(π/3) = −1/2.

91. It is not an identity. If y = π/6 thencos2(π/6)− sin2(π/6) =(√

3/2)2 − (1/2)2 = 3/4− 1/4 = 1/2and sin(2 · π/6) = sinπ/3 =

√3/2.

92. It is not an identity. If x = π/6 thencos(2 · π/6) = cosπ/3 = 1/2 and2 cos(π/6) sin(π/6) = 2(

√3/2)(1/2) =

√3/2.

93. 1− 1cos2(x)

= 1− sec2(x) = − tan2(x)

94.sin2 x(sin2 x− 1)

1/ cosx= sin2 x(− cos2 x) · cosx =

− sin2(x) cos3(x)

95.−(tan2 t+ 1)

sec2 t=− sec2 t

sec2 t= −1

96.cosw(sin2w + cos2w)

secw=

cos(w) · 1secw

= cos2w

97.(1− cos2w)− cos2w

1− 2 cos2w=

1− 2 cos2w

1− 2 cos2w= 1

98.− sin3 θ

sin θ(sin2 θ − 1)=− sin2 θ

(− cos2 θ)= tan2 θ

99.tanx(tan2 x− sec2 x)

− cotx=

tanx(−1)− cotx

= tan2 x

100.sin2 x+ cos2 x

sinx=

1sinx

= cscx

101.1

sin3 x− cos2(x)/ sin2(x)

sinx=

1sin3 x

− cos2(x)sin3 x

=

1− cos2 x

sin3 x=

sin2 x

sin3 x=

1sinx

= cscx

102.

1− 1/ cos2 x

sin2 x/ cos2 x= 1− 1

cos2 x· cos2 x

sin2 x=

1− 1sin2 x

= 1− csc2 x = − cot2 x

103. (sin2 x− cos2 x)(sin2 x+ cos2 x) =(sin2 x− cos2 x)(1) = sin2 x− cos2 x

104. (csc2 x+ cot2 x)(csc2 x− cot2 x) =(csc2 x+ cot2 x)(1) = csc2 x+ cot2 x

105. cos θ = ±√

1− sin2 θ = ±√

1− (1/3)2 =

±√

1− 1/9 = ±√

89

= ±2√

23

106. Note sec θ =54

.

Then tan θ = ±√

sec2 θ − 1 =

±√

(5/4)2 − 1 = ±√

25/16− 1 =

±√

916

= ±34

.

107. cos θ = ±√

1− sin2 θ = ±√

1− u2

108. Note sec θ =1u

.

Then tan θ = ±√

sec2 θ − 1 =

±

√(1u

)2

− 1 = ±

√1− u2

u2= ±√

1− u2

√u2

=



±√

1− u2

±u= ±√

1− u2

u.

109. Note, tanx =sinxcosx

is not valid if cosx = 0.

Thus, the identity is not valid if x =π

2+ kπ

where k is an integer.

110. Note, cotx =cosxsinx

is not valid if sinx = 0.

Thus, the identity is not valid if x = kπ wherek is an integer.

113. Let h be the height of the building. Usingright triangle trigonometry, we find

h = 2000 tan 30◦ ≈ 1155 ft.

114. Let α be the central angle. Using the formulas = rα, we obtain

5 = 60α560

radians = α

560· 180◦

πdegrees = α

4.8◦ = α

115. The amplitude is 5.

Since B = 2, the period is2πB

=2π2

= π.

Since 2x− π = 2(x− π

2

), phase shift is

π

2.

The range is the interval [−5+3, 5+3] = [−2, 8]

116. The period and frequency are reciprocals ofeach other. Then the frequency is

10.125

= 8 cycles/sec

117. cosβ = 0, for cos2 β + sin2 β = 1

118. a) sin−1

(−1

2

)= −30◦

b) cos−1

(−1

2

)= 120◦

c) tan−1 (−1) = −45◦

Thinking Outside the Box LII

Let r be the radius of the small circle, and let x bethe distance from the center of the small circle tothe point of tangency of any two circles with radius1.

By the Pythagorean theorem, we find

1 + (x+ r)2 = (1 + r)2

and1 + (1 + 2r + x)2 = 22.

The second equation may be written as

1 + (r + 1)2 + 2(r + 1)(r + x) + (r + x)2 = 4.

Using the first equation, the above equation sim-plifies to

(r + 1)2 + 2(r + 1)(r + x) + (1 + r)2 = 4

or(r + 1)2 + (r + 1)(r + x) = 2.

Since (from first equation, again)

x+ r =√

(1 + r)2 − 1

we obtain

(r + 1)2 + (r + 1)(√

(1 + r)2 − 1)

= 2.

Solving for r, we find

r =2√

3− 33

.

6.1 Pop Quiz

1.cosxsinx

1cosx

=1

sinx= cscx

2. cosα =

√1−

(13

)2

=√

89

=2√

23

cotα =cosαsinα

=2√

2/31/3

= 2√

2

3. Even, since cos(3(−x)) = cos(3x)



4. Note, arcsin(w) is an angle in quadrant 1 or 4.Since cosine is positive in quadrant 1 or 4, wefind cos(arcsin(w)) =

√1− sin2(arcsin(w)) =√

1− w2.

5.2

1/ cos2 α+

21/ sin2 α

= 2(cos2 α+ sin2 α

)=

2(1) = 2

6.1 Linking Concepts

a) Since x = 100(4) cos 60◦ = 200 andy = −16(4)2 + 100(4) sin 60◦ = 200

√3− 256,

after t = 4 the coordinates are(200, 200

√3− 256) ≈ (200, 90.4).

b) Note, y = −16t2+100t sin 60◦ = −16t2+50√

3t.Set y = 0.

−t(16t− 50√

3) = 0

t = 0,50√

316

t = 0,25√

38

The projectile is in the air for25√

38

seconds.

c) Using the answer from part b), we get

x = 100t cos 60◦ = 50t = 50

(25√

38

)≈ 270.6.

The projectile lands 270.6 feet from the gun.

d) The vertex of the function y = −16t2 + 50√

3t

(given by the height) can be shown

to be

(25√

316

,187516

)≈ (2.7, 117.2).

The maximum height is 117.2 feet.

e) If A is the time in the air which is the sameas the number of seconds before the projectilelands, then

−16A2 + voA sin θ = 0−A(16A− vo sin θ) = 0.

Then A =vo sin θ

16.

The distance, d, from the gun to the pointwhere the projectile lands is given by

d = x = vot cos θ = voA cos θ =

vo

(vo sin θ

16

)cos θ =

v2o

16sin θ cos θ,

i.e., d =v2o

16sin θ cos θ.

The t-coordinate of the vertex ofy = −16t2 + vo sin(θ)t (given by the height) is−b2a

=−vo sin θ−32

=vo sin θ

32. The maximum

height, ymax, is given by

ymax = −16(vo sin θ

32

)2

+ vo sin(θ)(vo sin θ

32

)

= −v2o sin2 θ

64+v2o sin2 θ

32

=v2o sin2 θ

64

f) Since x = vot cos θ, we find t =x

vo cos θ. Then

y = −16t2 + vot sin θ

= −16(

x

vo cos θ

)2

+ vo

(x

vo cos θ

)sin θ

y = −16 sec2 θ

v2o

x2 + x tan θ.

For Thought

1. True,sinx

1/ sinx= sinx · sinx

1= sin2 x.

2. False, if x = π/3 thencot(π/3)tan(π/3)

=√

3/3√3

=13

and tan2(π/3) = (√

3)2 = 3.

3. True,1/ cosx1/ sinx

=1

cosx· sinx

1=

sinxcosx

= tanx.

4. True, sinx · 1cosx

=sinxcosx

= tanx .

5. True,cosxcosx

+sinxcosx

= 1 + tanx.


6.2 VERIFYING IDENTITIES 353

6. False, if x = π/4 then

sec(π/4) +sin(π/4)cos(π/4)

=√

2 + 1 and

1 + sin(π/4) cos(π/4)cos(π/4)

=1 + (

√2/2)(

√2/2)√

2/2=

1 + (1/2)√2/2

= (3/2)(2/√

2) = 3/√

2.

7. True,1 + sinx1− sin2 x

=1 + sinx

(1− sinx)(1 + sinx)=

11− sinx

.

8. True, since tanx · cotx = tanx · 1tanx

= 1.

9. False, if x = π/3 then (1− cos(π/3))2 =(1− 1/2)2 = (1/2)2 = 1/4 andsin2(π/3) = (

√3/2)2 = 3/4.

10. False, if x = π/6 then(1− csc(π/6))(1+csc(π/6)) = (1−2)(1+2) =−3 and cot2(π/6) = (

√3)2 = 3.

6.2 Exercises

1. D, cosx tanx = cosx · sinxcosx

= sinx .

2. I, secx cotx =1

cosx· cosx

sinx=

1sinx

= cscx .

3. A, csc2 x− cot2 x = 1 .

4. J,sinxsinx

+cosxsinx

= 1 + cotx .

5. B, 1− sec2 x = − tan2 x .

6. C, csc2 x− 1 = cot2 x .

7. H,cscxcscx

− sinxcscx

= 1− sin2 x = cos2 x .

8. E,cosxsecx

− secxsecx

= cos2 x− 1 = − sin2 x .

9. G, csc2 x = 1 + cot2 x .

10. F,sin2 x+ cos2 x

cosx sinx=

1cosx sinx

.

11. 2 cos2 β − cosβ − 1

12. 2 csc2 β − 7 cscβ + 3

13. csc2 x+2 cscx sinx+sin2 x = csc2 x+2+sin2 x

14. 4 cos2 x− 4 cosx secx+ sec2 x =4 cos2 x− 4 + sec2 x

15. 4 sin2 θ − 1 16. 9 sec2 θ − 4

17. 9 sin2 θ+12 sin θ+4 18. 9 cos2 θ−12 cos θ+4

19. 4 sin4 y − 4 sin2 y csc2 y + csc4 y =4 sin4 y − 4 + csc4 y

20. tan4 y + 2 tan2 y cot2 y + cot4 y =tan4 y + 2 + cot4 y

21. Note the factorization of a difference of twosquares: (1 − sinα)(1 + sinα) = 1 − sin2 α =cos2 α.

22. Note the factorization of a difference of twosquares: (1− cosα)(1 + cosα) = 1− cos2 α =sin2 α.

23. Note the factorization of a difference of twosquares: (cscα − 1)(cscα + 1) = csc2 α − 1 =cot2 α.

24. Note the factorization of a difference of twosquares: (secα − 1)(secα + 1) = sec2 α − 1 =tan2 α.

25. Note the factorization of a difference of twosquares: (tanα − secα)(tanα + secα) =tan2 α− sec2 α = −1.

26. Note the factorization of a difference of twosquares: (cotα − cscα)(cotα + cscα) =cot2 α− csc2 α = −1.

27. (2 sin γ + 1) (sin γ − 3)

28. (cos γ − 3) (cos γ + 2)

29. (tanα− 4) (tanα− 2)

30. (2 cotα+ 3) (cotα− 1)

31. (2 secβ + 1)2 32. (3 csc θ − 2)2

33. (tanα− secβ) (tanα+ secβ)

34.(sin2 y − cos2 x

) (sin2 y + cos2 x

)=

(sin y − cosx) (sin y + cosx)(sin2 y + cos2 x

)Copyright 2013 Pearson Education, Inc.


35. cosβ(sin2 β + sinβ − 2

)=

cosβ (sinβ + 2) (sinβ − 1)

36. tan θ(cos2 θ − 2 cos θ − 3

)=

tan θ (cos θ − 3) (cos θ + 1)

37.(2 sec2 x− 1

)238.

(cos2 x− 1

)2 =[(cosx− 1)(cosx+ 1)]2 =(cosx− 1)2(cosx+ 1)2

39. cosα(sinα+ 1) + (sinα+ 1) =(sinα+ 1) (cosα+ 1)

40. sin θ (2 sin θ + 1)− cos θ (2 sin θ + 1) =(sin θ − cos θ) (2 sin θ + 1)

41. Combining, we get1− cos2 x

a=

sin2 x

a.

42. Combining, we get1− sin2 x

cosx=

cos2 x

cosx=

cosx.

43. We obtainsin(2x)

2+

2 sin(2x)2

=3 sin(2x)

2.

44. We obtain2 cos(2x)

2− cos(2x)

2=

cos(2x)2

.

45. Since 6 is the LCD, we get2 tanx

6+

3 tanx6

=5 tanx

6.

46. Since 3b is the LCD, we get3 sinx

3b+

sinx3b

=4 sinx

3b.

47. Separating the fraction, we obtain

sinxsinx

− sin2 x

sinx= 1− sinx.

48. Factoring, we getcosx(cos2 x− 1)

− cosx=

cosx(− sin2 x)− cosx

= sin2 x.

49. Factoring:(sinx− cosx)(sinx+ cosx)

sinx− cosx= sinx+ cosx.

50. Factoring:(1− cosx)(1 + cosx)

1− cosx= 1 + cosx.

51. Factoring:(sinx− 2)(sinx+ 1)(sinx− 2)(sinx+ 2)

=sinx+ 1sinx+ 2

.

52. Note, tan(−x) = − tan(x) andtanx− 11− tanx

= −1. Then

(tanx− 1)2

1− tanx= −(tanx− 1) = 1− tanx.

53. Note, sin(−x) = − sin(x). Factoring, we

obtainsin2 x+ sinx

1 + sinx=

sinx(sinx+ 1)1 + sinx

=

sinx.

54. Note, cos(−x) = cos(x). Factoring, we obtain

cos2 x− cosx1− cosx

=cosx(cosx− 1)

1− cosx=

cosx(cosx− 1)−(cosx− 1)

= − cosx.

55.

sinx cotx =

sinxcosxsinx

=

cosx

56.

cos2 x tan2 x =

cos2 xsin2 x

cos2 x=

sin2 x

57.

1− secx cos3 x =

1− 1cosx

cos3 x =

1− cos2 x =

sin2 x



58.

1− cscx sin3 x =

1− 1sinx

sin3 x =

1− sin2 x =

cos2 x

59.

1 + sec2 x sin2 x =

1 +1

cos2 xsin2 x =

1 + tan2 x =

sec2 x

60.

1 + csc2 x cos2 x =

1 +1

sin2 xcos2 x =

1 + cot2 x =

csc2 x

61.

sin3 x+ sinx cos2 x

cosx=

sinx(sin2 x+ cos2 x)cosx

=

(sinx)(1)cosx

=

tanx

62.

cosx sin2 x+ cos3 x

sinx=

cosx(sin2 x+ cos2 x)sinx

=

(cosx)(1)sinx

=

cotx

63.

sinxcscx

+cosxsecx

=

sinx1/ sinx

+cosx

1/cosx=

sin2 x+ cos2 x =

1

64.

sin3 x cscx+ cos3 x secx =

sin3 1sinx

+ cos3 x1

cosx=

sin2 x+ cos2 x =

1

65.

1csc θ − cot θ

· sin θsin θ

=

sin θ1− cos θ

· 1 + cos θ1 + cos θ

=

sin θ(1 + cos θ)1− cos2 θ

=

sin θ(1 + cos θ)sin2 θ

=

1 + cos θsin θ

66.

−1tan θ − sec θ

· cos θcos θ

=

− cos(θ)sin(θ)− 1

· sin θ + 1sin θ + 1

=

(− cos θ)(sin θ + 1)sin2 θ − 1

=

(− cos θ)(sin θ + 1)− cos2 θ

=

1 + sin(θ)cos θ



67.secx− cosx

secx=

1− cosxsecx

=

1− cos2 x =sin2 x

68.secx− cosx

cosx=

secxcosx

− 1 =

sec2 x− 1 =tan2 x

69.

=1− (− sinx)2

1 + sinx

=1− sin2 x

1 + sinx

=(1− sinx)(1 + sinx)

1 + sinx

1− sin(x)

70.

=1− 1 + sin2(x)

cos2(x)

=sin2(x)cos2(x)

tan2(x)

71.

=1− cot2w

(1− cos2w

)csc2w

=1− cot2w sin2w

csc2w

=1− cos2w

csc2w

=sin2w

csc2w

sin4w

72.

=sec2 z − csc2 z

(1− cos2 z

)cot2 z

=sec2 z − csc2 z sin2 z

cot2 z

=sec2 z − 1

cot2 z

=tan2 z

cot2 z

tan4 z

73.

=cosx+ cscx

cosx

=cosxcosx

+cscxcosx

1 + cscx secx

74.

tan2(−x)− − sinxsinx

=

tan2 x+ 1 =sec2 x

75. Rewrite the left side of the equation.

tan(x) cos(x) + csc(x) sin2(x) =sinx+ sinx =

2 sinx

76.

cot(x) sin(x)− cos2(x) sec(x) =cosx− cosx =

0

77.

(1 + sinα)2 + cos2 α =1 + 2 sinα+ sin2 α+ cos2 α =

2 + 2 sinα



78. (1 + 2 cotα+ cot2 α

)− 2 cotα =

1 + cot2 α =csc2 α =

1sin2 α

=

11− cos2 α

=

1(1− cosα)(1 + cosα)

79.

sin2 β + sinβ − 22 sinβ − 2

=

(sinβ + 2)(sinβ − 1)2(sinβ − 1)

=

sinβ + 22

80.

4 sec2 β + 4 secβ + 12 secβ + 1

=

(2 secβ + 1)2

secβ + 1=

2 secβ + 1 =

2cosβ

+ 1

81.

2− csc(β) sin(β) =2− 1 =

1 =sin2(β) + cos2(β)

82. (1− sin2 β

) (1 + sin2 β

)=

cos2(β)(2− cos2(β)

)=

2 cos2 β − cos4 β

83.

sinxcosx

+cosxsinx

=

sin2 x+ cos2 x

sin(x) cos(x)=

1sin(x) cos(x)

=

sec(x) csc(x)

84.

cscxcotx

− cotxcscx

=

csc2 x− cot2 xcot(x) csc(x)

=

1cot(x) csc(x)

=

tanxcscx

85.

sec(x)tan(x)

− tan(x)sec(x)

=

sec2(x)− tan2(x)tan(x) sec(x)

=

1tan(x) sec(x)

=

cot(x) cos(x)

86.

(1− sinx)(1 + sinx)1− sinx

=

1 + sin(x) =cscxcscx

+1

cscx=

cscx+ 1cscx

87. Rewrite the right side of the equation.

=cscx

cscx− sinx· sinx

sinx

=1

1− sin2 x



=1

cos2 x

sec2 x

88.

=cscx− 1

cot2 x

=cscx− 1csc2 x− 1

=cscx− 1

(cscx− 1)(cscx+ 1)

=1

cscx+ 1· sinx

sinxsinx

sinx+ 1

89.

=1 + sin(y)1− sin(y)

· csc(y)csc(y)

csc(y) + 1csc(y)− 1

90.

=sin y + cos ysin y − cos y

· sin y − cos ysin y − cos y

=sin2 y − cos2 y

sin2 y − 2 sin y cos y + cos2 y

=(1− cos2 y)− cos2 y

1− 2 sin y cos y

1− 2 cos2 y

1− 2 cos y sin y

91.

ln(sec θ) =

ln((cos θ)−1) =

− ln(cos θ)

92.

ln(tan θ) =

ln(

sin θcos θ

)=

ln(sin θ)− ln(cos θ) =

ln(sin θ) + ln((cos θ)−1) =

ln(sin θ) + ln(sec θ)

93.

ln∣∣∣∣(secα+ tanα) · secα− tanα

secα− tanα

∣∣∣∣ =

ln

∣∣∣∣∣sec2 α− tan2 α

secα− tanα

∣∣∣∣∣ =

ln∣∣∣∣ 1secα− tanα

∣∣∣∣ =

− ln |secα− tanα|

94.

ln∣∣∣∣(cscα+ cotα) · cscα− cotα

cscα− cotα

∣∣∣∣ =

ln

∣∣∣∣∣csc2 α− cot2 αcscα− cotα

∣∣∣∣∣ =

ln∣∣∣∣ 1cscα− cotα

∣∣∣∣ =

− ln |cscα− cotα|

95. It is an identity since

sin θsin θ

+cos θsin θ

=

1 + cot θ.

The graphs of y =sin θ + cos θ

sin θand

y = 1 + cot θ are shown to be identical.

Pi/2 Pix

-2

2

y

96. It is not an identity since the graphs of

y =sin θ + cos θ

cos θand y = 1 + cot θ



do not coincide as shown below.

Pi/2 Pix

-2

2

y


y = (sinx+ cscx)2 and y = sin2 x+ csc2 x

do not coincide as shown.

Pi/2 Pix

-2

2

y


y = tanx+ secx and y =sin2 x+ 1

cosxdo not coincide as shown.

Pi/2 Pix

-2

2

y

99. It is an identity. Re-arranging the numeratorof the right-hand side one finds

=1− cos2 x+ cosx

sinx

=sin2 x+ cosx

sinx

=sin2 x

sinx+

cosxsinx

sinx+ cotx.

The graphs of y = cotx+ sinx and

y =1 + cosx− cos2 x

sinxare shown to

be identical.

