Pre-Calc 1e SSM Chapter 6 part1 FINAL

361

CHAPTER 6

Section 6.1 Solutions--------------------------------------------------------------------------------

1. ( ) ( ) ( )sin 60 cos 90 60 cos 30= − =� � � � 3. ( ) ( )cos sin 90x x= −�

5. ( ) ( ) ( )csc 30 sec 90 30 sec 60= − =� � � �

7. ( ) ( )( ) ( )sin cos 90 cos 90x y x y x y+ = − + = − −� �

9. ( ) ( )( ) ( )cos 20 sin 90 20 sin 70A A A+ = − + = −� � � �

11. ( ) ( )( ) ( )cot 45 tan 90 45 tan 45x x x− = − − = +� � � �

13. ( ) ( )( ) ( )csc 60 sec 90 60 sec 30θ θ θ− = − − = +� � � �

15. sin csc sinx x x=1

sin x1

=

17. 1 cos 1

sec( ) cotcos( ) sin cos( )

xx x

x x x

− = =

−

cos x 1csc

sin sin( )x

x x

= =

19. 1 1

csc( )sin sinsin( ) sin

x x xx x

− = =− −

sin x 1= −

21. 2 1

sec cos( ) tancos

x x xx

− + = cos x 2 2 2tan 1 tan secx x x+ = + =

23. ( )2 2 2 2 2sin cot 1 sin csc sinx x x x x+ = =2

1

sin x1

=

25. ( )( ) 2 2sin cos sin cos sin cosx x x x x x− + = −

27.

1

csc sin

cot

x x

x=

cos

sin

x

x

1sec

cosx

x= =

29. 1 cot( ) 1 cot

11 cot 1 cot

x x

x x

− − += =

+ +

31. ( ) ( )2 2

4

2

1 cos 1 cos1 cos

1 cos

x xx

x

− +−=

+ 21 cos x+

2 21 cos sinx x= − =

33 ( ) ( )2 2

4

2

1 cot 1 cot1 cot

1 cot

x xx

x

+ −−=

− 21 cot x−

2 21 cot cscx x= + =

35.

( )2 2 1 cossin 1 cos1 1 1

1 cos 1 cos

xx x

x x

−−− = − = −

− −

( )1 cos

1 cos

x

x

+

−( )1 1 cos 1 1 cos cosx x x= − + = − − = −

362

37. 2 2

2 2

sin cossin cos

tan cot sin coscos sin2cos 2cossin costan cot

cos sin

x xx x

x x x xx xx xx xx x

x x

−−

−+ = + =

+ +2 2sin cos

sin cos

x x

x x

+2

2 22 2 2 2

2 2

1

2 2

2cos

sin cos2cos sin cos 2cos

sin cos

sin cos 1

x

x xx x x x

x x

x x

=

+

−= + = − +

+

= + =

��

39.

( ) ( )2 2

2

sin cos sin cos

sin 2sin cos

x x x x

x x x

+ + −

= +( )2 2

What remains after cancellation = 1

cos sin 2sin cosx x x x+ + −��

( )2

What remains after cancellation = 1

cos 2,x+ =��

as claimed.

41. ( )( ) ( )2 2 2csc 1 csc 1 csc 1 1 cot 1 cotx x x x x+ − = − = + − = , as claimed.

43. 2 2sin cos sin cos 1 1 1

tan cot csc seccos sin sin cos sin cos sin cos

x x x xx x x x

x x x x x x x x

+ + = + = = = =

,

as claimed.

45. ( )22 2 21 1 sin2 sin 1 cos 1 cos

sec coscos cos cos cos cos

xx x xx x

x x x x x

+ −− += = = + = +

47. [ ][ ] [ ][ ] 2 2cos( ) 1 1 cos cos 1 cos 1 cos 1 sinx x x x x x− − + = − + = − = −

49.

1 cos

sec( )cot sin( ) cos sin coscos( ) sin1

1csc( ) cos( ) sin cos sin

sin( )

x

x x x x x xx x

x x x x x

x

− −−= = = − = −

− −

−

51. 2 2

2 2

2 2

1 1 1 1sin cos 1

1 1csc sec

sin cos

x xx x

x x

+ = + = + = , as claimed.