Pi/2 Pix

-2

2

y

100. It is an identity. To see this, factor theleft-hand side as follows(

1− cos2 x)2

=(sin2 x

)2=

sin4 x.

The graphs of y = 1− 2 cos2 x+ cos4 x andy = sin4 x are shown to be identical.

Pi/2 Pix

-2

2

y


y =sinxcosx

− cosxsinx

and y =2 cos2 x− 1sinx cosx

are not the same as shown.

Pi 2Pix

-1

1

y



102. It is an identity.

(1 + sinx) + (1− sinx)(1− sinx)(1 + sinx)

=

21− sin2(x)

=

2cos2 x

=

The graphs of y =1

1− sinx+

11 + sinx

and

y =2

cos2 xare shown to be identical.

Pi 2Pix

-1

2

y

103. It is an identity.

cosx1− sin(x)

· 1 + sinx1 + sinx

=

cosx(1 + sinx)1− sin2 x

=

cosx(1 + sinx)cos2 x

=

1 + sinxcosx

=

1− sin(−x)cosx

=

The graphs of y =cos(−x)1− sinx

and

y =1− sin(−x)

cosxare shown to be identical.

-3Pi/2 2PiPi/2x

1

-1

y


y =sin2 x

1− cosxand y = 0.99 + cosx

do not coincide as shown below:

Pi/2x

2

y

107. sin2 x+ cos2 x = 1, 1 + cot2 x = csc2 x,

tan2 x+ 1 = sec2 x

108.1

cos2 x− tan2 x = sec2 x− tan2 x = −1

109.cscxsecx

=1/ sinx1/ cosx

=cosxsinx

= cotx

110. The midpoint is(π/3 + π/2

2,1 + 1

2

)=(

5π/62

,22

)=(

5π12, 1)

111. Amplitude is 4. Since B = 2π/3, the period

is2πB

=2π

2π/3= 3.

Since2πx

3− π

3=

2π3

(x− 1

2

), the phase shift

is12

.

112. Since B = 1/4, the period is2πB

=2π1/4

= 8π.

Solve for x inx

4=π

2+ kπ. Then the asymp-

totes arex = 2π + 4kπ

where k is an integer.

The range is (−∞,−2] ∪ [2,∞).

113. Solve for v0 in ft/sec:

132v20 sin 2(33◦) = 200

v0 =

√200(32)sin 66◦

ft/sec



v0 =

√200(32)sin 66◦

· 52803600

mph

v0 ≈ 57.1 mph.

Thinking Outside the Box LIII

The amplitude of the sine wave is 1/2 since theheight of the sine wave is 1. We use a coordinatesystem such that the sine wave begins at the originand extends to the right side and the first quadrant.Note, the period of the sine wave is π, which is thediameter of the tube. Then the highest point onthe sine wave is (π/2, 1). Thus, an equation of thesine wave is

y = −12

cos(2x) +12.

6.2 Pop Quiz

1. 2 sin2 x− sinx− 1

2. (2 cosx− 1)(cosx+ 1)

3.1

cosx− sin2 x

cosx=

1− sin2 x

cosx=

cos2 x

cosx= cosx

4.

cos(−x)− sec(−x)sec(x)

=

cosx− secxsecx

=

cosx− secxsecx

· cosxcosx

=

cos2 x− 11

=

− sin2 x

1=

− sin2 x


a) Assume the circle is given by x2 + y2 = r2 and(w

2, r − h

)is a point on the circle

(corresponding to the upper right hand cornerof the rectangular window). Substituting thispoint into the equation of the circle, we obtain

w2

4+ (r − h)2 = r2

w2

4+ (r2 − 2rh+ h2) = r2

w2

4− 2rh+ h2 = 0

2rh =w2 + 4h2

4

r =w2 + 4h2

8h.

If w = 36 and h = 10, then the radius of the

circle is r =362 + 4(10)2

80= 21.2 inches.

b) Consider the right triangle with vertices at the

point A(0, 0), B(w

2, r − h

), and C(0, r − h).

Let θ be the angle at point A. Then

tan θ =w/2r − h

tan θ =w

2(r − h)

θ = tan−1(

w

2(r − h)

).

If w = 36 and h = 10, then the

length of the circular arc is L = 2rθ =

2(21.2) tan−1

(36

2(21.2− 10)

)≈ 43.0 in.

c) r =w2 + 4h2

8has derived in part a)

d) In part b), we obtained θ = tan−1

(w

2(r − h)

).

A formula for the arclength L is given by

L = 2rθ

=w2 + 4h2

4htan−1

(w

2(r − h)

).



Equivalently, by using the fact that

r =w2 + 4h2

8h(see part c) we can rewrite L as

L =w2 + 4h2

4hsin−1

(4hw

4h2 + w2

).

If we interpret arcsin in degrees, then

L =w2 + 4h2

4hsin−1

(4hw

4h2 + w2

)· π

180

or

L =πw2 + 4πh2

720hsin−1

(4hw

4h2 + w2

).

For Thought

1. False, the right-hand side should be cos(5◦).

2. True, by the sum identity for cosine.

3. True, cos(t− π/2) = cos(π/2− t) = sin t .

4. False, sin(α− π/2) = − sin(π/2− α) = − cosα.

5. True, sec(π/3) = sec(π/2− π/6) = csc(π/6) .

6. False, since sin(5π/6) = 1/2 andsin(2π/3) + sin(π/6) =

√3/2 + 1/2 .

7. True, since the sum identity for sine is appliedto 5π/12 = π/6 + π/4 .

8. True, since the cofunction identity for tangentis applied to 90◦ − 68◦29′55” = 21◦30′5” .

9. False, the equation fails when x = π/2 .

10. True, since both sides of the equation (bythe sum identity for tangent) are equal totan(−7◦).

6.3 Exercises

1. cosine

2. cofunction

3. cos(π/3 + π/4) =cos(π/3) cos(π/4)− sin(π/3) sin(π/4) =

12·√

22−√

32·√

22

=√

2−√

64

4. cos(2π/3 + π/4) =cos(2π/3) cos(π/4)− sin(2π/3) sin(π/4) =

−12·√

22−√

32·√

22

=−√

2−√

64

5. cos(60◦ − 45◦) =cos(60◦) cos(45◦) + sin(60◦) sin(45◦) =

12·√

22

+√

32·√

22

=√

2 +√

64


−12·√

22

+√

32·√

22

=−√

2 +√

64

7. Since sin(20◦) = cos(90◦ − 20◦) = cos(70◦),the answer is 70◦.

8. Since cos(15◦) = cos(90◦ − 150◦) or cos(75◦),the answer is 75◦.

9. Since tan(π

6

)= cot

(π

2− π

6

)= cot

(π

3

),

the answer is π/3.

10. Since cot(π

3

)= tan

(π

2− π

3

)= tan

(π

6

),

the answer is π/6.

11. Since sec(90◦−6◦) = csc(6◦), the answer is 6◦.

12. Since csc(90◦ − 17◦) = sec(17◦), the answer is17◦.

13. sin(π/3 + π/4) =sin(π/3) cos(π/4) + cos(π/3) sin(π/4) =√

32·√

22

+12·√

22

=√

6 +√

24

14. sin(π/4 + 2π/3) =sin(π/4) cos(2π/3) + cos(π/4) sin(2π/3) =√

22· −1

2+√

22·√

32

=−√

2 +√

64

15. sin(60◦ − 45◦) =sin(60◦) cos(45◦)− cos(60◦) sin(45◦) =√

32·√

22− 1

2·√

22

=√

6−√

24


6.3 SUM AND DIFFERENCE IDENTITIES 363

16. sin(45◦ − 120◦) =sin(45◦) cos(120◦)− cos(45◦) sin(120◦) =√

22· −1

2−√

22·√

32

=−√

2−√

64

17. tan(

3π4

+π

3

)=

tan(3π/4) + tan(π/3)1− tan(3π/4) tan(π/3)

=

−1 +√

31− (−1)(

√3)

=√

3− 1√3 + 1

=√

3− 1√3 + 1

·√

3− 1√3− 1

=

4− 2√

32

= 2−√

3

18. tan(π

4+π

3

)=

tan(π/4) + tan(π/3)1− tan(π/4) tan(π/3)

=

1 +√

31− (1)(

√3)

=1 +√

31−√

3=

1 +√

31−√

3· 1 +

√3

1 +√

3=

4 + 2√

3−2

= −2−√

3

19. tan(210◦ − 45◦) =tan(210◦)− tan(45◦)

1 + tan(210◦) tan(45◦)=

√3/3− 1

1 + (√

3/3)(1)=

√3/3− 1

1 + (√

3/3)(1)· 3

3=

√3− 3

3 +√

3=√

3− 33 +√

3· 3−

√3

3−√

3=−12 + 6

√3

6=

√3− 2

20. tan(45◦ − 150◦) =tan(45◦)− tan(150◦)

1 + tan(45◦) tan(150◦)=

1− (−√

3/3)1 + (1)(−

√3/3)

=1 +√

3/31−√

3/3· 3

3=

3 +√

33−√

3=

3 +√

33−√

3· 3 +

√3

3 +√

3=

12 + 6√

36

=

2 +√

3

21.7π12

22.5π12

23.13π12

24.5π12

25. sin(23◦ + 67◦) = sin(90◦) = 1

26. sin(55◦ − 10◦) = sin(45◦) =√

2/2

27. cos(π

6+π

3

)= cos

(π

2

)= 0

28. cos(

7π12− π

3

)= cos

(π

4

)=√

2/2

29. tan(π

12+π

6

)= tan

(π

4

)= 1

30. tan(

5π12− π

12

)= tan

(π

3

)=√

3

31. sin(2k + k) = sin(3k)

32. cos(3y + y) = cos(4y)

33. 30◦ + 45◦ 34. 45◦ − 30◦

35. 120◦ + 45◦ 36. 150◦ + 45◦

37. cos(2π/3− π/4) =

cos(2π/3) cos(π/4) + sin(2π/3) sin(π/4) =

−12·√

22

+√

32·√

22

=√

6−√

24

38. cos(π/3 + π/4) =cos(π/3) cos(π/4)− sin(π/3) sin(π/4) =

12·√

22−√

32·√

22

=√

2−√

64

39. sin(π/3 + π/4) =sin(π/3) cos(π/4) + cos(π/3) sin(π/4) =√

32·√

22

+12·√

22

=√

6 +√

24

40. sin(2π/3− π/4) =sin(2π/3) cos(π/4)− cos(2π/3) sin(π/4) =√

32·√

22− −1

2·√

22

=√

6 +√

24

41. tan(45◦ + 30◦) =tan(45◦) + tan(30◦)

1− tan(45◦) tan(30◦)=

1 +√

3/31− 1 ·

√3/3· 3

3=

3 +√

33−√

3· 3 +

√3

3 +√

3=

12 + 6√

39− 3

= 2 +√

3

42. tan(30◦ − 45◦) =tan(30◦)− tan(45◦)

1 + tan(30◦) tan(45◦)=

√3/3− 1

1 +√

3/3· 3

3=√

3− 33 +√

3· 3−

√3

3−√

3=



−12 + 6√

39− 3

= −2 +√

3

43. sin(30◦ − 45◦) =sin(30◦) cos(45◦)− cos(30◦) sin(45◦) =

12·√

22−√

32·√

22

=√

2−√

64

44. sin(120◦ + 45◦) =sin(120◦) cos(45◦) + cos(120◦) sin(45◦) =√

32·√

22

+−12·√

22

=√

6−√

24

45. cos(135◦ + 60◦) =cos(135◦) cos(60◦)− sin(135◦) sin(60◦) =

−√

22· 1

2−√

22·√

32

=−√

2−√

64

46. cos(−75◦) = cos(75◦) = cos(30◦ + 45◦) =cos(30◦) cos(45◦)− sin(30◦) sin(45◦) =√

32·√

22− 1

2·√

22

=√

6−√

24

47. tan(−13π/12) = − tan(13π/12) =

− tan(

3π4

+π

3

)=

− tan(3π/4) + tan(π/3)1− tan(3π/4) tan(π/3)

=

− −1 +√

31− (−1)

√3

=1−√

31 +√

3· 1−

√3

1−√

3=

4− 2√

3−2

= −2 +√

3

48. tan(7π/12) = tan(π

4+π

3

)=

tan(π/4) + tan(π/3)1− tan(π/4) tan(π/3)

=

1 +√

31− 1 ·

√3

=1 +√

31−√

3· 1 +

√3

1 +√

3=

4 + 2√

3−2

= −2−√

3

49. sin(3◦) cos(−87◦) + cos(3◦) sin(87◦) =sin(3◦) cos(87◦) + cos(3◦) sin(87◦) =sin(3◦ + 87◦) = sin(90◦) = 1

50. sin(34◦) cos(13◦)− cos(34◦) sin(13◦) =sin(34◦ − 13◦) = sin(21◦)

51. cos(π/2) cos(π/5) + sin(π/2) sin(π/5) =

cos(π

2− π

5

)= cos(3π/10)

52. cos(12◦) cos(3◦) + sin(12◦) sin(3◦) =cos (12◦ − 3◦) = cos(9◦)

53.tan(π/7) + tan(π/6)

1− tan(π/7) tan(π/6)= tan

(π

7+π

6

)=

tan(13π/42)

54.tan(π/3)− tan(π/6)

1 + tan(π/3) tan(π/6)= tan

(π

3− π

6

)=

tan(π/6) =√

3/3

55. sin(14◦) cos(35◦) + cos(14◦) sin(35◦) =sin(14◦ + 35◦) = sin(49◦)

56. cos(10◦) cos(20◦)− cos(80◦) sin(20◦) =cos(10◦) cos(20◦)− sin(10◦) sin(20◦) =cos(10◦ + 20◦) = cos(30◦) =

√3/2

57. G, cos(44◦) = sin(90◦ − 44◦) = sin(46◦)

58. B, − sin(46◦) = − cos(90◦ − 46◦) = − cos(44◦)

59. H, cos(46◦) = sin(90◦ − 46◦) = sin(44◦)

60. H, sin(136◦) = cos(90◦ − 136◦) = cos(−46◦)= cos(46◦) = sin(90◦ − 46◦) = sin(44◦)

61. F, sec(1) = csc(π

2− 1

)= csc

(π − 2

2

)

62. D, tan(π

7

)= cot

(π

2− π

7

)= cot

(5π14

)63. A, csc(π/2) = 1 = cos(0)

64. E, − sin(44◦) = − cos(90◦ − 44◦) = − cos(46◦)

65. Since α is in quadrant II and β is in

quadrant I, cosα = −

√1−

(35

)2

=

−√

1− 925

= −√

1625

= −45

and cosβ =√1−

(513

)2

=√

1− 25169

=√

144169

=1213

.

So sin(α+ β) = sinα cosβ + cosα sinβ =35· 12

13+−45· 5

13=

1665

.



66. Since α is in quadrant III and β is in

quadrant IV, we obtain

cosα = −

√1−

(−45

)2

= −√

1− 1625

=

−√

925

= −35

and sinβ = −

√1−

(1213

)2

=

−√

1− 144169

= −√

25169

= − 513.

So sin(α− β) = sinα cosβ − cosα sinβ =−45· 12

13− −3

5· −5

13= −63

65.

67. Since α is in quadrant I and β is in

quadrant III, we obtain

cosα =

√1−

(23

)2

=√

1− 49

=

√59

=√

53

and cosβ = −

√1−

(−12

)2

=

−√

1− 14

= −√

34

= −√

32.

So cos(α+ β) = cosα cosβ − sinα sinβ =√

53· −√

32− 2

3· −1

2=

2−√

156

.


quadrant II, we find

sinα =

√√√√1−(√

34

)2

=√

1− 316

=

√1316

=√

134

and sinβ =

√√√√1−(−√

23

)2

=

√1− 2

9=√

79

=√

73.

So cos(α− β) = cosα cosβ + sinα sinβ =√

34· −√

23

+√

134·√

73

=−√

6 +√

9112

69. Since α is in quadrant III and β is in


cosα = −

√1−

(−2425

)2

= − 725

and sinβ =

√1−

(−817

)2

=1517

.

Then sin(α− β) = sinα cosβ − cosα sinβ =−2425· −8

17− −7

25· 15

17=

297425

.


quadrant III, we find

cosα = −

√1−

(725

)2

= −2425

and cosβ = −

√1−

(−817

)2

= −1517

.

Then sin(α+ β) = sinα cosβ + cosα sinβ =725· −15

17+−2425· −8

17=

87425

.


quadrant IV, we find

cosα = −

√1−

(2425

)2

= − 725

and sinβ = −

√1−

(817

)2

= −1517

.

Then cos(α− β) = cosα cosβ + sinα sinβ =−725· 8

17+

2425· −15

17= −416

425.

72. Since α is in quadrant IV and β is in


cosα =

√1−

(−725

)2

=2425

and cosβ = −

√1−

(817

)2

= −1517

.

Then cos(α+ β) = cosα cosβ − sinα sinβ =2425· −15

17− −7

25· 8

17= −304

425.

73. cos(π/2− (−α)) = sin(−α) = − sinα

74. sinα cosπ − cosα sinπ =sinα · (−1)− cosα · 0 = − sinα

75. cos 180◦ cosα+ sin 180◦ sinα =(−1) · cosα+ 0 · sinα = − cosα



76. sin 180◦ cosα− cos 180◦ sinα =0 · cosα− (−1) · sinα = sinα

77. The period is 360◦, so sin(360◦ − α) =sin(−α) = − sinα

78. cosα cosπ + sinα sinπ =cosα · (−1) + sinα · 0 = − cosα

79. sin(90◦ − (−α)) = cos(−α) = cosα

80. The period is 360◦, so cos(360◦ − α) =cos(−α) = cosα

81.

sin(180◦ − α) =sin(180◦) cosα− cos(180◦) sinα =

sinα =

sin2 α

sinα=

1− cos2 α

sinα

82. We rewrite both sides:

cos(x− π/2) =cosx cos(π/2) + sinx sin(π/2) =

sinx =

cosx · sinxcosx

=

cosx tanx

83.

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

=

cos(x) cos(y)− sin(x) sin(y)cos(x) cos(y)

=

cos(x) cos(y)cos(x) cos(y)

− sin(x) sin(y)cos(x) cos(y)

=

1− tan(x) tan(y)

84.

sin(x+ y)sinx cos y

=

sinx cos y + cosx sin ysinx cos y

=

sinx cos ysinx cos y

+cosx sin ysinx cos y

=

1 + cotx tan y

85. Substitute the sum and difference sineidentities into the left-hand side to geta difference of two squares.

sin(α+ β) sin(α− β) =(sinα cosβ)2 − (cosα sinβ)2 =

sin2 α(1− sin2 β)− (1− sin2 α) sin2 β =sin2 α− sin2 α sin2 β − sin2 β + sin2 α sin2 β =

sin2 α− sin2 β

86. Substitute the sum and difference cosineidentities into the left-hand side to geta difference of two squares.

cos(α+ β) cos(α− β) =(cosα cosβ)2 − (sinα sinβ)2 =

(1− sin2 α) cos2 β − sin2 α(1− cos2 β) =cos2 β − sin2 α cos2 β − sin2 α+ sin2 α cos2 β =

cos2 β − sin2 α

87. Using the sum identity for cosine, we obtain

cos(x+ x) =cosx cosx− sinx sinx =

cos2 x− sin2 x

88. Applying the sum identity for sine, we get

sin(x+ x) =sinx cosx+ cosx sinx =

2 sinx cosx

89.

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

2 sin(x− y) =2(sinx cos y − cosx sin y) =2 sinx cos y − 2 cosx sin y



90.

cos(x− y) + cos(y − x) =cos(x− y) + cos(x− y) =

2 cos(x− y) =2(cosx cos y + sinx sin y) =2 cosx cos y − 2 sinx sin y

91.

tan(s+ t) tan(s− t) =

tan s+ tan t1− tan(s) tan(t)

· tan s− tan t1 + tan(s) tan(t)

=

tan2 s− tan2 t

1 + tan2(s) tan2(t)

92. Using the cofunction identity for tangent, weget

tan(π/4 + x) =cot(π/2− (π/4 + x)) =

cot(π/2− π/4− x) =cot(π/4− x)

93. In the proof, divide each term by cosα cosβ.

cos(α+ β)sin(α− β)

=

cosα cosβcosα cosβ

− sinα sinβcosα cosβ

sinα cosβcosα cosβ

− cosα sinβcosα cosβ

=

1− tan(α) tan(β)tan(α)− tan(β)

94. In the proof, divide each term by cosα cosβ.

cos(α− β)sin(α+ β)

=

cosα cosβcosα cosβ

+sinα sinβcosα cosβ

sinα cosβcosα cosβ

+cosα sinβcosα cosβ

=

1 + tan(α) tan(β)tan(α) + tan(β)

95. In the proof, multiply each term by cos(v− t).Also, the sum and difference identities forcosine expresses cos(v + t) cos(v − t) as adifference of two squares.

sec(v + t) =1

cos(v + t)=

cos(v − t)cos(v + t) cos(v − t)

=

cos(v − t)cos2(v) cos2(t)− sin2(v) sin2(t)

=

cos(v − t)cos2(v) cos2(t)− (1− cos2 v)(1− cos2 t)

=

cos(v − t)÷[cos2(v) cos2(t)−

(1− cos2 v − cos2 t+ cos2(v) cos2(t))]

=

cos(v − t)−1 + cos2 v + cos2 t

=

cos(v − t)cos2 v − sin2 t

=

cos(v) cos(t) + sin(v) sin(t)cos2 v − sin2 t

96. In the proof, multiply each term by sin(v+ t).Also, the sum and difference identities for sineexpresses sin(v − t) sin(v + t) as a differenceof two squares.

csc(v − t) =

1sin(v − t)

=

sin(v + t)sin(v − t) sin(v + t)

=

sin(v + t)sin2(v) cos2(t)− cos2(v) sin2(t)

=

sin(v + t)sin2 v(1− sin2 t)− (1− sin2 v) sin2 t

=

sin(v + t)÷[sin2 v − sin2(v) sin2(t)



− sin2(t) + sin2(v) sin2(t)]

=

sin(v + t)sin2 v − sin2 t

=

sin(v) cos(t) + cos(v) sin(t)sin2 v − sin2 t

=

97. In the proof, divide each term by cosx sin y.

cos(x+ y)cos(x− y)

=

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

=

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

− sin(x) sin(y)cos(x) sin(y)

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

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

=

cot(y)− tan(x)cot(y) + tan(x)

98. In the proof, divide each term by sinx sin y.

sin(x+ y)sin(x− y)

=

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

=

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

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

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

− cos(x) sin(y)sin(x) sin(y)

=

cot(y) + cot(x)cot(y)− cot(x)

99. In the proof, multiply each term by sin(α−β).Also, the sum and difference identities for sineexpresses sin(α+ β) sin(α− β) as a differenceof two squares.

sin(α+ β)sinα+ sinβ

=

sin(α+ β)sinα+ sinβ

· sin(α− β)sin(α− β)

=

sin2 α cos2 β − cos2 α sin2 β

(sinα+ sinβ) sin(α− β)=

sin2 α(1− sin2 β)− (1− sin2 α) sin2 β


sin2 α− sin2 α sin2 β − sin2 β + sin2 α sin2 β


sin2 α− sin2 β


(sinα− sinβ)(sinα+ sinβ)(sinα+ sinβ) sin(α− β)

=

sinα− sinβsin(α− β)

100. In the proof, we smultiply each term bycos(α−β). Also, the sum and difference iden-tities for cosine expresses cos(α+β) cos(α−β)as a difference of two squares.

cos(α+ β)cosα+ sinβ

=

cos(α+ β)cosα+ sinβ

· cos(α− β)cos(α− β)

=

cos2 α cos2 β − sin2 α sin2 β

(cosα+ sinβ) cos(α− β)=

cos2 α(1− sin2 β)− (1− cos2 α) sin2 β


cos2 α− cos2 α sin2 β − sin2 β + cos2 α sin2 β


cos2 α− sin2 β


(cosα− cosβ)(cosα+ cosβ)(cosα+ sinβ) cos(α− β)

=

cosα− sinβcos(α− β)

=

cosα− sinβcos(β − α)

103. If α = β = π/6, then sin(α+β) 6= sinα+sinβ

104. The following formulas will be useful

sin(90◦ − α) = cosα



andcos(90◦ − α) = sinα.

In particular, cos(89◦) = sin(1◦), cos(88◦) =sin(2◦), sin(89◦) = cos(1◦), and so on. Thus,for k = 1◦, ..., 44◦ we have

sin2(k◦) + sin2((90− k)◦) = 1.

Since sin2(45◦) = 1/2, we find

sin2(1◦) + sin2(2◦) + ...+ sin2(90◦) =44 + sin2(45◦) + sin2(90◦) =

45 +12.

Similarly, we obtain

cos2(1◦) + cos2(2◦) + ...+ cos2(90◦) = 44 +12.

Finally, we obtain

sin2(1◦) + ...+ sin2(90◦)cos2(1◦) + ...+ cos2(90◦)

=

45 + 1/244 + 1/2

=9189.

105. 1− sin2 α = cos2 α.

106.sinx

cscx− sinx· sinx

sinx=

sin2 x

1− sin2 x=

sin2 x

cos2 x= tan2 x

107. Since B = 2, the period isπ

B=π

2.

Solve 2x = kπ where k is an integer. Then the

asymptotes are x =kπ

2.

108. We find g(f(h(x))) = g(f(3x)) = g(sin 3x) =

sin(3x) + 2.

And, h(g(f(x))) = h(g(sinx)) = h(sinx+2) =

3(sinx+ 2) = 3 sin(x) + 6.

109. cosα = −√

1− (1/4)2 = −√

154

tanα =sinαcosα

=1/4

−√

15/4= − 1√

15= −√

1515

cscα =1

sinα= −4

secα =1

cosα= − 4√

15= −4

√15

15

cotα =1

tanα= − 15√

15= −√

15

110.

a) − sinx b) cosx

c) − tanx d) − cscx

e) secx f) − cotx

Thinking Outside the Box LIV

The angle spanned by the first seventeenrectangles is

tan−1

(11

)+tan−1

(1√2

)+ ...+tan−1

(1√17

)≈ 365◦

while the angle spanned by the first sixteenrectangles is

tan−1

(11

)+tan−1

(1√2

)+ ...+tan−1

(1√16

)≈ 351◦.

Thus, the 17th rectangle is the first rectangle thatoverlaps with the first rectangle.

6.3 Pop Quiz


−√

22· −1

2+√

22·√

32

=√

2 +√

64

2. 80◦, since sin 10◦ = cos(90◦ − 10◦)

3. Using the sum identity for sine, the answer issin 3x.

4. Since α and β are in quadrant I, we obtain

cosα =

√1−

(45

)2

=35

and sinβ =

√1−

(12

)2

=√

32

.

Then sin(α− β) = sinα cosβ − cosα sinβ =

45· 1

2− 3

5·√

32

=4− 3

√3

10.



For Thought

1. True,sin(2 · 21◦)

2=

2 sin(21◦) cos(21◦)2

= sin(21◦) cos(21◦).

2. True, by a cosine double angle identitycos(2

√2) = 2 cos2(

√2)− 1.

3. False, sin(

300◦

2

)=

√1− cos(300◦)

2.

4. True, sin(

400◦

2

)= −

√1− cos(400◦)

2

= −

√1− cos(40◦)

2.

5. False, tan(

7π/42

)= −

√1− cos(7π/4)1 + cos(7π/4)

.

6. True, tan(−π/4

2

)=

1− cos(−π/4)sin(−π/4)

=

1− cos(π/4)sin(−π/4)

7. False, if x = π/4 thensin(2 · π/4)

2=

sin(π/2)2

=12

and sin(π/4) =√

2/2.

8. False, since cos(2π/3) = −1/2 while√1 + cos(2x)

2is a non-negative number.

9. True, since 1− cosx ≥ 0 we find√(1− cosx)2 = |1− cosx| = 1− cosx

10. True, α is in quadrant III or IV, whileα/2 is in quadrant II.

6.4 Exercises

1. sin(2 · 45◦) = 2 sin(45◦) cos(45◦) =

2 ·√

22·√

22

= 2 · 24

= 1.

2. cos(2 · 30◦) = 2 cos2(30◦)− 1 =

2(√

3/2)2 − 1 = 2 · 34− 1 =

32− 1 =

12

3. tan(2 · 30◦) =2 tan(30◦)

1− tan2(30◦)=

2(√

3/3)1− (

√3/3)2

=

2√

3/31− 1/3

=2√

3/32/3

=√

3

4. cos(2 · 90◦) = 2 cos2(90◦)− 1 = 2(0)2 − 1 =

0− 1 = −1

5. sin(

2 · 3π4

)= 2 sin(3π/4) cos(3π/4) =

2 ·√

22· −√

22

= 2 · −24

= −1

6. cos(

2 · 2π3

)= 2 cos2(2π/3)− 1 =

2(−1/2)2 − 1 = 2 · 14− 1 =

12− 1 = −1

2

7. tan(

2 · 2π3

)=

2 tan(2π/3)1− tan2(2π/3)

=2(−√

3)1− (−

√3)2

=−2√

31− 3

=−2√

3−2

=√

3

8. sin(

2 · π3

)= 2 sin(π/3) cos(π/3)

= 2 ·√

32· 1

2= 2 ·

√3

4=√

32

9. cos(

30◦

2

)=

√1 + cos(30◦)

2=√

1 +√

3/22

· 22

=

√2 +√

34

=

√2 +√

32

10. cos(π/42

)=

√1 + cos(π/4)

2=√

1 +√

2/22

· 22

=

√2 +√

24

=

√2 +√

22

11. sin(

30◦

2

)=

√1− cos(30◦)

2=√

1−√

3/22

· 22

=

√2−√

34

=

√2−√

32

12. sin(−π/3

2

)= −

√1− cos(−π/3)

2=

−

√1− 1/2

2· 2

2= −

√14

= −12


6.4 DOUBLE AND HALF-ANGLE IDENTITIES 371

13. tan(

30◦

2

)=

1− cos(30◦)sin(30◦)

=

1−√

3/21/2

· 22

= 2−√

3

14. tan(

3π/42

)=

1− cos(3π/4)sin(3π/4)

=

1− (−√

2/2)√2/2

· 22

=2 +√

2√2·√

2√2

=

2√

2 + 22

=√

2 + 1

15. sin(

45◦

2

)=

√1− cos(45◦)

2=√

1−√

2/22

· 22

=

√2−√

24

=

√2−√

22

16. tan(

150◦

2

)=

1− cos(150◦)sin(150◦)

=

1− (−√

3/2)1/2

· 22

= 2 +√

3

17. Positive, 118.5◦ is in quadrant II

18. Negative, 222.5◦ is in quadrant III

19. Negative, 100◦ is in quadrant II

20. Negative, 9π/7 is in quadrant III

21. Negative, −5π/12 is in quadrant IV

22. Positive, 17π/12 is in quadrant III

23. sin(2 · 13◦) = sin 26◦

24. − cos(

2 · π5

)= − cos(2π/5)

25. cos(2 · 22.5◦) = cos 45◦ =√

2/2

26. cos(2 · (−π/8)) = cos(−π/4) =√

2/2

27.12· 2 tan 15◦

1− tan2 15◦=

12· tan(2 · 15◦) =

12· tan 30◦ =

12·√

33

=√

36

28.12· 2 tan 30◦

1− tan2 30◦=

12· tan(2 · 30◦) =

12· tan 60◦ =

12·√

3 =√

32

29. tan(

12◦

2

)= tan 6◦

30.1− cos 8◦

sin 8◦= tan

(8◦

2

)= tan 4◦

31. 2 sin(π

9− π

2

)cos

(π

9− π

2

)=

sin(

2 ·(π

9− π

2

))= sin

(2π9− π

)=

sin(−7π/9) = − sin(7π/9).

32. cos(

2 ·(π

5− π

2

))= cos

(2π5− π

)=

cos(−3π/5) = cos(3π/5)

33. cos(2 · (π/9)) = cos(2π/9)

34.2 tan 5

1− tan2 5= tan(2 · 5) = tan 10

35. c, since sin2 x = 1− cos2 x

36. e, since tan(2x) =2 tanx

1− tan2 x

37. g, for cos(x

2

)= ±

√1 + cos(x)

2

38. i, for tanx

2=

sinx1 + cosx

and cotx is

the reciprocal of tanx.

39. a, for sin(2x) = 2 sinx cosx

40. j, for tan2(x) = sec2(x)− 1

41. h, for tan(x

2

)= ±

√1− cosx1 + cosx

42. b, since cos(2x) = cos2 x− sin2 x

43. f, since sin(x

2

)= ±

√1− cos(x)

2

44. d, since cos2 x = 1− sin2 x

45. Since cos(2α) = 2 cos2 α− 1, we get

2 cos2 α− 1 =35

2 cos2 α =85

cos2 α =45

cosα = ± 2√5.



But 0◦ < α < 45◦, so cosα =2√5

and

sinα =

√1−

(2√5

)2

=√

1− 45

=√15

=1√5.

Furthermore, secα =√

52

, cscα =√

5,

tanα =1/√

52/√

5=

12

, cotα = 2.

46. Since cos(2α) = 2 cos2 α− 1, we obtain

2 cos2 α− 1 =13

2 cos2 α =43

cos2 α =23

cosα = ±√

2√3

cosα = ±√

63.

But 180◦ < α < 225◦, so cosα = −√

63

and

sinα = −

√√√√1−(−√

63

)2

= −√

1− 69

=

−√

13

= − 1√3

= −√

33.

Furthermore, secα = − 3√6

, cscα = −√

3,

tanα =−√

3/3−√

6/3=

1√2

=√

22

, cotα =√

2.

47. Since 2α = sin−1(5/13) ≈ 22.6◦, we find

cos 2α =

√1−

(513

)2

=1213

. Then

sinα =√

1− cos 2α2

=

√1− 12/13

2=√

2626

,

cosα =√

1 + cos 2α2

=

√1 + 12/13

2=

5√

2626

,

tanα =sinαcosα

=√

26/265√

26/26=

15

cscα =1

sinα=√

26,

secα =1

cosα=√

265

,

and cotα =1

tanα= 5.

48. Since 2α lies in quadrant III, we obtain

cos 2α = −

√1−

(−817

)2

= −1517

.

Then sinα =√

1− cos 2α2

=√1− (−15/17)

2=

4√

1717

,

cosα = −√

1 + cos 2α2

= −

√1 + (−15/17)

2=

−√

1717

, tanα =sinαcosα

=4√

17/17−√

17/17= −4

cscα =1

sinα=√

174

,

secα =1

cosα= −√

17,

and cotα =1

tanα= −1

4.

49. By a half-angle identity, we have

−√

1 + cosα2

= −14

1 + cosα2

=116

1 + cosα =18

cosα = −78.

But π ≤ α ≤ 3π/2,

so sinα = −

√1−

(−7

8

)2

=

−√

1− 4964

= −√

1564

= −√

158.

Furthermore, secα = −87

, cscα = − 8√15

,



tanα =−√

15/8−7/8

=√

157

, cotα =7√15

.

50. By a half-angle identity, we obtain

−√

1− cosα2

= −13

1− cosα2

=19

1− cosα =29

cosα =79.

But α is in qudrant IV,

so sinα = −

√1−

(79

)2

= −√

1− 4981

=

−√

3281

= −√

329

= −4√

29

.

Furthermore, secα =97

, cscα = − 9√32

, or

cscα = −9√

28

, tanα =−√

32/97/9

= −√

327

, or

tanα = −4√

27

, and cotα = − 7√32

= −7√

28

.

51. By a half-angle identity, we find√1− cosα

2=

45

1− cosα2

=1625

1− cosα =3225

cosα = − 725.

Since (π/2 + 2kπ) ≤ α/2 ≤ (π+ 2kπ) for someinteger k, (π + 4kπ) ≤ α ≤ (2π + 4kπ).So α is in quadrant III because cosα < 0.

sinα = −

√1−

(− 7

25

)2

= −√

1− 49625

=

−√

576625

= −2425.

Furthermore, secα = −257

, cscα = −2524

,

tanα =−24/25−7/25

=247

, and cotα =724

.

52. By using the half-angle identity for sine,we get √

1− cosα2

=15

1− cosα2

=125

1− cosα =225

cosα =2325.

Since (π/2 + 2kπ) ≤ α/2 ≤ (π + 2kπ) forsome integer k, (π + 4kπ) ≤ α ≤ (2π + 4kπ).So α is in quadrant IV because cosα > 0.

Then sinα = −

√1−

(2325

)2

=

−√

1− 529625

= −√

96625

= −4√

625

.

Furthermore, secα =2523

, cscα = − 254√

6=

−25√

624

, tanα =−4√

6/2523/25

= −4√

623

,

and cotα = − 234√

6= −23

√6

24.

53.

cos4 s− sin4 s =(cos2 s− sin2 s)(cos2 s+ sin2 s) =

cos(2s) · (1) =cos(2s)

54.

= 2 sin(s) sin(π/2− s)= 2 sin(s) cos(s)

sin(2s)

55.

cos(2t+ t) =cos(2t) cos(t)− sin(2t) sin(t) =[

cos2 t− sin2 t]

cos t− [2 sin t cos t] sin t =

cos3 t− sin2 t cos t− 2 sin2 t cos t =cos3 t− 3 sin2 t cos t



56.

sin(4t)4

=

2 sin(2t) cos(2t)4

=

2 · 2 sin t cos t · (cos2 t− sin2 t)4

=

sin t cos t(cos2 t− sin2 t) =cos3 t sin t− sin3 t cos t

57.

cos(2x) + cos(2y)sin(x) + cos(y)

=

1− 2 sin2 x+ 2 cos2 y − 1sinx+ cos y

=

2cos2 y − sin2 x

sinx+ cos y=

2(cos y − sinx)(cos y + sinx)

sinx+ cos y=

2 cos(y)− 2 sin(x)

58.

(sinα− cosα)2 =sin2 α− 2 sinα cosα+ cos2 α =

1− 2 sinα cosα =1− sin(2α)

59.cos 2xsin2 x

=

1− 2 sin2 x

sin2 x=

1sin2 x

− 2 · sin2 x

sin2 x=

csc2 x− 2

60.

cos(2s)cos2 s

=

1− 2 sin2 s

cos2 s=

1cos2 s

− 2 · sin2 s

cos2 s=

sec2 s− 2 tan2 s

61.

=sin2 u

1 + cosu

=1− cos2 u

1 + cosu

=(1− cosu)(1 + cosu)

1 + cosu

= (1− cosu) · 22

= 2 · 1− cosu2

2 sin2(u/2)

62.

=1− tan2 y

1 + tan2 y

=1− tan2 y

sec2 y

=1

sec2 y− tan2 y

sec2 y

= cos2(y)− sin2 y/ cos2 y

1/ cos2 y

= cos2 y − sin2 y

cos(2y)

63. Multiply and divide by cosx.

=secx+ cosx− 2

secx− cosx· cosx

cosx

=1 + cos2 x− 2 cosx

1− cos2 x

=cos2 x− 2 cosx+ 1

1− cos2 x

=(1− cosx)2

(1 + cosx)(1− cosx)

=1− cosx1 + cosxtan2(x/2)

64. Multiply and divide by cosx.

=2 secx+ 2

secx+ 2 + cosx· cosx

cosx

=2 + 2 cosx

1 + 2 cosx+ cos2 x



=2(1 + cosx)(1 + cosx)2

=2

1 + cosx

=(

cos2(x

2

))−1

sec2(x

2

)

65.

1− sin2(x/2)1 + sin2(x/2)

=

1−(

1− cosx2

)1 +

(1− cosx

2

) · 22

=

2− (1− cosx)2 + (1− cosx)

=

1 + cosx3− cosx

66.

1− cos2(x/2)1− sin2(x/2)

=

1−(

1 + cosx2

)1−

(1− cosx

2

) · 22

=

2− (1 + cosx)2− (1− cosx)

=

1− cosx1 + cosx

67. It is not an identity. If x = π/4, thensin(2 · π/4) = sin(π/2) = 1 and2 sin(π/4) = 2 · (

√2/2) =

√2.

68. It is not an identity. If x = π, thencos(2π)

2=

12

and cos(π) = −1.

69. It is not an identity. If x = 2π/3, then

tan(

2π/32

)= tan(π/3) =

√3 and

12· tan(2π/3) =

12· (−√

3).

70. It is not an identity. If x = 4π/3, then

tan(

4π/32

)= tan(2π/3) = −

√3 while√

1− cos(4π/3)1 + cos(4π3)

is a positive number.

71. It is not an identity. If x = π/2, then

sin (2 · π/2) sin(π/22

)= sin(π) sin(π/4)

= 0 ·√

22

= 0 and sin2(π/2) = 1.

72. It is not an identity. If x = π/4, thentan (π/4) + tan (π/4) = 1 + 1 = 2 andtan(2 · π/4) = tan(π/2) = 0 .

73. It is an identity. The proof below usesthe double-angle identity for tangent.

cot(x/2)− tan(x/2) =1

tan(x/2)− tan(x/2) =

1− tan2(x/2)tan(x/2)

=

2 · 1− tan2(x/2)2 · tan(x/2)

=

2 · 1tanx

=

2 · cosxsinx

· sinxsinx

=

2 sinx cosxsin2 x

=

sin(2x)sin2 x

74. It is an identity. The proof uses thedouble-angle identity for sine.

csc2(x/2) + sec2(x/2) =1

sin2(x/2)+

1cos2(x/2)

=

cos2(x/2) + sin2(x/2)sin2(x/2) cos2(x/2)

=

1sin2(x/2) cos2(x/2)

=



44 sin2(x/2) cos2(x/2)

=

4[2 sin(x/2) cos(x/2)

]2 =

4[sin(2 · (x/2))]2

=

4sin2 x

=

4 csc2 x

75. Note, cosα = −

√1−

(35

)2

= −45

.