53. ( ) ( )( )( )

2

2 2

1 sin 1 sin1 1 2 22sec

1 sin 1 sin 1 sin 1 sin 1 sin cos

x xx

x x x x x x

+ + −+ = = = =

− + + − −, as claimed.

55. ( )2 2 1 cossin 1 cos

1 cos 1 cos

xx x

x x

−−= =

− −

( )1 cos

1 cos

x

x

+

−1 cos x= + , as claimed.

363

57.

( )( ) 2 2

2

sec tan sec tan sec tansec tan

sec tan sec tan

1 tan

x x x x x xx x

x x x x

x

+ − −+ = =

− −

+=

( ) 2tan x− 1,

sec tan sec tanx x x x=

− −

as claimed.

59.

2cos sin1 sin

csc tan sin cossin cos1 cossec cot

cos sin

x xx

x x x xx xxx x

x x

−−

−= =

+ +2sin cos

sin cos

x x

x x

+

2

2

cos sin

sin cos

x x

x x

−=

+, as claimed.

61.

( )( )

( ) ( )

2 22

2 22

1 sin 1 sin sin sin 2cos 1 sin

cos 3 sin 41 sin 3

sin 1 sin 2

x x x xx x

x xx

x x

− + + − − −+ + = =+ − −− +

− + −=

(sin 2)x− −

sin 1 1 sin,

sin 2 2 sin(sin 2)

x x

x xx

+ += =

+ ++

as claimed.

63.

( )

( )

1

2 2

2 2 2

1 sin cos 1 sin cos 1 1sec tan cot

cos cos sin cos cos sin cos cos sin

1 1 1 csccsc ,

cos sin cos cos

x x x xx x x

x x x x x x x x x

xx

x x x x

= +

+ = + = =

= = =

��

as claimed.

65. Observe that the left-side of the equation simplifies as follows:

( )( ) ( )

( )( )

2 2 2 2

2 2 2 2

cos tan sec tan sec cos tan sec

cos tan 1 tan cos

x x x x x x x x

x x x x

− + = −

= − + = −

Since 2cos x− is, in fact, never equal to 1 (being a non-positive quantity), the given

equation is a conditional.

67. Observe that

3

cot cot

1cot

csc cot 1 1 1 1 cos 1sin cot cot cos cot cot1 1sec tan sin sin tan sin tan

tan tancos cos x x

xx x xx x x x x xx x x x x x x

x xx x = =

= = = = =

So, the given equation is an identity.

69. The equation sin cos 2x x+ = is a conditional since, for instance, it does not hold if

x = 0 (since the left-side reduces to 1).

71. Observe that

( )2 2 2 2tan sec tan 1 tan 1 1x x x x− = − + = − ≠ .

So, the given equation is a conditional (which, incidentally, is never true).

364

73. The equation 2sin 1 cosx x= − is a conditional. To see this, observe that the right-

side reduces as follows: 2 21 cos sin sinx x x− = = . As such, the given equation is not

true for 32

x π= , for instance.

(Note: The equation IS an identity if one restricts attention to only those x-values for

which sin 0x > .)

75. Observe that 2 2sin cos 1 1x x+ = = . So, the given equation is an identity.

77. Conditional. Note that it is false when 0x = .

79. ( ) ( )2 2 2sec 1 tan tanx x xπ π π π= + = +

81. 2 2 2 2 2 21 tan 1 tan sec sech hθ θ θ θ+ = ⇒ = + = = .

83. Simplified the two fractions in Step 2 incorrectly. The correct computations are: 2

2

cos cos cos

sin cos sin cos sin1

cos cos

sin sin sin

cos sin cos sin cos1

sin sin

x x x

x x x x x

x x

x x x

x x x x x

x x

= =− −

−

= =− −−

85. Need to observe that 2tan 1x = is a conditional, which does not hold for all values of

x. Verifying the truth of the equation for a particular value of x (here, 4

x π= ) is not

enough to prove that it is an identity.