Then sin 2α = 2 sinα cosα =

2 · 35· −4

5= −24

25.

76. Since tanα = − 815

and α lies in quadrant IV,

we obtain cosα =15√

82 + 152=

1517

and sinα = − 8√82 + 152

= − 817

.

Then sin 2α = 2 sinα cosα =

2 · −817· 15

17= −240

289.

77. cos 2α = 1− 2 sin2 α = 1− 2(

817

)2

=161289

78. Note, cosα = −

√1−

(−45

)2

= −35

.

Then tanα =sinαcosα

=−4/5−3/5

=43

and

tan 2α =2 tanα

1− tan2 α=

2(4/3)1− (4/3)2

=

24/9−7/9

= −247

.

79. Since tanα =35

, sinα =3√34

and

cosα =5√34

. By a half-angle identity,

we obtain

tanα

2=

sinα1 + cosα

=3√34

1 + 5√34

=3

5 +√

34

and since tanα

2=BD

5then

BD =15

5 +√

34

=15(5−

√34)

25− 34

=15(√

34− 5)9

BD =5√

34− 253

.

80. Let CD be the distance between C and D.

"""""""""""""""

!!!!

!!!!

!!!!

!!!

β

β

C

D

BA

2

!!!!

!!!!

!!!!

!!!

10

Note, tanβ =15

and tan 2β =CD + 2

10.

Using the double angle identity for tangent,one finds

tan 2β =2 tanβ

1− tan2 β

=2(1/5)

1− (1/5)2

=2/5

24/25

CD + 210

=512

CD + 2 =5012



CD =136

81. Since the base of the TV screen is b = d cosαand its height is h = d sinα, then the area Ais given by

A = bh

= (d cosα)(d sinα)= d2 cosα sinα

A =d2

2sin(2α).

82. The area is A =322

2sin(2 · 37.2◦) ≈ 493.1 in.2

85.1

1− sinx+

11 + sinx

=1 + sinx+ 1− sinx

1− sin2 x=

2cos2 x

= 2 sec2 x

86. Factor cosx as follows:cosx

(cos2 x+ sin2 x

)sinx

=cosx · 1

sinx= cotx

87. a) cosx cos y − sinx sin y

b) cosx cos y + sinx sin y

88. Let A = 26◦, and let a and b be the sidesopposite and adjacent to A, respectively. SincesinA = a/38.6, we find

a = 38.6 sin 26◦ ≈ 16.9 in.

Since cosA = b/38.6, we find

b = 38.6 cos 26◦ ≈ 34.7 in.

89. a)12

b) −1 c) undefined

d) 2 e) −1 f) 1

90. The race car will lap the track in

circumferencespeed

=2π 1

4

180=

π

360hour =

π

6min

Then the angular speed is

2πtime around 1 lap

=2ππ/6

= 12 radians/min.

Thinking Outside the Box LV

An eighth of the region that gets watered by allsprinklers is region Ra below with vertices B, C,and D.

B

A

C

D

30o

The area of Ra is the area of the sector deter-mined by C, A , and D minus the area of triangle∆ABD. In the figure above, we have AB =

√2,

BC = 2−√

2, BD =√

3− 1, angle 〈ABD = 135◦,and 〈CBD = 45◦.

The area of the sector is

As =12

(22 π

12

)=π

6

and the area of ∆ABD is

At =12

(AB)(BD) sin 135◦

=12

√2(√

3− 1)√

22

At =√

3− 12

.

Thus, the area watered by all sprinklers is

Area = 8 (As −At)

= 8

(π

6−√

3− 12

)

Area =4π3

+ 4− 4√

3.



6.4 Pop Quiz

1. Since sinα = 1/4 and cosα = −√

15/4, we find

sin 2α = 2 sinα cosα = 2 · 14· −√

154

= −√

158.

2. Since cosα = −3/5 andπ

2<α

2<

3π4

,

we obtain

sinα

2=

√1− (−3/5)

2=√

45

=2√

55

3.

sin4 x− cos4 x =(sin2 x− cos2 x)(sin2 x+ cos2 x) =

−(cos2 x− sin2 x)(1) =− cos 2x


a) x =2 · v2

0 sin θ cos θ2 · 16

=v20 sin(2θ)

32.

b) A graph of x =502 sin(2θ)

32or x = 78.125 sin 2θ

is sketched below

45 135theta

x

c) For any v0, the maximum value of

x =v20 sin(2θ)

32is attained when sin(2θ) = 1,

i.e., when θ = 45◦. Yes, this value θ = 45◦

maximizes x for any velocity v0.

d) The initial velocity v0 is obtained by solving thefollowing equations (note: 55 yards = 165feet).

v20 sin(2 · 45◦)

32= 165

v20 · 132

= 165

v20 = (32)165

v0 =√

32(165) ft/secv0 ≈ 72.6636 ft/sec

v0 ≈ 72.6636 · 36005280

mph

v0 ≈ 49.5 mph

e) Taking air resistance into account, the actualinitial velocity is larger than the answer foundin part d).

For Thought

1. True, sin 45◦ cos 15◦ =(1/2) [sin(45◦ + 15◦) + sin(45◦ − 15◦)] =0.5 [sin 60◦ + sin 30◦].

2. False, cos(π/8) sin(π/4) =(1/2) [sin(π/8 + π/4)− sin(π/8− π/4)] =0.5 [sin(3π/8)− sin(−π/8)] =0.5 [sin(3π/8) + sin(π/8)] .

3. True, 2 cos(6◦) cos(8◦) =cos(6◦ − 8◦) + cos(6◦ + 8◦) =cos(−2◦) + cos(14◦) = cos(2◦) + cos(14◦).

4. False, sin(5◦)− sin(9◦) =

2 cos(

5◦ + 9◦

2

)sin(

5◦ − 9◦

2

)=

2 cos(7◦) sin(−2◦) = −2 cos(7◦) sin(2◦).

5. True, cos(4) + cos(12) =

2 cos(

4 + 122

)cos

(4− 12

2

)=

2 cos(8) cos(−4) = 2 cos(8) cos(4).

6. False, cos(π/3)− cos(π/2) =

−2 sin(π/3 + π/2

2

)sin(π/3− π/2

2

)=

−2 sin(5π/12) sin(−π/12) =2 sin(5π/12) sin(π/12).


6.5 PRODUCT AND SUM IDENTITIES 379

7. True,√

2 sin(π/6 + π/4) =√2 [sin(π/6) cos(π/4) + cos(π/6) sin(π/4)] =√

2[sin(π/6) · 1√

2+ cos(π/6) · 1√

2

]=

sin(π/6) + cos(π/6).

8. True,12

sin(π/6) +√

32

cos(π/6) =

12· 1

2+√

32·√

32

=14

+34

= 1 = sin(π/2).

9. True, y = cos(π/3) sinx+ sin(π/3) cosx =sin(x+ π/3).

10. True, since y = cos(π/4) sinx+ sin(π/4) cosx= sin(x+ π/4) by the sum identity for sine.

6.5 Exercises

1.12

[cos(13◦ − 9◦)− cos(13◦ + 9◦)] =

0.5 [cos 4◦ − cos 22◦]

2.12

[cos(34◦ − 39◦) + cos(34◦ + 39◦)] =

0.5 [cos(−5◦) + cos 73◦] = 0.5 [cos 5◦ + cos 73◦]

3.12

[sin(16◦ + 20◦) + sin(16◦ − 20◦)] =

0.5 [sin 36◦ + sin(−4◦)] = 0.5 [sin 36◦ − sin 4◦]

4.12

[sin(9◦ + 8◦)− sin(9◦ − 8◦)] =

0.5 [sin 17◦ − sin 1◦]

5.12

[sin(5◦ + 10◦) + sin(5◦ − 10◦)] =

0.5 [sin 15◦ + sin(−5◦)] = 0.5 [sin 15◦ − sin 5◦]

6.12

[cos(6◦ − 8◦) + cos(6◦ + 8◦)] =

0.5 [cos(−2◦) + cos 14◦] = 0.5 [cos 2◦ + cos 14◦]

7.12

[cos

(π

6− π

5

)+ cos

(π

6+π

5

)]=

0.5[cos

(−π30

)+ cos

(11π30

)]=

0.5[cos

(π

30

)+ cos

(11π30

)]8.

12

[cos(

2π9− 3π

4)− cos(

2π9

+3π4

)]

=

0.5 [cos(−19π/36)− cos(35π/36)] =0.5 [cos(19π/36)− cos(35π/36)]

9.12

[cos(5y2 − 7y2) + cos(5y2 + 7y2)

]=

0.5[cos(−2y2) + cos(12y2)

]=

0.5[cos(2y2) + cos(12y2)

]10.

12

[sin(3t+ 5t)− sin(3t− 5t)] =

0.5 [sin(8t)− sin(−2t)] =0.5 [sin(8t) + sin(2t)]

11.12

[sin((2s− 1) + (s+ 1))+

sin((2s− 1)− (s+ 1))]

=

0.5[sin(3s) + sin(s− 2)

]12.

12

[cos((3t− 1)− (2t+ 3))−

cos((3t− 1) + (2t+ 3))]

=

12

[cos(t− 4)− cos(5t+ 2)

]13.

12

[cos(52.5◦ − 7.5◦)− cos(52.5◦ + 7.5◦)] =

12

[cos 45◦ − cos 60◦] =

12

[√2

2− 1

2

]=√

2− 14

14.12

[cos(105◦ − 75◦) + cos(105◦ + 75◦)] =

12

[cos(30◦) + cos 180◦] =12

[√3

2+ (−1)

]=

√3− 24



15.12

[sin(

13π24

+5π24

)+ sin

(13π24− 5π

24

)]=

12

[sin(18π/24) + sin(8π/24)] =

12

[sin(3π/4) + sin(π/3)] =

12

[√2

2+√

32

]=√

2 +√

34

16.12

[sin(

5π24

+−π24

)− sin

(5π24− −π

24

)]=

12

[sin(4π/24)− sin(6π/24)] =

12

[sin(π/6)− sin(π/4)] =

12

[12−√

22

]=

1−√

24

17.2 cos

(12◦ + 8◦

2

)sin(

12◦ − 8◦

2

)=

2 cos 10◦ sin 2◦

18.2 sin

(7◦ + 11◦

2

)cos

(7◦ − 11◦

2

)=

2 sin 9◦ cos (−2◦) = 2 sin 9◦ cos 2◦

19.−2 sin

(80◦ + 87◦

2

)sin(

80◦ − 87◦

2

)=

−2 sin 83.5◦ sin (−3.5◦) = 2 sin 83.5◦ sin 3.5◦

20.2 cos

(44◦ + 31◦

2

)cos

(44◦ − 31◦

2

)=

2 cos 37.5◦ cos 6.5◦

21.2 cos

(3.6 + 4.8

2

)sin(

3.6− 4.82

)=

2 cos (4.2) sin (−0.6) = −2 cos(4.2) sin(0.6)

22.2 sin

(5.1 + 6.3

2

)cos

(5.1− 6.3

2

)=

2 sin(5.7) cos (−0.6) = 2 sin(5.7) cos(0.6)

23.

−2 sin(π/3 + π/5

2

)sin(π/3− π/5

2

)=

−2 sin(4π/15) sin (π/15) =

24.

2 cos(

1/2 + 2/32

)cos

(1/2− 2/3

2

)=

2 cos(7/12) cos(−1/12) =2 cos(7/12) cos(1/12)

25.

−2 sin(

(5y − 3) + (3y + 9)2

)·

sin(

(5y − 3)− (3y + 9)2

)=

−2 sin(4y + 3) sin (y − 6)

26.

2 cos

((6t2 − 1) + (4t2 − 1)

2

)·

cos

((6t2 − 1)− (4t2 − 1)

2

)=

2 cos(5t2 − 1) cos(t2)

27.2 cos

(5α+ 8α

2

)sin(

5α− 8α2

)=

2 cos (6.5α) sin (−1.5α) =−2 cos(6.5α) sin(1.5α)

28.2 sin

(3s+ 5s

2

)cos

(3s− 5s

2

)=

2 sin(4s) cos (−s) = 2 sin(4s) cos s

29.2 sin

(75◦ + 15◦

2

)cos

(75◦ − 15◦

2

)=

2 sin 45◦ cos (30◦) = 2 ·√

22

√3

2=√

62

30.2 cos

(285◦ + 15◦

2

)sin(

285◦ − 15◦

2

)=

2 cos 150◦ sin 135◦ = 2 · −√

32

√2

2= −√

62



31.

−2 sin

−π24

+7π24

2

sin

−π24− 7π

242

=

−2 sin(3π/24) sin (−4π/24) =−2 sin(π/8) sin (−π/6) =

−2 sin(π/42

)· −1

2= −2

√1− cos(π/4)

2· −1

2

=

√1−√

2/22

· 22

=

√2−√

24

=

√2−√

22

32.

2 cos(

5π/24 + π/242

)cos

(5π/24− π/24

2

)=

2 cos(3π/24) cos(2π/24) =

2 cos(π/8) cos(π/12) =

2 cos(π/42

)cos

(π/62

)=

2

√1 + cos(π/4)

2·

√1 + cos(π/6)

2=

2

√1 +√

2/22

· 22·

√1 +√

3/22

· 22

=

2

√2 +√

24

·

√2 +√

34

=

2

√4 + 2

√2 + 2

√3 +√

616

=√4 + 2

√2 + 2

√3 +√

62

33. Since a = 1 and b = −1, we obtain

r =√

12 + (−1)2 =√

2.

If the terminal side of α passes through(1,−1), then cosα = a/r = 1/

√2 and

sinα = b/r = −1/√

2. Choose α = −π/4.

Thus, sinx− cosx = r sin(x+ α) =√2 sin(x− π/4).

34. Since a = 2 and b = 2, r =√

22 + 22 = 2√

2.If the terminal side of α passes through (2, 2),

then cosα = a/r =2

2√

2=

1√2

and

sinα = b/r =2

2√

2=

1√2. Choose α = π/4.

Thus, 2 sinx+ 2 cosx = r sin(x+ α) =2√

2 sin(x+ π/4).

35. Since a = −1/2 and b =√

3/2, we obtain

r =√

(−1/2)2 + (√

3/2)2 = 1. If theterminal side of α passes through(−1/2,

√3/2), then cosα = a/r = a/1 =

a = −1/2 and sinα = b/r = b/1 = b =√

3/2.

Choose α = 2π/3. So −12

sinx+√

32

cosx =

r sin(x+ α) = sin(x+ 2π/3).

36. Since a =√

2/2 and b = −√

2/2, we find

r =√

(√

2/2)2 + (√

2/2)2 = 1. If the terminalside of α passes through (

√2/2,−

√2/2),

then cosα = a/r = a/1 = a =√

2/2 andsinα = b/r = b/1 = b = −

√2/2.

Choose α = −π/4. Thus,√

22

sinx−√

22

cosx = r sin(x+ α) =

sin(x− π/4).

37. Since a =√

3/2 and b = −1/2, we have

r =√

(√

3/2)2 + (−1/2)2 = 1.

If the terminal side of α passes through(√

3/2,−1/2), then

cosα = a/r = a/1 = a =√

3/2

and

sinα = b/r = b/1 = b = −1/2.

Choose α = −π/6. Thus,√

32

sinx− 12


sin(x− π/6).

38. Since a = −√

3/2 and b = −1/2, we get

r =√

(−√

3/2)2 + (−1/2)2 = 1. If theterminal side of α passes through(−√

3/2,−1/2), then cosα = a/r = a/1 =



a = −√

3/2 and sinα = b/r = b/1 =b = −1/2. Choose α = 7π/6. Thus,

−√

32

sinx− 12


sin(x+ 7π/6).

39. Since a = −1 and b = 1, we obtainr =

√(−1)2 + 12 =

√2. If the terminal side

of α passes through (−1, 1), thencosα = a/r = −1/

√2 and

sinα = b/r = 1/√

2. Choose α = 3π/4.Then y = − sinx+ cosx = r sin(x+ α) =√

2 sin(x+ 3π/4). Amplitude is√

2, periodis 2π, and phase shift is −3π/4.

-2Pi -Pi/4 3Pi/4 2Pix

-1

1

y

40. Since a = 1 and b =√

3, we find

r =√

12 +√

32

= 2. If the terminal side ofα passes through (1,

√3), then cosα = a/r =

1/2 and sinα = b/r =√

3/2. Choose α = π/3.Thus, y = sinx +

√3 cosx = r sin(x + α) =

2 sin(x + π/3). Amplitude is 2, period is 2π,and phase shift is −π/3.

-2Pi -5Pi/6 Pi/6 2Pix

-2

2

y

41. Since a =√

2 and b = −√

2, we obtain

r =√√

22

+ (−√

2)2 = 2. If the terminal sideof α passes through (

√2,−√

2), thencosα = a/r =

√2/2 and

sinα = b/r = −√

2/2. Choose α = −π/4.So y =

√2 sinx −

√2 cosx = r sin(x + α) =

2 sin(x − π/4). Amplitude is 2, period is 2π,and phase shift is π/4.

-5Pi/4 -Pi/4 3Pi/4 2Pix

2

y

42. Since a = 2 and b = −2, we getr =

√22 + (−2)2 =

√8 = 2

√2. If the terminal

side of α passes through (2,−2), then cosα =a/r = 2/(2

√2) = 1/

√2 and sinα = b/r =

−2/(2√

2) = −1/√

2. Choose α = −π/4.Then y = 2 sinx − 2 cosx = r sin(x + α) =2√

2 sin(x− π/4). Amplitude is 2√

2, period is2π, and phase shift is π/4.

-5Pi/4 3Pi/4 2Pix

3

y

43. Since a = −√

3 and b = −1, we findr =

√(−√

3)2 + (−1)2 = 2. If the terminalside of α passes through (−

√3,−1), then

cosα = a/r = −√

3/2 andsinα = b/r = −1/2. Choose α = 7π/6.Then y = −

√3 sinx− cosx = r sin(x+ α) =

2 sin(x + 7π/6). Amplitude is 2, period is 2π,and phase shift is −7π/6.

-2Pi/3 Pi/3 4Pi/3x

-2

2

y

44. Since a = −1/2 and b = −√

3/2, we obtain

r =√

(−1/2)2 + (−√

3/2)2 = 1. If theterminal side of α passes through(−1/2,−

√3/2), then cosα = a/r = −1/2

and sinα = −√

3/2 Choose α = 4π/3.Thus, y = −(1/2) sinx− (

√3/2) cosx =

r sin(x + α) = sin(x + 4π/3). Amplitude is 1,period is 2π, and phase shift is −4π/3.



-2Pi -5Pi/6 7Pi/6 2Pix

-1

1

y

45. Since a = 3 and b = 4, the amplitude is√32 + 42 =

√25 = 5. If the terminal side of

α passes through (3, 4), then tanα = 4/3 andα = tan−1(4/3) ≈ 0.9. Phase shift is −0.9.

46. Since a = 1 and b = 5, the amplitude is√12 + 52 =

√26. If the terminal side of

α passes through (1, 5), then tanα = 5/1 = 5and α = tan−1(5) ≈ 1.4. Phase shift is −1.4.

47. Since a = −6 and b = 1, amplitude is√(−6)2 + 12 =

√37.

If the terminal side of α passes through(−6, 1), then tanα = −1/6. Using a calcu-lator, one gets tan−1(−1/6) ≈ −0.165 whichis an angle in quadrant IV. Since (−6, 1) is inquadrant II and π is the period of tanx,

α ≈ −0.165 + π ≈ 3.0.

The phase shift is −3.0.

48. Since a = −√

5 and b = 2, amplitude is√(−√

5)2 + 22 = 3. If the terminal side of αpasses through (−

√5, 2), then

tanα = −2/√

5. Using a calculator, one getstan−1(−2/

√5) ≈ −0.73 which is an angle in

quadrant IV. Since (−√

5, 2) is in quadrant IIand π is the period of tanx, α ≈ −0.73 + π ≈2.4. The phase shift is −2.4.

49. Since a = −3 and b = −5, amplitude is√(−3)2 + (−5)2 =


of α passes through (−3,−5), thentanα = 5/3. Using a calculator, one getstan−1(5/3) ≈ 1.03 which is an angle inquadrant I. Since (−3,−5) is in quadrant IIIand π is the period of tanx, α ≈ 1.03+π ≈ 4.2.Phase shift is −4.2.

50. Since a = −√

2 and b = −√

7, amplitude is√(−√

2)2 + (−√

7)2 = 3. If the terminal sideof α passes through (−

√2,−√

7), thentanα =

√7/√

2. Using a calculator, one getstan−1(

√7/√

2) ≈ 1.08 which is an angle inquadrant I. Since (−

√2,−√

7) is in quadrantIII and π is the period of tanx, α ≈ 1.08+π ≈4.2. Phase shift is −4.2.

51. By using a sum-to-product identity, we get

sin(3t)− sin(t)cos(3t) + cos(t)

=

2 cos(

3t+ t

2

)sin(

3t− t2

)2 cos

(3t+ t

2

)cos

(3t− t

2

) =

2 cos(2t) sin t2 cos(2t) cos t

=

tan t

52. By using a sum-to-product identity, we obtain

sin(3x) + sin(5x)sin(3x)− sin(5x)

=

2 sin(

3x+ 5x2

)cos

(3x− 5x

2

)2 cos

(3x+ 5x

2

)sin(

3x− 5x2

) =

2 sin(4x) cos(−x)2 cos(4x) sin(−x)

=

−2 sin(4x) cosx2 cos(4x) sinx

=

−tan(4x)tanx

53. By using a sum-to-product identity, we find

cosx− cos(3x)cosx+ cos(3x)

=

−2 sin(x+ 3x

2

)sin(x− 3x

2

)2 cos

(x+ 3x

2

)cos

(x− 3x

2

) =

−2 sin(2x) sin(−x)2 cos(2x) cos(−x)

=



2 sin(2x) sinx2 cos(2x) cosx

=

tan(2x) tan(x)

54.cos(5y) + cos(3y)cos(5y)− cos(3y)

=

2 cos(

5y + 3y2

)cos

(5y − 3y

2

)−2 sin

(5y + 3y

2

)sin(

5y − 3y2

) =

2 cos(4y) cos y−2 sin(4y) sin y

=

− cot(4y) cot y

55. By using a product-to-sum identity, we get

= − sin(x+ y) sin(x− y)

= −12

[cos ((x+ y)− (x− y))−

cos ((x+ y) + (x− y))]

= −12

[cos (2y)− cos (2x)

]= −1

2

[(2 cos2 y − 1)− (2 cos2 x− 1)

]= −1

2

[2 cos2 y − 2 cos2 x

]cos2 x− cos2 y

56. By using a product-to-sum identity, we obtain

= sin(x+ y) sin(x− y)

=12

[cos ((x+ y)− (x− y))−

cos ((x+ y) + (x− y))]

=12

[cos(2y)− cos(2x)

]=

12

[(1− 2 sin2 y)− (1− 2 sin2 x)

]=

12

[−2 sin2 y + 2 sin2 x)

]sin2 x− sin2 y.

57. Let A =x+ y

2and B =

x− y2

.

Note, A+B = x and A−B = y. Expandthe left-hand side and use product-to-sumidentities.

(sinA+ cosA) (sinB + cosB) =

sinA sinB + sinA cosB+cosA sinB + cosA cosB =

12

[cos(A−B)− cos(A+B)

]+

12

[sin(A+B) + sin(A−B)

]+

12

[sin(A+B)− sin(A−B)

]+

12

[cos(A−B) + cos(A+B)

]=

12

[cos y − cosx

]+

12

[sinx+ sin y

]+

12

[sinx− sin y

]+

12

[cos y + cosx

]=

12

[2 cos y + 2 sinx

]=

sinx+ cos y

58. Factor the right-hand side as a differenceof two squares and use sum-to-productidentities.

= sin2(A+B)− sin2(A−B)

=[sin(A+B)− sin(A−B)

]·[

sin(A+B) + sin(A−B)]

= 2 cos(

(A+B) + (A−B)2

)·

sin(

(A+B)− (A−B)2

)·

2 sin(

(A+B) + (A−B)2

)·

cos(

(A+B)− (A−B)2

)= 4 cosA sinB sinA cosB= (2 sinA cosA)(2 sinB cosB)



= sin(2A) sin(2B)

59. Use a sum-to-product identity in the 2nd line,and a product-to-sum identity in the 5th line.

= sin2(A+B)− sin2(A−B)= sin(2A) sin(2B)= (2 sinA cosA)(2 sinB cosB)= [2 cosA cosB] · [2 sinA sinB]

=[cos(A−B) + cos(A+B)

]·[

cos(A−B)− cos(A+B)]

= cos2(A−B)− cos2(A+B)

60. Expand the left-hand side. Proof uses sumand difference identities.

(sinA+ cosA) (sinB + cosB) =

sinA sinB + sinA cosB+cosA sinB + cosA cosB =[

sinA cosB + cosA sinB]+[

cosA cosB + sinA sinB]

=

sin(A+B) + cos(A−B) =

61. Note that x can be written in the formx = a sin(t+ α). The maximum displacementof x =

√3 sin t+ cos t is

a =√√

32

+ 12 = 2.

Thus, 2 meters is the maximum distancebetween the block and its resting position.

Since the terminal side of α goes through(√

3, 1), we get tanα = 1/√

3 and one canchoose α = π/6. Then x = 2 sin(t+ π/6).

62. The maximum distance ofx = −0.3 sin t+ 0.5 cos t is its amplitudewhich is

√(0.3)2 + (0.5)2 ≈ 0.58 meters

67. a)tanx+ tan y

1− tanx tan yb)

tanx− tan y1 + tanx tan y

68. sinα cosβ + cosα sinβ =

13

√3

2+

(−2√

23

)12

=√

3− 2√

26

69.1− cosx

sinx=

1− (−1/3)−√

8/3=

4/3−2√

2/3=

2−√

2= −√

2

70.√

1− cosx2

=

√1− (−1/4)

2=

√5/42

=√58

=√

1016

=√

104

71. Since sin y = −45

and cos y =35

, we obtain

sin 2y = 2 sin y cos y = 2(−4

5

)35

= −2425

72. Let A be the angle opposite 5 miles, and B theangle opposite 12 miles. Then

A = arcsin(

513

)≈ 22.6◦

B = arcsin(

1213

)≈ 67.4◦

Thinking Outside the Box

LVI. Given below are two circles with radii a.Consider the triangle with a vertex at the toppoint of intersection and with the centers ofthe circles as the other two vertices. This tri-angle is an equilateral triangle.



By subtracting the area of a triangle from thearea of a sector, we find that the area of theshaded region above is

12a2π

3− 1

2a2

√3

2or

a2(2π − 3√

3)12

.

Then the area of the region inside the circle onthe right that is outside the circle on the leftis

2[

12a2 2π

3

]− 2

[a2(2π − 3

√3)

12

]or equivalently

a2(2π + 3√

3)6

.

If we add the area of the circle on the left tothe above expression we obtain the total areasprinkled, that is,

πa2 +a2(2π + 3

√3)

6=a2(8π + 3

√3)

6.

LVII. Suppose L represents a move to the leftsquare, R denotes a move to the right square,U represents a move to the upper square, andD represents a move to the square below.

A path from the upper left square to the lowerright square may be represented as an orderedsequence using the letters L, R, U , and D. Ifwe let l, r, u and d denote the number of L’s,R’s, U ’s, and D’s, respectively, then

l + r + u+ d = 63

r − l = 7 and

d− u = 7.

When you add all the three equations, we ob-tain

2r + 2d = 77

which is impossible since the left side of theabove equation is an even number but the rightside is an odd number. Thus, there is no pathfrom the upper left square to the lower rightsquare.

6.5 Pop Quiz

1. Using the sum-to-product identity

sinx− sin y = 2 cos(x+ y

2

)sin(x− y

2

)we find

12

[sin(2α+ β)− sin(2α− β)] =

cos(

2α+ β + 2α− β2

)sin(

2α+ β − (2α− β)2

)=

cos (2α) sin (β) .

2. Using the sum-to-product identity

cosx+ cos y = 2 cos(x+ y

2

)cos

(x− y

2

)we find

cos 2α+ cos 4α =

2 cos(

2α+ 4α2

)cos

(2α− 4α

2

)=

2 cos 3α cos (−α) =2 cos 3α cosα

3. Dividing by 2, we obtain

y

2=√

32

sinx− 12

cosx

y

2= sin

(x− π

6

)

y = 2 sin(x− π

6

).

4. The amplitude is√

32 + 52 =√

34.


a) Let Pc be the profit function (in thousands ofdollars) for the Christmas Store.Assume Pc = A sin(Bx+ C) +D.



Since 10 ≤ Pc ≤ 50, Pc = 20 sin(Bx+C) + 30.Since (0, 50) is a point on the graph, then

20 sin(C) + 30 = 50sin(C) = 1

C =π

2

Likewise, since (6, 10) is a point on the graphwe obtain

20 sin(

6B +π

2

)+ 30 = 10

sin(

6B +π

2

)= −1

6B +π

2=

3π2

B =π

6

The profit function for the Christmas Store is

Pc = 20 sin(π

6x+

π

2

)+ 30

(in thousands of dollars) or equivalently

Pc = 20, 000 sin(π

6x+

π

2

)+ 30, 000

(in dollars) or

Pc = 20, 000 sin(π

6x+

π

2− 2π

)+ 30, 000

= 20, 000 sin(π

6x− 3π

2

)+ 30, 000

Pc = 20, 000 sin(π

6[x− 9]

)+ 30, 000.

A sketch of the graph of Pc is provided.

6 12x

10

50

y

b) Let Pp be the profit function (in thousandsof dollars) for the Pool Store.

Assume Pp = A sin(Bx+ C) +D.Since 20 ≤ Pp ≤ 80, Pc = 30 sin(Bx+C) + 50.Since (8, 80) is a point on the graph, we obtain

30 sin(8B + C) + 50 = 80sin(8B + C) = 1

8B + C =π

2.

Likewise, since (2, 20) is a point on the graphwe get

30 sin (2B + C) + 50 = 20sin (2B + C) = −1

2B + C =−π2.

Then (8B + C)− (2B + C) =π

2− −π

2or

equivalently 6B = π. Solving, we find B =π

6

and C =−5π

6.

The profit function for the Pool Store is

Pp = 30 sin(π

6x− 5π

6

)+ 50

(in thousands of dollars) or equivalently

Pp = 30, 000 sin(π

6(x− 5)

)+ 50, 000

(in dollars). A sketch of the graph of Pp isprovided on the next page.

2 8x

20

80

y

c) The total profit is

y = 20, 000 sin(π

6[x− 9]

)+

30, 000 sin(π

6(x− 5)

)+ 80, 000.



The period is 12 and a sketch of the graph of

the total profit function is given below.

3.4 9.4x

53.5

106.5

y

d) Maximum profit is $106,458. This occursin September (corresponding to x ≈ 9.363).

e) Minimum profit is $53,542. This occursin March (corresponding to x ≈ 3.363).

f) One cannot express y = sin(x) + sin(2x) in theform y = A sin(Bx+C) +D where A,B,C,Dare real constants. To see this, simply sketchthe graph of y = sin(x) + sin(2x).

g) No, the sum of two periodic is not necessarilyperiodic. For instance, y = sin(x) + sin(

√2x)

is not a periodic function.

For Thought

1. False, the only solutions are 45◦ and 315◦.

2. False, there is no solution in [0, π).

3. True, since −29◦ and 331◦ are coterminalangles.

4. True 5. True, since the right-side isa factorization of the left-side.

6. False, x = 0 is a solution to the first equationand not to the second equation.

7. False, cos−1 2 is undefined. 8. True

9. False, x = 3π/4 is not a solution to thefirst equation but is a solution to thesecond equation.

10 . False, rather{x|3x =

π

2+ 2kπ

}={

x|x =π

6+

2kπ3

}.

6.6 Exercises

1. {x | x = π + 2kπ, k an integer}

2.{x | x =

π

2+ kπ, k an integer

}3. {x | x = kπ, k an integer}

4.{x | x =

π

2+ 2kπ, k an integer

}

5.{x | x =

3π2

+ 2kπ, k an integer}

6. {x | x = 2kπ, k an integer}

7. Solutions in [0, 2π) are x =π

3,5π3

. So solution

set is{x | x =

π

3+ 2kπ or x =

5π3

+ 2kπ}.


4,7π4

. So solution

set is{x | x =

π

4+ 2kπ or x =

7π4

+ 2kπ}.


4,3π4

. So solution

set is{x | x =

π

4+ 2kπ or x =

3π4

+ 2kπ}.


3,2π3

. So solution

set is{x | x =

π

3+ 2kπ or x =

2π3

+ 2kπ}.

11. Solution in [0, π) is x =π

4.

The solution set is{x | x =

π

4+ kπ

}.

12. Solution in [0, π) is x =π

6.


π

6+ kπ

}.

13. Solutions in [0, 2π) are x =5π6,7π6

.

Then the solution set is{x | x =

5π6

+ 2kπ or x =7π6

+ 2kπ}.


6.6 CONDITIONAL TRIGONOMETRIC EQUATIONS 389


.

Then the solution set is{x | x =

3π4

+ 2kπ or x =5π4

+ 2kπ}.


.


5π4

+ 2kπ or x =7π4

+ 2kπ}.


.

So the solution set is{x | x =

4π3

+ 2kπ or x =5π3

+ 2kπ}.

17. Solution in [0, π) is x =3π4

. The solution

set is{x | x =

3π4

+ kπ

}.

18. Solution in [0, π) is x =2π3

. The solution

set is{x | x =

2π3

+ kπ

}.

19. Solutions in [0, 360◦) are α = 90◦, 270◦.

So solution set is {α | α = 90◦ + k · 180◦} .

20. Solution in [0, 360◦) is α = 180◦. So the

solution set is {α | α = 180◦ + k · 360◦} .





23. Solution in [0, 180◦) is α = 0◦.

The solution set is {α | α = k · 180◦}.

24. Solution in [0, 180◦) is α = 135◦.

The solution set is {α | α = 135◦ + k · 180◦} .

25. One solution is cos−1(0.873) ≈ 29.2◦. Anothersolution is 360◦ − 29.2◦ = 330.8◦. Solution setis {α | α = 29.2◦ + k360◦ or α = 330.8◦ + k360◦} .

26. One solution is cos−1(−0.158) ≈ 99.1◦.Another solution is 360◦ − 99.1◦ = 260.9◦.The solution set is{α | α = 99.1◦ + k360◦ or α = 260.9◦ + k360◦} .

27. One solution is sin−1(−0.244) ≈ −14.1◦.This is coterminal with 345.9◦. Anothersolution is 180◦ + 14.1◦ = 194.1◦. Solution setis {α | α = 345.9◦ + k360◦ or α = 194.1◦ + k360◦} .

28. One solution is sin−1(0.551) ≈ 33.4◦. Anothersolution is 180◦ − 33.4◦ = 146.6◦. Solution setis {α | α = 33.4◦ + k360◦ or α = 146.6◦ + k360◦} .

29. One solution is tan−1(5.42) ≈ 79.5◦.Solution set is {α | α = 79.5◦ + k · 180◦} .

30. One solution is tan−1(−2.31) ≈ −66.6◦.Since 180◦ is the period of tanx, anothersolution is 180◦ − 66.6◦ = 113.4◦

Solution set is {α | α = 113.4◦ + k · 180◦}.

31. Values of x/2 in [0, 2π) are π/3 and 5π/3.Then we get

x

2=π

3+ 2kπ or

x

2=

5π3

+ 2kπ

x =2π3

+ 4kπ or x =10π3

+ 4kπ.


2π3

+ 4kπ or x =10π3

+ 4kπ}.

32. Since cos(2x) = −√

22

, values of 2x in [0, 2π)

are 3π/4 and 5π/4. Then

2x =3π4

+ 2kπ or 2x =5π4

+ 2kπ

x =3π8

+ kπ or x =5π8

+ kπ.


3π8

+ kπ or x =5π8

+ kπ

}.

33. Value of 3x in [0, 2π) is 0. Thus, 3x = 2kπ.


2kπ3

}.



34. Values of 2x in [0, 2π) are π/2 and 3π/2. So

2x =π

2+ 2kπ or 2x =

3π2

+ 2kπ

x =π

4+ kπ or x =

3π4

+ kπ.


π

4+kπ

2

}.

35. Since sin(x/2) = 1/2, values of x/2 in[0, 2π) are π/6 and 5π/6. Then

x

2=π

6+ 2kπ or

x

2=

5π6

+ 2kπ

x =π

3+ 4kπ or x =

5π3

+ 4kπ.


π

3+ 4kπ or x =

5π3

+ 4kπ}.

36. Values of 2x in [0, 2π) are 0 and π. So

2x = 0 + 2kπ or 2x = π + 2kπ

x = kπ or x =π

2+ kπ.


kπ

2

}.

37. Since sin(2x) = −√

2/2, values of 2x in[0, 2π) are 5π/4 and 7π/4. Thus,

2x =5π4

+ 2kπ or 2x =7π4

+ 2kπ

x =5π8

+ kπ or x =7π8

+ kπ.


5π8

+ kπ or x =7π8

+ kπ

}.

38. Since sin(x/3) = −1, value of x/3 in[0, 2π) is 3π/2 . So

x

3=

3π2

+ 2kπ.


9π2

+ 6kπ}.

39. Value of 2x in [0, π) is π/3. Then

2x =π

3+ kπ.


π

6+kπ

2

}.

40. Since tan(3x) = −1/√

3, value of 3x in[0, π) is 5π/6. Thus,

3x =5π6

+ kπ.


5π18

+kπ

3

}.

41. Value of 4x in [0, π) is 0. Then

4x = kπ.


kπ

4

}.

42. Value of 3x in [0, π) is 3π/4. Then

3x =3π4

+ kπ.


π

4+kπ

3

}.

43. The values of πx in [0, 2π) are π/6 and 5π/6.Then

πx =π

6+ 2kπ or πx =

5π6

+ 2kπ

x =16

+ 2k or x =56

+ 2k.


16

+ 2k or x =56

+ 2k}.

44. Value of πx/4 in [0, π) is π/4. So

πx

4=π

4+ kπ

πx = π + 4kπ.

The solution set is {x | x = 1 + 4k} .



45. Values of 2πx in [0, 2π) are π/2 and 3π/2. So

2πx =π

2+ 2kπ or 2πx =

3π2

+ 2kπ

x =14

+ k or x =34

+ k.


14

+k

2

}.

46. Value of 3πx in [0, 2π) is π/2. Then

3πx =π

2+ 2kπ.


16

+2k3

}.

47. Since sinα = −√

3/2, the solution setis {240◦, 300◦} . 48. {120◦, 300◦} .

49. Since cos 2α = 1/√

2, values of 2α in[0, 360◦) are 45◦ and 315◦. Thus,

2α = 45◦ + k · 360◦ or 2α = 315◦ + k · 360◦

α = 22.5◦ + k · 180◦ or α = 157.5◦ + k · 180◦.

Then let k = 0, 1. The solution set is

{22.5◦, 157.5◦, 202.5◦, 337.5◦} .

50. The value of 6α in [0, 360◦) is 90◦. Then

6α = 90◦ + k · 360◦

α = 15◦ + k · 60◦.

By choosing k = 0, 1, ..., 5, one obtains thesolution set {15◦, 75◦, 135◦, 195◦, 255◦, 315◦}.

51. Values of 3α in [0, 360◦) are 135◦ and 225◦.Then

3α = 135◦ + k · 360◦ or 3α = 225◦ + k · 360◦

α = 45◦ + k · 120◦ or α = 75◦ + k · 120◦.

By choosing k = 0, 1, 2, one obtains thesolution set {45◦, 75◦, 165◦, 195◦, 285◦, 315◦}.

52. Since csc 5α = −2, the values of 5α in [0, 360◦)are 210◦ and 330◦. Then

5α = 210◦ + k · 360◦ or 5α = 330◦ + k · 360◦

α = 42◦ + k · 72◦ or α = 66◦ + k · 72◦.

Choosing k = 0, 1, 2, 3, 4, the solution set is{42◦, 66◦, 114◦, 138◦, 186◦, 210◦, 258◦, 282◦,330◦, 354◦} .

53. The value of α/2 in [0, 180◦) is 30◦. Then

α

2= 30◦ + k · 180◦

α = 60◦ + k · 360◦.

By choosing k = 0, the solution set is {60◦} .

54. The values of α/2 in [0, 360◦) are 45◦ and 315◦.So

α

2= 45◦ + k · 360◦ or

α

2= 315◦ + k · 360◦

α = 90◦ + k · 720◦ or α = 630◦ + k · 720◦.

Let k = 0 in the first case.Then the solution set is {90◦} .

55. A solution is 3α = sin−1(0.34) ≈ 19.88◦.Another solution is 3α = 180◦ − 19.88◦ =160.12◦. Then

3α = 19.88◦+k ·360◦ or 3α = 160.12◦+k ·360◦

α ≈ 6.6◦ + k · 120◦ or α ≈ 53.4◦ + k · 120◦.

Solution set is

{α | α = 6.6◦ + k · 120◦ or α = 53.4◦ + k · 120◦} .

56. A solution is 2α = cos−1(−0.22) ≈ 102.71◦.Another solution is 2α = 360◦ − 102.71◦ =257.29◦. Therefore,

2α = 102.71◦+k·360◦ or 2α = 257.29◦+k·360◦

α ≈ 51.4◦ + k · 180◦ or α ≈ 128.6◦ + k · 180◦.

Solution set is

{α | α = 51.4◦ + k180◦ or α = 128.6◦ + k180◦} .



57. A solution is 3α = sin−1(−0.6) ≈ −36.87◦.This is coterminal with 323.13◦. Anothersolution is 3α = 180◦+36.87◦ = 216.87◦. Then

3α = 323.13◦+k·360◦ or 3α = 216.87◦+k·360◦

α ≈ 107.7◦ + k · 120◦ or α ≈ 72.3◦ + k · 120◦.

The solution set is

{α | α = 107.7◦ + k120◦ or α = 72.3◦ + k120◦} .

58. A solution is 4α = tan−1(−3.2) ≈ −72.65◦.Another solution is 180◦ − 72.65◦ = 107.35◦.

4α = 107.35◦ + k · 180◦

α = 26.8◦ + k · 45◦

Solution set is {α | α = 26.8◦ + k · 45◦} .

59. A solution is 2α = cos−1(1/4.5) ≈ 77.16◦.Another solution is 2α = 360◦ − 77.16◦ =282.84◦. Thus,

2α = 77.16◦+k ·360◦ or 2α = 282.84◦+k ·360◦

α ≈ 38.6◦ + k · 180◦ or α ≈ 141.4◦ + k · 180◦.

The solution set is

{α | α = 38.6◦ + k180◦ or α = 141.4◦ + k180◦} .

60. A solution is 3α = sin−1(−1/1.4) ≈ −45.58◦.This is coterminal with 314.42◦.Another solution is 3α = 180◦ + 45.58◦ =225.58◦. Thus,

3α = 314.42◦+k·360◦ or 3α = 225.58◦+k·360◦

α ≈ 104.8◦ + k · 120◦ or α ≈ 75.2◦ + k · 120◦.

The solution set is

{α | α = 104.8◦ + k120◦ or α = 75.2◦ + k120◦} .

61. A solution is α/2 = sin−1(−1/2.3) ≈ −25.77◦.This is coterminal with 334.23◦. Anothersolution is α/2 = 180◦ + 25.77◦ = 205.77◦.Thus,α

2= 334.23◦+k ·360◦ or

α

2= 205.77◦+k ·360◦

α ≈ 668.5◦ + k · 720◦ or α ≈ 411.5◦ + k · 720◦.

The solution set is

{α | α = 668.5◦ + k720◦ or α = 411.5◦ + k720◦} .

62. A solution is α/2 = tan−1(1/4.7) ≈ 12.01◦.Then

α

2= 12.01◦ + k · 180◦

α = 24.0◦ + k · 360◦.

Solution set is {α | α = 24.0◦ + k · 360◦} .

63. Set the right-hand side to zero and factor.

3 sin2 x− sinx = 0sinx(3 sinx− 1) = 0

Set each factor to zero.

sinx = 0 or sinx = 1/3x = 0, π or x = sin−1(1/3) ≈ 0.3

Another solution to sinx = 1/3 isx = π − 0.3 ≈ 2.8.The solution set is {0, 0.3, 2.8, π}.


2 tan2 x− tanx = 0tanx(2 tanx− 1) = 0

Set each factor to zero.

tanx = 0 or tanx = 1/2x = 0, π or x = tan−1(1/2) ≈ 0.5

Another solution to tanx = 1/2 is x = π+ 0.5≈ 3.6. The solution set is {0, 0.5, π, 3.6}.


2 cos2 x+ 3 cosx+ 1 = 0(2 cosx+ 1)(cosx+ 1) = 0

Set the factors to zero.

cosx = −1/2 or cosx = −1x = 2π/3, 4π/3 or x = π

The solution set is {π, 2π/3, 4π/3}.


2 sin2 x+ sinx− 1 = 0(2 sinx− 1)(sinx+ 1) = 0



Set each factor to 0.

sinx = 1/2 or sinx = −1x = π/6, 5π/6 or x = 3π/2

The solution set is {π/6, 5π/6, 3π/2}.

67. Substitute cos2 x = 1− sin2 x.

5 sin2 x− 2 sinx = 1− sin2 x

6 sin2 x− 2 sinx− 1 = 0

Apply the quadratic formula.

sinx =2±√

2812

sinx =1±√

76

Then

x = sin−1

(1 +√

76

)or x = sin−1

(1−√

76

)x ≈ 0.653 or x ≈ −0.278.

Another solution is π−0.653 ≈ 2.5 . An anglecoterminal with −0.278 is 2π − 0.278 ≈ 6.0 .Another solution is π + 0.278 ≈ 3.4 .The solution set is {0.7, 2.5, 3.4, 6.0}.

68. Divide the equation by cos2 x.

tan2 x− 1 = 0tan2 x = 1tanx = ±1

Solution set is {π/4, 3π/4, 5π/4, 7π/4}.

69. Squaring both sides of the equation, we obtain

tan2 x = sec2 x− 2√

3 secx+ 3sec2 x− 1 = sec2 x− 2

√3 secx+ 3

−4 = −2√

3 secxsecx = 2/

√3

x = π/6, 11π/6.

Checking x = π/6, one gets tan(π/6) = 1/√

3and sec(π/6)−

√3 = 2/

√3−√

3 = −1/√

3.Then x = π/6 is an extraneous root and thesolution set is {11π/6}.

70. Squaring both sides, we find

csc2 x− 2√

3 cscx+ 3 = cot2 xcsc2 x− 2

√3 cscx+ 3 = csc2 x− 1−2√

3 cscx = −4cscx = 2/

√3

x = π/3, 2π/3.

Checking x = π/3, one gets cot(π/3) = 1/√

3and csc(π/3)−

√3 = 2/

√3−√

3 = −1/√

3.So x = π/3 is an extraneous root andthe solution set is {2π/3}.

71. Square both sides of the equation.

sin2 x+ 2√

3 sinx+ 3 = 27 cos2 x

sin2 x+ 2√

3 sinx+ 3 = 27(1− sin2 x)28 sin2 x+ 2

√3 sinx− 24 = 0

14 sin2 x+√

3 sinx− 12 = 0

By the quadratic formula, we get

sinx =−√

3±√

67528

sinx =−√

3± 15√

328

sinx =√

32,−4√

37

.

Thus,

x =π

3,2π3

or x = sin−1

(−4√

37

)

x =π

3,2π3

or x ≈ −1.427.

Checking x = 2π/3, one findssin(2π/3) +

√3 =√

3/2 +√

3 and3√

3 cos(2π/3) is a negative number.Then x = 2π/3 is an extraneous root.

An angle coterminal with −1.427 is2π−1.427 ≈ 4.9 . In a similar way, one checksthat π + 1.427 ≈ 4.568 is an extraneous root.Thus, the solution set is {π/3, 4.9}.

72. Substitute sin2 x = 1− cos2 x.

6(1− cos2 x)− 2 cosx = 56− 6 cos2 x− 2 cosx− 5 = 0

6 cos2 x+ 2 cosx− 1 = 0



Apply the quadratic formula.

cosx =−2±

√28

12

cosx =−1±

√7

6

Then

x = cos−1

(−1 +

√7

6

)or x = cos−1

(−1−

√7

6

)x ≈ 1.3 or x ≈ 2.2.

Two other solutions are 2π − 1.3 ≈ 5.0 and2π − 2.2 ≈ 4.1. The solution set is{1.3, 2.2, 4.1, 5.0}.

73. Express the equation in terms of sinxand cosx.

sinxcosx

· 2 sinx cosx = 0

2 sin2 x = 0sinx = 0

Solution set is {0, π}.


3 sec2 x tanx− 4 tanx = 0tanx(3 sec2 x− 4) = 0

Then

tanx = 0 or secx = ± 2√3

x = 0, π or x = π/6, 5π/6, 7π/6, 11π/6.

Solution set is {0, π/6, 5π/6, π, 7π/6, 11π/6}.

75. Substitute the double-angle identity for sinx.

2 sinx cosx− sinx cosx = cosxsinx cosx− cosx = 0

cosx(sinx− 1) = 0cosx = 0 or sinx = 1

x = π/2, 3π/2 or x = π/2

Solution set is {π/2, 3π/2}.

76. Apply double-angle identities.

2 cos2(2x)− 2(4 sin2 x cos2 x) = −12 cos2(2x)− 2 sin2(2x) = −12(cos2(2x)− sin2(2x)) = −1

2 cos(4x) = −1cos(4x) = −1/2

Then

4x =2π3

+ 2kπ or 4x =4π3

+ 2kπ

x =π

6+kπ

2or x =

π

3+kπ

2.

Let k = 0, 1, 2, 3. Then the solution set is

{π/6, 5π/6, 7π/6, 11π/6, π/3, 2π/3, 4π/3, 5π/3}.

77. Use the sum identity for sine.

sin(x+ π/4) = 1/2

x+π

4=π

6+ 2kπ or x+

π

4=

5π6

+ 2kπ

x =−π12

+ 2kπ or x =7π12

+ 2kπ

By choosing k = 1 in the first case andk = 0 in the second case, one findsthe solution set is {23π/12, 7π/12}.

78. Multiply both sides by −1 and use thedifference identity for sine.

sinx cos(π/6)− cosx sin(π/6) = 1/2sin(x− π/6) = 1/2

Then

x− π

6=π

6+ 2kπ or x− π

6=

5π6

+ 2kπ

x =π

3+ 2kπ or x = π + 2kπ.

Choose k = 0, and so the solution setis {π/3, π}.

79. Apply the difference identity for sine.

sin(2x− x) = −1/2sinx = −1/2

The solution set is {7π/6, 11π/6}.



80. By the sum identity for cosine, we get

cos(2x+ x) = 1/2cos(3x) = 1/2.

Then

3x =π

3+ 2kπ or 3x =

5π3

+ 2kπ

x =π

9+

2kπ3

or x =5π9

+2kπ

3.

By choosing k = 0, 1, 2, one finds the solutionset is {π/9, 5π/9, 7π/9, 11π/9, 13π/9, 17π/9}.

81. Since 4 ·42 sin2 x = 43 sinx, we set the exponentsequal to each other. Then

2 sin2 x+ 1 = 3 sinx2 sin2 x− 3 sinx+ 1 = 0

(2 sinx− 1)(sinx− 1) = 0

sinx = 1,12.

Thus, the solution set is {π/6, π/2, 5π/6}.

82. Since 2−1 · 22 cos2 x = 2cosx, we set the expo-nents equal to each other. Then

2 cos2 x− 1 = cosx2 cos2 x− cosx− 1 = 0

(2 cosx+ 1)(cosx− 1) = 0

cosx = 1,−12.

Thus, the solution set is {0, 2π/3, 4π/3}.

83. Use a half-angle identity for cosine andexpress equation in terms of cos θ.

1 + cos θ2

=1

cos θcos θ + cos2 θ = 2

cos2 θ + cos θ − 2 = 0(cos θ + 2)(cos θ − 1) = 0

cos θ = −2 or cos θ = 1no solution or θ = 0◦

Solution set is {0◦}.

84. By the half-angle identity for sine, we find

1− cos θ = cos θ1 = 2 cos θ

1/2 = cos θ.

Solution set is {60◦, 300◦}.

85. Dividing the equation by 2 cos θ, we get

sin θcos θ

=12

tan θ = 0.5θ = tan−1(0.5) ≈ 26.6◦.

Another solution is 180◦ + 26.6◦ = 206.6◦ .Solution set is {26.6◦, 206.6◦}.

86. Dividing equation by 3 cos 2θ, we obtain

sin 2θcos 2θ

=13

tan 2θ = 1/32θ = tan−1(1/3) ≈ 18.43◦ + k · 180◦

θ ≈ 9.2◦ + k · 90◦.

By choosing k = 0, 1, 2, 3, one gets thatthe solution set is {9.2◦, 99.2◦, 189.2◦, 279.2◦}.

87. Express equation in terms of sin 3θ.

sin 3θ =1

sin 3θsin2 3θ = 1sin 3θ = ±1

Then

3θ = 90◦ + k · 360◦ or 3θ = 270◦ + k · 360◦

θ = 30◦ + k · 120◦ or θ = 90◦ + k · 120◦.

By choosing k = 0, 1, 2, one finds that thesolution set is{30◦, 90◦, 150◦, 210◦, 270◦, 330◦}.

88. Express equation in terms of tan θ.

tan2 θ − 1tan2 θ

= 0

tan4 θ − 1 = 0tan θ = ±1

The solution set is {45◦, 135◦, 225◦, 315◦}.



89. By the method of completing the square,we get

tan2 θ − 2 tan θ = 1tan2 θ − 2 tan θ + 1 = 2

(tan θ − 1)2 = 2tan θ − 1 = ±

√2

θ = tan−1(1 +√

2) or θ = tan−1(1−√

2)θ ≈ 67.5◦ or θ = −22.5◦.

Other solutions are 180◦ + 67.5◦ = 247.5◦,180◦ − 22.5◦ = 157.5◦, and180◦ + 157.5◦ = 337.5◦. The solution setis {67.5◦, 157.5◦, 247.5◦, 337.5◦} .

90. By the method of completing the square,we find

cot2 θ − 4 cot θ = −2cot2 θ − 4 cot θ + 4 = 2

(cot θ − 2)2 = 2cot θ − 2 = ±

√2

θ = tan−1(

12 +√

2

)or θ = tan−1

(1

2−√

2

)θ ≈ 16.3◦ + k · 180◦ or θ = 59.6◦ + k · 180◦.

By choosing k = 0, 1, one obtains that the

solution set is {16.3◦, 59.6◦, 196.3◦, 239.6◦}.

91. Factor as a perfect square.

(3 sin θ + 2)2 = 0sin θ = −2/3

θ = sin−1 (−2/3) ≈ −41.8◦

An angle coterminal with −41.8◦ is360◦ − 41.8◦ = 318.2◦. Another solutionis 180◦ + 41.8◦ = 221.8◦. The solution setis {221.8◦, 318.2◦}.

92. Factoring, we obtain

(4 cos θ + 3)(3 cos θ − 2) = 0cos θ = −3/4 or cos θ = 2/3

θ = cos−1(−3/4) or θ = cos−1(2/3)θ ≈ 138.6◦ or θ ≈ 48.2◦.

Other solutions are 360◦ − 138.6◦ = 221.4◦

and 360◦ − 48.2◦ = 311.8◦. The solution setis {48.2◦, 138.6◦, 221.4◦, 311.8◦} .

93. By using the sum identity for tangent, we get

tan(3θ − θ) =√

32θ = 60◦ + k · 180◦

θ = 30◦ + k · 90◦.

By choosing k = 1, 3, one obtains that thesolution set is {120◦, 300◦}. Note, 30◦ and210◦ are not solutions.

94. By using the sum identity for tangent,we find

tan(3θ + 2θ) = 1tan 5θ = 1

5θ = 45◦ + k · 180◦

θ = 9◦ + k · 36◦.

The solution set is{9◦, 81◦, 117◦, 153◦, 189◦, 261◦, 297◦, 333◦}.Note, 45◦ and 225◦ are not solutions.

95. Factoring, we get

(4 cos2 θ − 3)(2 cos2 θ − 1) = 0.

Then

cos2 θ = 3/4 or cos2 θ = 1/2cos θ = ±

√3/2 or cos θ = ±1/

√2.

The solution set is

{30◦, 45◦, 135◦, 150◦, 210◦, 225◦, 315◦, 330◦}.

96. Factoring, we have

(4 sin2 θ − 1)(sin2 θ − 1) = 0.

Then

sin2 θ = 1/4 or sin2 θ = 1sin θ = ±1/2 or sin θ = ±1.

The solution set is{30◦, 90◦, 150◦, 210◦, 270◦, 330◦}.


(sec2 θ − 1)(sec2 θ − 4) = 0sec2 θ = 1 or sec2 θ = 4

sec θ = ±1 or sec θ = ±2.

Solution set is {0◦, 60◦, 120◦, 180◦, 240◦, 300◦}.




(cot2 θ − 1)(cot2 θ − 3) = 0cot2 θ = 1 or cot2 θ = 3

cot θ = ±1 or cot θ = ±√

3.

Solution set is

{30◦, 45◦, 135◦, 150◦, 210◦, 225◦, 315◦, 330◦}.

99. Multiplying the equation by LCD, we get

13.7 sin 33.2◦ = a · sin 45.6◦

13.7 sin 33.2◦

sin 45.6◦= a

10.5 ≈ a.

100. Multiplying by the LCD, we find

b · sin 49.6◦ = 55.1 sin 88.2◦

b =55.1 sin 88.2◦

sin 49.6◦

b ≈ 72.3.

101. Multiplying by the LCD, we get

25.9 sinα = 23.4 sin 67.2◦

sinα =23.4 sin 67.2◦

25.9sinα ≈ 0.833

α ≈ sin−1(0.833)α ≈ 56.4◦.

102. Multiplying by the LCD, we obtain

52.9 sin 9.7◦ = 15.4 sinβ

52.9 sin 9.7◦

15.4= sinβ

0.579 ≈ sinβsin−1(0.579) ≈ β

35.4◦ ≈ β.

Since 90◦ < β < 180◦, we findβ = 180◦ − 35.4◦ = 144.6◦.

103. Isolate cosα on one side.

2(5.4)(8.2) cosα = 5.42 + 8.22 − 3.62

cosα =5.42 + 8.22 − 3.62

2(5.4)(8.2)

cosα ≈ 0.942α ≈ cos−1(0.942)α ≈ 19.6◦

104. Isolate cosα on one side.

2(3.2)(4.6) cosα = 3.22 + 4.62 − 6.82

cosα =3.22 + 4.62 − 6.82

2(3.2)(4.6)

cosα ≈ −0.504α ≈ cos−1(−0.504)α ≈ 120.3◦

105. Given below is the graph of

y = sin(x/2)− cos(3x).

The intercepts or solutions on [0, 2π) areapproximately {0.4, 1.9, 2.2, 4.0, 4.4, 5.8} .

1 2 3 4 5x

-1

1

y

106. Given below is the graph of

y = 2 sin(x)− csc(x+ 0.2).

The intercepts or solutions on [0, 2π)are approximately {0.7, 2.2, 3.8, 5.4}.

1 2 3 4 5 6x

-1

1

y

107. The graph of y =x

2− π

6+√

32− sinx



is shown. The solution set is {π/3}.

0.5 2x

1

-1

y

108. Below is the graph of y = x2 − sinx.The intercepts or solutions on [0, 2π) areapproximately {0, 0.9}.

1 2 3 4 5 6x

-1

1

y

109. Since a =√

3 and b = 1, we obtain

r =√√

32

+ 12 = 2. If the terminal side ofα goes through (

√3, 1), then tanα = 1/

√3.

Then one can choose α = π/6 andx = 2 sin(2t+ π/6). The times when x = 0are given by

sin(

2t+π

6

)= 0

2t+π

6= k · π

2t = −π6

+ k · π

t = − π

12+k · π

2

t = − π

12+π

2+k · π

2

t =5π12

+k · π

2

where k is a nonnegative integer.

110. Since a = −0.3 and b = 0.5, we obtainr =

√(−0.3)2 + (0.5)2 =

√0.34. If the

terminal side of α goes through (−0.3, 0.5),then tanα = −0.5/0.3. Since tan−1(−5/3) ≈−1.03, one can choose α = π − 1.03 ≈ 2.11

and x =√

0.34 sin(3t+ 2.11). The times whenx = 0 are given by

sin(3t+ 2.11) = 03t+ 2.11 = k · π

3t = −2.11 + k · π

t = −0.703 +k · π

3

t = −0.703 +π

3+k · π

3

t = 0.34 +k · π

3


111. First, find the values of t when x =√

3.

2 sin(πt

3

)=√

3

sin(πt

3

)=√

32

πt

3=π

3+ 2kπ or

πt

3=

2π3

+ 2kπ

πt = π + 6kπ or πt = 2π + 6kπt = 1 + 6k or t = 2 + 6k

Then the ball is√

3 ft above sea level forthe values of t satisfying

1 + 6k < t < 2 + 6k


112. First, find the values of t when x = 9.3 .

6.2 + 3.1 sin(π

6(t− 9)

)= 9.3

sin(π

6(t− 9)

)= 1

π

6(t− 9) =

π

2+ 2kπ

t− 9 = 3 + 12kt = 12 + 12k

Since the values of t are limited from 1 to 12,then one must choose k = 0 and t = 12. InDecember, the store anticipates selling 9300units.



113. Since vo = 325 and d = 3300, we have

3252 sin 2θ = 32(3300)

sin 2θ =32(3300)

3252

sin 2θ ≈ 0.999762θ ≈ sin−1(0.99976)2θ ≈ 88.74◦

θ ≈ 44.4◦.

Another angle is given by 2θ = 180◦ − 88.74◦

= 91.26◦ or θ = 91.26◦/2 ≈ 45.6◦.The muzzle was aimed at 44.4◦ or 45.6◦ .

114. We will use the equation from Exercise 113

and sketch the graph d =3252 sin(2θ)

32. We

find the maximum distance is d ≈ 3300.8 ft.

10.5theta

3300.8

d

115. Note, 90 mph= 90 · 52803600

ft/sec= 132 ft/sec.

In v2o sin 2θ = 32d, let vo = 132 and d = 230.

1322 sin 2θ = 32(230)

sin 2θ =32(230)

1322

sin 2θ ≈ 0.42242θ = sin−1(0.4224) ≈ 25.0◦ or 155◦

θ ≈ 12.5◦ or 77.5◦

The two possible angles are 12.5◦ and 77.5◦.The time it takes the ball to reach home platecan be found by using x = vot cos θ.(See Example 11). For the angle 12.5◦, it takes

t =230

132 cos 12.5◦≈ 1.78 sec

while for 77.5◦ it takes

t =230

132 cos 77.5◦≈ 8.05 sec.

The difference in time is 8.05− 1.78 ≈ 6.3 sec.

116. In v2o sin 2θ = 32d, let d = 3(18, 500) =

55, 500 ft. The muzzle velocity vo is given by

v2o sin(2 · 45◦) = 32(55, 500)

v2o · 1 = 32(55, 500)

vo =√

32(55, 500)vo ≈ 1332.7 ft/sec.

117. Observe,

y =√

2(

(sinx)1√2− (cosx)

1√2

)=√

2 sin(x+

7π4

).

The amplitude is√

2, period is 2π, and phaseshift −7π

4

118.

sinα cosβ + cosα sinβ =

13

(−√

32

)+

(2√

23

)12

=

2√

2−√

36

119. Using a cofunction relationship,

sin(π

2− x

)= cosx =

34.

120. Using a half-angle identity, we find

tan(x

2

)=

1− cosxsinx

=1− (−1/3)

2√

2/3

=4/3

2√

2/3

tan(x

2

)=√

2

121. Using a half-angle identity, we obtain

sin(x

2

)=

√1− cosx

2



=

√1− 1

4

2=

√38

=√

616

sin(x

2

)=√

64

122. a) sin(3.5 + 2.1) = sin 5.6

b) sin(2x− x) = sinx

c) sin(2 · (4.8)) = sin 9.6

Thinking Outside the Box LVIII

The angles in the shaded triangle are 15◦, 60◦, and105◦. The side opposite the 105◦-angle is one unitl long. Using the sine law (see Chapter 7), we find

x =sin 60◦

sin 105◦.

Recall, the area of a triangle is one-half times theproduct of the length of any two sides and the sineof the included angle of the two sides. Since theincluded angle between x and 1 is 15◦, the area ofthe triangle is

Area =12

sin 60◦

sin 105◦sin 15◦

=√

34

sin 15◦

sin 105◦

=√

34

√(2−

√3/2)/2√

(2 +√

3/2)/2

=√

34

√2−√

3√2 +√

3

=√

34

(2−√

3)

Area =2√

3− 34

.

6.6 Pop Quiz

1. 45◦ + k360◦ or 135◦ + k360◦

2. 60◦ + k360◦ or 300◦ + k360◦

3. 135◦ + k180◦

4. Sincex

2=π

6+ 2kπ or

x

2=

5π6

+ 2kπ

we obtain x =π

3+ 4kπ or x =

5π3

+ 4kπ.

Since x lies in [0, 2π], we find

x =π

3,5π5.

5. 0, 2π

6. Since 2x =π

4+ kπ, we obtain

x =π

8+ k

π

2.

Let k = 0, 1, 2, 3. The solutions in [0, 2π] are

x =π

8,5π8,9π8,13π8.


a) Since first base is 150 feet away, we find

x = 130t cos θ = 150, i.e., t =150

130 cos θ.

Substituting into y we find that the angle θsatisfies

−16t2 + 130t sin θ + 5 = 5−16t2 + 130t sin θ = 0

−16(

150130 cos θ

)2

+ 130(

150130 cos θ

)sin θ = 0

−16(

1513

)2 1cos2 θ

+ 150sin θcos θ

= 0.

Then multiply the previous equation by cos2 θ.

150 sin θ = 16(

1513

)2 1cos θ

150 sin θ cos θ = 16(

1513

)2


CHAPTER 6 REVIEW EXERCISES 401

75 sin(2θ) = 16(

1513

)2

sin(2θ) =16(

1513

)2

75

2θ = sin−1

16(

1513

)2

75

2θ ≈ 16.5◦

θ ≈ 8.25◦

b) It takes t =150

130 cos 8.25◦≈ 1.1659 sec

for the ball to reach first base without skipping(see part a)) .

c) Setting 130t cos θ = 75, we find t =75

130 cos θ.

Substituting into y = 0 and by using acalculator (such as the solver in a TI-83),we obtain the angle θ for which the ball mustbe thrown so that it hits the ground after 75feet. Thus,

−16(

75130 cos θ

)2

+130(

75130 cos θ

)sin θ+5 = 0

θ ≈ 0.2487◦.

d) It takes t =75

130 cos 0.2487◦≈ 0.5769 sec

for the ball to reach the skip point.(See part c)).

e) Using the symmetry and the work shown inpart d), the ball reaches first base in

2 · 75130 cos 0.2487◦

≈ 1.1539 sec.

f) Time saved is 0.012 sec. (= 1.1659− 1.1539).

g) If θ = 0, the ball reaches first base in

t =150

130 cos 0◦≈ 1.1538 seconds.

h) We are assuming that the ball is onlysubjected to gravity, and not to air resistancefor example.

Review Exercises

1. 1− sin2 α = cos2 α

2.1

sinx· sinx

cosx+ secx =

1cosx

+ secx = 2 secx

3. (1− cscx)(1 + cscx) = 1− csc2 x = − cot2 x

4.cos 2xsin 2x

= cot 2x

5.1

1 + sinα+

sinαcos2 α

=1

1 + sinα+

sinα1− sin2 α

=

(1− sinα) + sinα1− sin2 α

=1

cos2 α= sec2 α

6. By using cofunction identities, the expressioncan be simplified to 2 cosα sinα = sin 2α.

7. tan(4s), by the double angle idenity for tangent

8. tan(2w − 4w) = tan(−2w) = − tan(2w)

9. sin(3θ − 6θ) = sin(−3θ) = − sin(3θ)

10. tan(

2y2

)= tan y, by a double-angle identity

for tangent

11. tan(

2z2

)= tan z, by a double-angle identity

for tangent

12. cos(

2 · x2

)= cosx, by a double-angle identity

for cosine

13. e 14. h 15. c 16. d

17. a 18. b 19. g 20. f

21. Note, sinα =

√1−

(−513

)2

=√

1− 25169

=√144169

=1213

. Then tanα =12/13−5/13

= −125

,

cotα = − 512

, cscα =1312

, secα = −135

.

22. Note, secα = −

√1 +

(512

)2

= −√

1 +25144

=



−√

169144

= −1312

. Then cosα = −1213

,

sinα = −

√1−

(1213

)2

= −√

1− 144169

=

−√

25169

= − 513. cotα =

125

, cscα = −135

.

23. By using a cofunction identity,

we get cosα =−35.

Then sinα = −

√1−

(−35

)2

= −√

1− 925

=

−√

1625

= −45

, secα = −53

, cscα = −54

,

tanα =−4/5−3/5

=43

, cotα =34

.

24. By applying a cofunction identity,we find secα = 3.

Then cosα =13

and sinα =

√1−

(13

)2

=√1− 1

9=√

89

=√

83

=2√

23

, cscα =3√8

,

tanα =2√

2/31/3

= 2√

2, cotα =1√8

.

25. By the half-angle identity for sine, we find√1− cosα

2=

35

1− cosα2

=925

1− cosα =1825

cosα =725.

Since3π2< α < 2π, α is in quadrant IV

and sinα = −

√1−

(725

)2

= −√

1− 49625

=

−√

576625

= −2425

.

Then tanα =−24/25

7/25= −24

7,

cotα = − 724

, secα =257

, cscα = −2524.

26. By the half-angle identity for cosine, we get

−√

1 + cosα2

= −13

1 + cosα2

=19

1 + cosα =29

cosα = −79.

Since π < α <3π2

, α is in quadrant III and

sinα = −

√1−

(−79

)2

= −√

1− 4981

=

−√

3281

= −√

329. Then tanα =

−√

32/9−7/9

=

√327

, cotα =7√32

, secα = −97

,

and cscα = − 9√32

.

27. It is an identity as shown below.

(sinx+ cosx)2 =sin2 x+ 2 sinx cosx+ cos2x =

1 + 2 sinx cosx =1 + sin(2x)

28. It is not an identity since the right-handside is equal to cos(A+B) and is notequal to the left-hand side.

29. It is not an identity since csc2 x− cot2 x = 1and tan2− sec2 x = −1.

30. It is an identity. Apply a half-angle identityand simplify the right-hand side.

sin2 x

2=

1− cos2 x

2 + 2 sinx cosx cscx1− cosx

2=

1− cos2 x

2 + 2 cosx

=(1− cosx)(1 + cosx)

2(1 + cosx)

=1− cosx

2



31. Odd, since f(−x) =sin(−x)− tan(−x)

cos(−x)=

− sinx+ tanxcosx

= −sinx− tanxcosx

= −f(x)

32. Even, since f(−x) = 1 + sin2(−x) =

1 + (− sinx)2 = 1 + sin2 x = f(x)

33. It is neither even nor odd. Since f(π/4) =

cos(π/4)− sin(π/4)sec(π/4)

=√

2/2−√

2/2√2

= 0

and f(−π/4) =cos(−π/4)− sin(−π/4)

sec(−π/4)=

√2/2 +

√2/2√

2=√

2√2

= 1, we see

that f(π/4) 6= ±f(−π/4)

34. Odd, since f(−x) = csc3(−x)− tan3(−x) =(− cscx)3 − (− tanx)3 = − csc3 x+ tan3 x= −f(x)

35. Even, since f(−x) =sin(−x) tan(−x)

cos(−x) + sec(−x)=

(− sinx)(− tanx)cosx+ secx

=sinx tanx

cosx+ secx= f(x)

36. It is neither even nor odd. Since f(π/4) =sin(π/4) + cos(π/4) =

√2/2 +

√2/2 =

√2

and f(−π/4) = sin(−π/4) + cos(−π/4) =−√

2/2+√

2/2 = 0, then f(π/4) 6= ±f(−π/4).

37. f, since sin(π/2− α) = cosα

38. g, since sin(−x) = − sinx

39. e, for sinx+sin y = 2 sin(x+ y

2

)cos

(x− y

2

)

40. a, for sinx−sin y = 2 cos(x+ y

2

)sin(x− y

2

)41. b, since sin 2x = 2 sinx cosx

42. d, since cos 2x = cos2 x− sin2 x

43. h, for cosx+ cos y =

2 cos(x+ y

2

)cos

(x− y

2

)44. i, for cosx− cos y =

−2 sin(x+ y

2

)sin(x− y

2

)

45. c, for tan(x

2

)=

sinx1 + cosx

46. j, for cos2 x

2=

1 + cosx2

47. Rewrite the right side.

=1 + tan2 θ

1− tan2 θ

=sec2 θ

1− sin2 θ

cos2 θ

· cos2 θ

cos2 θ

=1

cos2 θ − sin2 θ

=1

cos 2θ= sec 2θ


=1− cos 2θ1 + cos 2θ

=1− (1− 2 sin2 θ)1 + (2 cos2 θ − 1)

=2 sin2 θ

2 cos2 θ

= tan2 θ

49. Rewrite the right side as follows:

=csc2 x− cot2 x

2 csc2 x+ 2 cscx cotx

=1

2sin2 x

+ 2 · 1sinx

· cosxsinx

=1

2sin2 x

+2 cosxsin2 x

· sin2 x

sin2 x

=sin2 x

2 + 2 cosx

=1− cos2 x

2(1 + cosx)

=(1− cosx)(1 + cosx)

2(1 + cosx)



=1− cosx

2

= sin2(x

2

)

50. Rewrite the right side as follows:

=1− sin2 x

cos(−x) sin(−x)

=cos2 x

− cosx sinx

=cosx− sinx

= cot(−x)

51. Rewrite the left side as follows:

cot(α− 45◦) =(tan(α− 45◦))−1 =(

tanα− tan 45◦

1 + tanα tan 45◦

)−1

=

(tanα− 11 + tanα

)−1

=

1 + tanαtanα− 1

=

52. Rewrite the left side as follows:

cos(α+ 45◦) =cosα cos 45◦ − sinα sin 45◦ =

cosα · 1√2− sinα · 1√

2=

cosα− sinα√2

=

53. Rewrite the left side.

sin 2β2 cscβ

=

2 sinβ cosβ2/ sinβ

· sinβsinβ

=

sin2 β cosβ =


=cos 2β√

2(cosβ + sinβ)

=cos2 β − sin2 β√2(cosβ + sinβ)

=(cosβ − sinβ)(cosβ + sinβ)√

2(cosβ + sinβ)

=cosβ − sinβ√

2

=1√2· cosβ − 1√

2· sinβ

= sin 45◦ cosβ − cos 45◦ sinβ

= sin(45◦ − β)

55. Factor the numerator on the left-hand sideas a difference of two cubes.Note, cotw tanw = 1.

cot3 y − tan3 y

sec2 y + cot2 y=

(cot y − tan y)(cot2 y + 1 + tan2 y

)sec2 y + cot2 y

=

(cot y − tan y)(cot2 y + sec2 y

)sec2 y + cot2 y

=

cot y − tan y =

1tan y

− tan y =

1− tan2 y

tan y=

2 · 1− tan2 y

2 tan y=

2 · (tan 2y)−1 =

2 cot(2y) =

56. Factor the numerator on the left-hand sideas a difference of two cubes.

sin3 y − cos3 y

sin y − cos y=

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

)sin y − cos y

=



(sin y − cos y) (1 + sin y cos y)sin y − cos y

=

(1 + sin y cos y) · 22

=

2 + 2 sin y cos y2

=

2 + sin(2y)2

=

57. By using double-angle identities, we obtain

cos(2 · 2x) =1− 2 sin2(2x) =

1− 2 (2 sinx cosx)2 =1− 8 sin2 x cos2 x =

1− 8 sin2 x(1− sin2 x) =8 sin4 x− 8 sin2 x+ 1.

58. By the sum identity for cosine, we have

cos(x+ 2x) =cosx cos 2x− sinx sin 2x =

cosx(1− 2 sin2 x

)− sinx(2 sinx cosx) =

cosx− 2 cosx sin2 x− 2 sin2 x cosx =cosx− 4 sin2 x cosx =cosx

(1− 4 sin2 x

).

59. By the double-angle identity for sine, we get

sin4(2x) =(2 sinx cosx)4 =16 sin4 x cos4 x =

16 sin4 x(1− sin2 x)2 =16 sin4 x(1− 2 sin2 x+ sin4 x) =

16 sin4 x− 32 sin6 x+ 16 sin8 x =

60. By using the identity cos2 x = 1− sin2 x,we get

1− cos6 x =

1−(1− sin2 x

)3=

1−(1− 3 sin2 x+ 3 sin4 x− sin6 x

)=

3 sin2 x− 3 sin4 x+ sin6 x.

61. tan(−π/6

2

)=

1− cos(−π/6)sin(−π/6)

=

1−√

3/2−1/2

· 22

=2−√

3−1

=√

3− 2

62. sin(−π/4

2

)= −

√1− cos(−π/4)

2=

−

√1−√

2/22

· 22

= −

√2−√

24

=

−

√2−√

22

63. sin(−150◦

2

)= −

√1− cos(−150◦)

2=

−

√1− (−

√3/2)

2· 2

2= −

√2 +√

34

=

−

√2 +√

32

64. cos(

210◦

2

)= −

√1 + cos 210◦

2=

−

√1 + (−

√3/2)

2· 2

2= −

√2−√

34

=

−

√2−√

32

65. Let a = 4, b = 4, and r =√

42 + 42 = 4√

2.If the terminal side of α goes through (4, 4),then tanα = 4/4 = 1 and one can chooseα = π/4. So y = 4

√2 sin(x+ π/4), amplitude

is 4√

2, and phase shift is −π/4.

-9Pi/4 -3Pi/4 Pi/4 7Pi/4x

-6

6

y

66. Let a =√

3, b = 3, and

r =√√

32

+ 32 = 2√

3. If the terminal sideof α goes through (

√3, 3), then

tanα = 3/√

3 =√

3 and one can chooseα = π/3. So y = 2

√3 sin(x+ π/3), amplitude

is 2√

3, and phase shift is −π/3.



-7Pi/3 -5Pi/6 Pi/6 8Pi/3x

-4

4

y

67. Let a = −2, b = 1, andr =

√(−2)2 + 12 =


of α goes through (−2, 1), then tanα = −1/2.Since tan−1(−1/2) ≈ −0.46 and (−2, 1) is inquadrant II, one can choose α = π − 0.46 =2.68. So y =

√5 sin(x + 2.68), amplitude is√

5, and phase shift is −2.68.

-2 2 2Pix

-3

3

y

68. Let a = −2, b = −1, andr =

√(−2)2 + (−1)2 =

√5. If the terminal

side of α goes through (−2,−1), thentanα = 1/2. Since (−2,−1) is in quadrant III,one can choose α = π + tan−1(1/2) ≈ 3.61.Thus, y =

√5 sin(x+ 3.61), amplitude is

√5,

and phase shift is −3.61.

-2Pi -2 2 2Pix

-3

3

y

69. Isolate cos 2x on one side.

2 cos 2x = −1

cos 2x = −12

2x =2π3

+ 2kπ or 2x =4π3

+ 2kπ

x =π

3+ kπ or x =

2π3

+ kπ


π

3+ kπ or x =

2π3

+ kπ

}.

70. Isolate sin 2x on one side.

2 sin 2x = −√

3

sin 2x = −√

32

2x =4π3

+ 2kπ or 2x =5π3

+ 2kπ

x =2π3

+ kπ or x =5π6

+ kπ


2π3

+ kπ or x =5π6

+ kπ

}.

71. Set each factor to zero.

(√

3 cscx− 2)(cscx− 2) = 0

cscx =2√3

or cscx = 2

Thus, x =π

3,2π3,π

6,5π6

plus multiples of 2π.


π

3+ 2kπ,

2π3

+ 2kπ,π

6+ 2kπ,

5π6

+ 2kπ}.

72. Set each factor to zero.

(secx−√

2)(√

3 secx+ 2) = 0

secx =√

2 or secx = − 2√3

Then x =π

4,7π4,5π6,7π6



π

4+ 2kπ,

7π4

+ 2kπ,5π6

+ 2kπ,7π6

+ 2kπ}.


2 sin2 x− 3 sinx+ 1 = 0(2 sinx− 1)(sinx− 1) = 0

sinx =12

or sinx = 1

The x =π

6,5π6,π

2plus multiples of 2π.


π

6+ 2kπ,

5π6

+ 2kπ,π

2+ 2kπ

}.




4 sin2 x− sinx− 3 = 0(4 sinx+ 3)(sinx− 1) = 0

sinx = −34

or sinx = 1

Since sin−1

(−3

4

)≈ −0.848 and 2π− 0.848 ≈

5.44 and π + 0.848 ≈ 3.99, the solution set is{x | x = 5.44 + 2kπ, 3.99 + 2kπ,

π

2+ 2kπ

}.

75. Isolate sinx

2on one side.

sinx

2=

128√

3

sinx

2=

32√

3

sinx

2=√

32

x

2=π

3+ 2kπ or

x

2=

2π3

+ 2kπ

x =2π3

+ 4kπ or x =4π3

+ 4kπ


2π3

+ 4kπ or x =4π3

+ 4kπ}.

76. Isolate cosx

2on one side.

−2 cosx

2=√

2

cosx

2= −

√2

2x

2=

3π4

+ 2kπ orx

2=

5π4

+ 2kπ

x =3π2

+ 4kπ or x =5π2

+ 4kπ


3π2

+ 4kπ or x =5π2

+ 4kπ}.

77. By using the double-angle identity for sine,we get

cosx

2− sin

(2 · x

2

)= 0

cosx

2− 2 sin

x

2cos

x

2= 0

cosx

2(1− 2 sin

x

2) = 0

cosx

2= 0 or sin

x

2=

12.

Thenx

2=π

2,3π2,π

6,5π6

plus mutiples of 2π.

Or x = π, 3π,π

3,5π3


The solution set is{x | x = π + 2kπ,

π

3+ 4kπ,

5π3

+ 4kπ}.

78. Apply the double-angle identity for sineand set the right-hand side to zero.

sin 2x =sinxcosx

2 sinx cosx− sinxcosx

= 0

2 sinx cos2 x− sinx = 0sinx(2 cos2 x− 1) = 0

sinx = 0 or cosx = ± 1√2

x = kπ or x =π

4+kπ

2

The solution set is{x | x = kπ or x =

π

4+kπ

2

}.

79. By the double-angle identity for cosine,we find

cos 2x+ sin2 x = 0cos2 x− sin2 x+ sin2 x = 0

cos2 x = 0x =

π

2+ kπ.


π

2+ kπ

}.



80. By the double-angle identity for sine, we have

tanx

2= sin

(2 · x

2

)sin

x

2cos

x

2

= 2 sinx

2cos

x

2

sinx

2= 2 sin

x

2cos2 x

2

sinx

2

(1− 2 cos2 x

2

)= 0

sinx

2= 0 or cos

x

2= ± 1√

2x

2= kπ or

x

2=π

4+kπ

2x = 2kπ or x =

π

2+ kπ.

The solution set is{x | x = 2kπ or x =

π

2+ kπ

}.

81. By factoring, we obtain

sinx(cosx+ 1) + (cosx+ 1) = 0(sinx+ 1)(cosx+ 1) = 0.

Then

sinx = −1 or cosx = −1

x =3π2

+ 2kπ or x = π + 2kπ.


3π2

+ 2kπ or x = π + 2kπ}.

82. By factoring, we find

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

(cos 2x+ 1)(sin 2x− 1) = 0cos 2x = −1 or sin 2x = 1.

Then 2x = π+2kπ or 2x =π

2+2kπ. Solution

set is{x | x =

π

2+ kπ or x =

π

4+ kπ

}.

83. By multiplying the equation by 2, we obtain

2 sinα cosα = 1sin 2α = 1

2α = 90◦ + k360◦

α = 45◦ + k180◦.

By choosing k = 0, 1, one gets the solutionset {45◦, 225◦} .

84. By using a double-angle identity for cosine,we find

2 cos2 α− 1 = cosα2 cos2 α− cosα− 1 = 0

(2 cosα+ 1)(cosα− 1) = 0

cosα = −12

or cosα = 1.

The solution set is {0◦, 120◦, 240◦} .

85. Suppose 1 + cosα 6= 0. Dividing the equationby 1 + cosα, we get

sinα1 + cosα

= 1

tanα

2= 1

α

2= 45◦ + k180◦

α = 90◦ + k360◦.

One solution is 90◦. On the other hand if1 + cosα = 0, then cosα = −1 and α = 180◦.Note α = 180◦ satisfies the given equation.The solution set is {90◦, 180◦} .

86.

cosα · 1sinα

= cot2 α

cotα = cot2 αcotα− cot2 α = 0

cotα(1− cotα) = 0cotα = 0, 1

The solution set is {45◦, 90◦, 225◦, 270◦}.

87. No solution since the left-hand side is equalto 1 by an identity. The solution set is ∅.



88. No solution since the left-hand side is equalto 1 by an identity. The solution set is ∅.

89. Isolate sin 2α on one side.

sin4 2α =14

sin 2α = ± 4

√14

sin 2α = ± 1√2

2α = 45◦ + k90◦

α = 22.5◦ + k45◦

By choosing k = 0, 1, ..., 7, one gets thesolution set {22.5◦, 67.5◦, 112.5◦, 157.5◦, 202.5◦,247.5◦, 292.5◦, 337.5◦} .

90. By using the double angle identity for sine,we find

2 sinα cosα =sinαcosα

2 sinα cos2 α = sinαsinα(2 cos2 α− 1) = 0

sinα = 0 or cosα = ± 1√2

α = 0◦, 180◦ or α = 45◦ + k90◦.

By choosing k = 0, 1, 2, 3, one gets the solutionset {0◦, 45◦, 135◦, 180◦, 225◦, 315◦}.

91. Suppose tanα 6= 0. Divide the equationby tanα.

2 tanα1− tan2 α

= tanα

21− tan2 α

= 1

2 = 1− tan2 α

tan2 α = −1

The last equation is inconsistent since tan2 αis nonnegative. But if tanα = 0, thenα = 0◦, 180◦ and these two values of αsatisfy the given equation.The solution set is {0◦, 180◦} .

92. Multiply the equation by tanα and notethat tanα · cotα = 1. Then

tan2 α = 1tanα = ±1

α = 45◦ + k90◦.

By choosing k = 0, 1, 2, 3, one gets the solutionset {45◦, 135◦, 225◦, 315◦} .

93. By using the sum identity for sine, we obtain

sin(2α+ α) = cos 3αsin 3α = cos 3αtan 3α = 1

3α = 45◦ + k180◦

α = 15◦ + k60◦.

By choosing k = 0, 1, ..., 5, one gets thesolution set {15◦, 75◦, 135◦, 195◦, 255◦, 315◦} .

94. By appplying the sum identity for cosine,we get

cos(2α+ α) =cos 3αsin 3α

cos 3α =cos 3αsin 3α

.

Suppose cos 3α 6= 0. Divide the equationby cos 3α.

1 =1

sin 3αsin 3α = 1

3α = 90◦ + k360◦

α = 30◦ + k120◦

On the other hand if cos 3α = 0, then3α = 90◦ + k180◦ and α = 30◦ + k60◦.Note the values for α = 30◦+k60◦ include thevalues from α = 30◦ + k120◦. By choosingk = 0, 1, ..., 5 in α = 30◦ + k60◦, one obtainsthat the solution set is{30◦, 90◦, 150◦, 210◦, 270◦, 330◦}.

95. cos 15◦ + cos 19◦ =

2 cos(

15◦ + 19◦

2

)cos

(15◦ − 19◦

2

)=

2 cos 17◦ cos(−2◦) = 2 cos 17◦ cos 2◦



96. cos 4◦ − cos 6◦ =

−2 sin(

4◦ + 6◦

2

)sin(

4◦ − 6◦

2

)=

−2 sin 5◦ sin(−1◦) = 2 sin 5◦ sin 1◦

97. sin(π/4)− sin(−π/8) = sin(π/4) + sin(π/8) =

2 sin(π/4 + π/8

2

)cos

(π/4− π/8

2

)=

2 sin(

3π/82

)cos

(π/82

)=

2 sin(3π/16) cos(π/16)

98. sin(π/12)− sin(π/6) =

2 cos(π/12 + π/6

2

)sin(π/12− π/6

2

)=

2 cos(

3π/122

)sin(−π/12

2

)=

2 cos(3π/24) sin(−π/24) =

−2 cos(π/8) sin(π/24)

99. 2 sin 11◦ cos 13◦ =sin(11◦ + 13◦) + sin(11◦ − 13◦) =sin 24◦ + sin(−2◦) = sin 24◦ − sin 2◦

100. 2 sin 8◦ sin 12◦ =cos(8◦ − 12◦)− cos(8◦ + 12◦) =cos(−4◦)− cos 20◦ = cos 4◦ − cos 20◦

101. 2 cosx

4cos

x

3=

cos(x

4− x

3

)+ cos

(x

4+x

3

)=

cos(− x

12

)+ cos

(7x12

)=

cos(x

12

)+ cos

(7x12

)102. 2 cos(s) sin(3s) = sin(s+ 3s)− sin(s− 3s) =

sin 4s− sin(−2s) = sin 4s+ sin 2s



√√√√1−(√

32

)2

=

−12

and sinβ =

√√√√1−(√

22

)2

=√

22

.

So sin(α− β) = sinα cosβ − cosα sinβ =

√3

2·√

22− −1

2·√

22

=√

6 +√

24

.

104. Since α is in quadrant IV and β is in

quadrant II, cosα =

√√√√1−(−√

22

)2

=

√2

2and cosβ = −

√1−

(12

)2

= −√

32

.

So sin(α+ β) = sinα cosβ + cosα sinβ =

−√

22· −√

32

+√

22· 1

2=√

6 +√

24

.


quadrant II, cosα =

√√√√1−(√

32

)2

=

12

and sinβ =

√√√√1−(−√

22

)2

=√

22

.

So cos(α− β) = cosα cosβ + sinα sinβ =

12· −√

22

+√

32·√

22

=√

6−√

24

.



√√√√1−(√

22

)2

=

−√

22

and cosβ =

√1−

(12

)2

=√

32

.

So cos(α+ β) = cosα cosβ − sinα sinβ =

−√

22·√

32−√

22· 1

2= −√

6 +√

24

.

107. Let a = 0.6, b = 0.4, andr =√

0.62 + 0.42 ≈ 0.72. If theterminal side of α goes through (0.6, 0.4),then tanα = 0.4/0.6 and one can chooseα = tan−1(2/3) ≈ 0.588.Thus, x = 0.72 sin(2t+ 0.588).The values of t when x = 0 are given by

sin(2t+ 0.588) = 02t+ 0.588 = kπ

2t = −0.588 + kπ

t = −0.294 +kπ

2When k = 1, 2 , one gets t ≈ 1.28, 2.85.


CHAPTER 6 TEST 411

108. Let v0 = 400 and d = 3000.

4002 sin 2θ = 32(3000)

sin 2θ =32(3000)

4002

sin 2θ = 0.62θ = sin−1(0.6) ≈ 36.869◦

θ ≈ 18.4◦

Another solution is 2θ = 180◦ − 36.869◦

= 143.131◦. Then θ ≈ 143.13◦

2≈ 71.6◦.

The angles are 18.4◦ and 71.6◦.

Thinking Outside the Box LIX

WRONG = 25938 and RIGHT = 51876

Chapter 6 Test

1.1

cosx· cosx

sinx· 2 sinx cosx = 2 cosx

2. sin(2t+ 5t) = sin 7t

3.1

1− cos y+

11 + cos y

=1 + cos y + 1− cos y

1− cos2y=

2sin2 y

= 2 csc2 y

4. tan(π/5 + π/10) = tan(3π/10)

5.sinβ cosβsinβ/ cosβ

=

sinβ cosβ · cosβsinβ

=

cos2 β =1− sin2 β =

6.1

sec θ − 1− 1

sec θ + 1=

sec θ + 1− (sec θ − 1)sec2 θ − 1

=

2tan2 θ

=

2 cot2 θ =

7. Using the cofunction identity for cosine,we get

cos(π/2− x) cos(−x) =sinx cosx =

2 sinx cosx2

=

sin(2x)2

.

8. Factor the left-hand side and use a half-angleidentity for tangent. Then

tan(t/2) · (cos2 t− 1) =

1− cos tsin t

· (− sin2 t) =

(1− cos t) · (− sin t) =

(cos t− 1) sin t =

cos t sin t− sin t =sin tsec t

− sin t.

9. Since − sin θ = 1, we get sin θ = −1 and the

solution set is{θ | θ =

3π2

+ 2kπ}

.

10. Since cos 3s =12

, we obtain

3s =π

3+ 2kπ or 3s =

5π3

+ 2kπ

s =π

9+

2kπ3

or s =5π9

+2kπ

3.

The solution set is{s | s =

π

9+

2kπ3

or s =5π9

+2kπ

3

}.

11. Since tan 2t = −√

3, we have

2t =2π3

+ kπ

t =π

3+kπ

2.

The solution set is{t | t =

π

3+kπ

2

}.



12.

2 sin θ cos θ = cos θcos θ(2 sin θ − 1) = 0

cos θ = 0 or sin θ = 1/2

The solution set is{θ | θ =

π

2+ kπ,

π

6+ 2kπ,

5π6

+ 2kπ}.

13. By factoring, we obtain

(3 sinα− 1)(sinα− 1) = 0sinα = 1/3 or sinα = 1

α = sin−1(1/3) ≈ 19.5◦ or α = 90◦.

Another solution is α = 180◦−19.5◦ = 160.5◦.The solution set is {19.5◦, 90◦, 160.5◦} .

14.

tan(2α− 7α) = 1tan(−5α) = 1− tan 5α = 1

tan 5α = −15α = 135◦ + k180◦

α = 27◦ + k36◦

The solution set is{27◦, 63◦, 99◦, 171◦, 207◦, 243◦, 279◦, 351◦}.Note, 135◦ and 315◦ are not solutions.

15. Let a = 1, b = −√

3, r =√

12 + (−√

3)2 = 2.

If the terminal side of α goes through(1,−

√3), then tanα = −

√3 and one can

choose α = 5π/3. Then y = 2 sin(x + 5π/3),the period is 2π, amplitude is 2, and phaseshift is −5π/3.

-5Pi/3 -Pi Pi 2Pix

2

y

16. If cscα = 2, then sinα = 1/2.Since α is in quadrant II, we obtain

cosα = −√

1− (1/2)2 =

−√

1− 1/4 = −√

3/4 = −√

3/2,

secα = −2/√

3, tanα =1/2−√

3/2= −1/

√3,

and cotα = −√

3.

17. Even, f(−x) = (−x) sin(−x) =(−x)(− sinx) = x sinx = f(x).

18. By using a half-angle identity, we obtain

sin(−π/6

2

)= −

√1− cos(−π/6)

2=

−

√1−√

3/22

· 22

= −

√2−√

34

=

−

√2−√

32

.

19. If x = y = π/6, then tanx+ tan y =

2 tan(π/6) = 2 ·√

33

=2√

33

and

tan(x+y) = tan(π/6+π/6) = tan(π/3) =√

3.Thus, it is not an identity.

20. Let a = 2, b = −4, r =√

22 + (−4)2 =√

20.If the terminal side of α goes through (2,−4),then one can choose α = tan−1(−4/2) ≈−1.107. Then d =

√20 sin(3t− 1.107).

The values of t when d = 0 are given by

sin(3t− 1.107) = 03t− 1.107 = kπ

3t = 1.107 + kπ

t ≈ 0.4 +kπ

3.

By choosing k = 0, 1, 2, 3, one obtains thevalues of t in [0, 4], namely, 0.4 sec, 1.4 sec,2.5 sec, and 3.5 sec.


TYING IT ALL TOGETHER 413

Tying It All Together

1. Odd, since f(−x) = 3(−x)3 − 2(−x) =−3x3 + 2x = −f(x)

2. Even, f(−x) = 2| − x| = 2|x| = f(x)

3. Odd, f(−x) = (−x)3 + sin(−x) =−x3 − sinx = −f(x)

4. Even, f(−x) = (−x)3 sin(−x) =(−x3)(− sinx) = x3 sinx = f(x)

5. Even, f(−x) = (−x)4 − (−x)2 + 1 =x4 − x2 + 1 = f(x)

6. Odd, f(−x) =1−x

= −1x

= −f(x)

7. Even, f(−x) =sin(−x)−x

=− sinx−x

=

sinxx

= f(x)

8. Even, f(−x) = | sin(−x)| = | − sinx| =| sinx| = f(x)

9. It is not an identity. If α = β = π/6,then sin(α+ β) = sin(π/3) =

√3/2 and

sin(π/6) + sin(π/6) = 2 · (1/2) = 1.

10. It is not an identity. If α = β = 1, then(α+ β)2 = 22 = 4 and α2 + β2 = 12 + 12 = 2.

11. Identity

12. Identity

13. It is not an identity; for if x = 1, thensin−1(1) = π/2 ≈ 1.57 and 1/ sin 1 ≈ 1.19 .

14. It is not an identity; for if x =√

7π/6,

then sin2√

7π/6 is a positive number and

sin((√

7π/6)2)

= sin(7π/6) = −1/2.

15. Form a right triangle with α = 30◦, a = 4.

"""""""""""""""

30◦

4

b

cβ

Since tan 30◦ = 4/b and sin 30◦ = 4/c, we getb = 4/ tan 30◦ = 4

√3 and c = 4/ sin 30◦ = 8.

Also β = 90◦ − 30◦ = 60◦.

16. Form a right triangle with a =√

3, b = 1.

"""""""""""""""

α

√3

1

cβ

Then c =√

(√

3)2 + 12 = 2 by thePythagorean Theorem.Since tanα =

√3/1 =

√3, we get

α = 60◦ and β = 30◦ .

17. Form a right triangle with b = 5 asshown below.

"""""""""""""""

β

5

a

cα

Since cosβ = 0.3, we findβ = cos−1(0.3) ≈ 72.5◦ and α = 17.5◦.Since sin 72.5◦ = 5/c and tan 72.5◦ = 5/a,we obtain c = 5/ sin 72.5◦ ≈ 5.2 anda = 5/ tan 72.5◦ ≈ 1.6.



18. Form a right triangle with a = 2.

"""""""""""""""

α

2

b

cβ

Since sinα = 0.6, we obtainα = sin−1(0.6) ≈ 36.9◦ and β = 53.1◦.Since sin 36.9◦ = 2/c and tan 36.9◦ = 2/b,we get c = 2/ sin 36.9◦ ≈ 3.3 andb = 2/ tan 36.9◦ ≈ 2.7.

19.12

+ i

√3

2

20. 2

(−1

2+ i ·

√3

2

)= −1 + i

√3

21. cos2 225◦ + 2i cos 225◦ sin 225◦ − sin2 225◦ =cos2 225◦ − sin2 225◦ + 2i cos 225◦ sin 225◦ =cos(2 · 225◦) + i · sin(2 · 225◦) =cos 450◦ + i · sin 450◦ =cos 90◦ + i · sin 90◦ = i

22. cos2 3◦ − (i sin 3◦)2 = cos2 3◦ + sin2 3◦ = 1

23. (2 + i)2(2 + i) = (4 + 4i− 1)(2 + i) =(3 + 4i)(2 + i) = 6 + 3i+ 8i− 4 = 2 + 11i

24. (√

2(1− i))4 =√

24(1− i)4 = 4

((1− i)2

)2 =4(1− 2i− 1)2 = 4(−2i)2 = 4(−4) = −16

25. minutes 26. seconds

27. unit

28. π

29. αr

30. rω

31. y, x

32. fundamental

33. amplitude

34. phase shift

Concepts of Calculus

1. Consider the isosceles triangle below.

��SSSSSSSSSSSS

θ/2

h

b

r r

A pentagon inscribed in a circle of radius rconsists of 5 triangles each one like the oneshown above with radius r, θ = 72◦ or θ/2 =36◦, and where b is the base.

Since h = r cos(36◦) and sin(36◦) =b/2r

,

we get b = 2r sin(36◦) and the area of thetriangle is

12bh =

12

(2r sin(36◦))(r cos(36◦))

= r2 sin(36◦) cos(36◦)12bh =

r2

2sin(72◦).

Hence, the area of the pentagon is

5 sin(72◦)2

r2.

2. An n-gon consists of n triangles like the one

shown in Exercise 1 where θ/2 =360◦/n

2.

For this n-gon, h = r cos(

360◦

2n

),

sin(

360◦

2n

)=b/2r

or b = 2r sin(

360◦

2n

),

and the area of the triangle is given by


TYING IT ALL TOGETHER 415

12bh =

12

(2r sin

(360◦

2n

))(r cos

(360◦

2n

))= r2 sin

(360◦

2n

)cos

(360◦

2n

)= r2 sin

(180◦

n

)cos

(180◦

n

)12bh =

r2

2sin(

360◦

n

).

Thus, the area An of an n-gon inscribed in acircle of radius r is

An =nr2

2sin(

360◦

n

).

Another formula for An can be obtained byusing the above calculations and the fact that

sin(90◦ − α) = cosα

andcos(90◦ − α) = sinα.

Thus, an equivalent formula for the area of then-gon is

An = nr2 sin(

180◦

n

)cos

(180◦

n

)An = nr2 sin

(90◦ − 180◦

n

)cos

(90◦ − 180◦

n

).

3. For an n-gon, the constant of proportionality is

n

2sin(

360◦

n

).

For a decagon (n = 10), kilogon n = 1000, andmegagon (n = 106), the constants of propor-tionality are

2.938926261,

3.141571983, and

3.141592654, respectively.

4. The shape of the n-gon as n increasesapproaches the shape of a circle.

5. Note, when n = 106, the megagon is almosta circle. We will use the constant for themegagon calculated in Exercise 3. Thus, thearea of a circle of radius r could be approxi-mated by the area of the magagon which

3.141592654r2.

6. As derived in part Exercise 2, the base of thetriangle is

b = 2r sin(

180◦

n

).

Thus, the perimeter P of an n-gon is P = nb,or

P = 2nr sin(

180◦

n

)or equivalently

P = 2nr cos(

90◦ − 180◦

n

).

7. When n is a large number, the shape of ann-gon approximates the shape of a circle.Consequently, the circumference C of acircle of radius r is approximately

C ≈ 2nr sin(

180◦

n

).

Note, if n = 106 then

n sin(

180◦

n

)≈ 3.141592654.

Thus, the circumference is

C ≈ 2r(3.141592654).



8. Note, π is the ratio

π =circumference

diameter.

From Exercise 6, the circumference of an n-gonis

P = 2nr cos(

90◦ − 180◦

n

)which is approximately the circumference of acircle. Thus, when n = 106 we obtain

π =circumference

diameter

π ≈ P

2r

π ≈2nr cos

(90◦ − 180◦

n

)2r

π ≈ n cos(

90◦ − 180◦

n

).

In particular, when n = 106 we find

π ≈ 106 cos(

90◦ − 180◦

106

)π ≈ 3.141592654

Alternatively, we can find π by using the for-mula

π =area of a circle with radius r

r2.

From Exercise 2, the area of an n-gon is

An = nr2 sin(

90◦ − 180◦

n

)cos

(90◦ − 180◦

n

)which is approximately the area of the circle.Thus, when n = 106 we obtain

π =area of circle with radius r

r2

π ≈nr2 sin

(90◦ − 180◦

n

)cos

(90◦ − 180◦

n

)r2

π ≈ n sin(

90◦ − 180◦

n

)cos

(90◦ − 180◦

n

)π ≈ 106 sin

(90◦ − 180◦

106

)cos

(90◦ − 180◦

106

)π ≈ 3.141592654.


For Thought x - TopCatMath · + sin2 x= cos2 x+ sin2 x= 1 15. sin(x) 1=sin(x) + cos(x) 1=cos(x) =...

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Transcript of For Thought x - TopCatMath · + sin2 x= cos2 x+ sin2 x= 1 15. sin(x) 1=sin(x) + cos(x) 1=cos(x) =...