87. False. The equation 2sin 1x = is not an identity since even though it is true for

2x nπ π= + (n is an integer), it is false for all other values of x for which sin x is defined.

89. The equation 2cos 1 sinθ θ= − is true whenever cos 0θ > (since 21 sin θ− =

2cos cosθ θ= .) This occurs for those angles θ whose terminal side is in QI or QIV.

91. 2csc cscθ θ= − occurs for those angles θ whose terminal side is in QIII or QIV.

93. The equation sin( ) sin sinA B A B+ = + is not true in general. For instance, let

30A = � and 60B = �

. Then, ( ) ( )sin 30 60 sin 90 1+ = =� � � , whereas

( ) ( ) 1 3 1 3sin 30 sin 60 1

2 2 2

++ = + = ≠� �

.

95. No. For instance, take 4

A π= . Note that tan(2 )A is not defined, whereas

( )42 tan 2π = .

97.

( ) ( )2 2

2 2

sin cos sin cos

sin 2 sin cos

a x b x b x a x

a x ab x x

+ + −

= + 2 2 2 2cos sin 2 sin cosb x b x ab x x+ + −

( ) ( ) ( )( )

2 2

2 2 2 2 2 2 2 2 2 2 2 2

1

cos

sin cos sin cos

a x

a b x a b x a b x x a b

=

+

= + + + = + + = +��

365

99. Observe that for any integer n,

( )( ) ( )2

2 2

1 1 1csc 2 sec .

sin 2 sin cosn

n

π

π πθ π θ

θ π θ θ+ + = = = =

+ + +

101. Observe that

[ ]

1 1csc 2 sec sin( ) sin( )

2 2sin 2 cos

2 2

1 1sin( )

sin cos2 2

1 1sin

sin cos2 2

1 1 1sin

cos sin

π ππ θ θ θ θ

π πθ π θ

θπ π

θ θ

θπ π

θ θ

θθ θ

− − ⋅ − ⋅ − = ⋅ ⋅ −

− + + − −

= ⋅ ⋅ −

− + − −

= − ⋅ −

+ −

= ⋅ ⋅ = seccos

θθ

=

103. The correct identity is cos( ) cos cos sin sinA B A B A B+ = − .

Consider the graphs of the following functions. Note that the graphs of 1y (BOLD) and

3y are the same. (The graph of 2

y is dotted.)

1 cos4

y xπ

= +

( ) ( )2 cos cos sin sin4 4

y x xπ π

= +

( ) ( )3 cos cos sin sin4 4

y x xπ π

= −

105. The correct identity is sin( ) sin cos cos sinA B A B A B+ = + .

Consider the graphs of the following functions. Note that the graphs of 1y (BOLD) and

2y are the same. (The graph of 3

y is dotted.)

1 sin4

y xπ

= +

( ) ( )2 sin cos cos sin4 4

y x xπ π

= +

( ) ( )3 sin cos cos sin4 4

y x xπ π

= −

366

107.

( )22 2

2

sin 1 sin

cos

cos ,

a a a

a

a

θ θ

θ

θ

− = −

=

=

since 2 2π πθ− ≤ ≤ .

109.

( )2 2 2

2

sec sec 1

tan

tan ,

a a a

a

a

θ θ

θ

θ

− = −

=

=

since 2

0 πθ≤ < .

Section 6.2 Solutions --------------------------------------------------------------------------------

1. Using the addition formula sin( ) sin cos cos sinA B A B A B− = − yields

sin sin sin cos cos sin12 3 4 3 4 3 4

3 2 1 2 2 3 1 6 2

2 2 2 2 2 2 2 4

π π π π π π π = − = −

− = − = − =

3. Using the even identity for cosine, along with the addition formula

cos( ) cos cos sin sinA B A B A B+ = − yields

5 5cos cos cos cos cos sin sin

12 12 6 4 6 4 6 4

3 2 1 2 2 3 1 6 2

2 2 2 2 2 2 2 4

π π π π π π π π − = = + = −

− = − = − =

5. First, we need to compute both sin12

π

and cos12

π

. Then, we use the definition

of tangent to compute tan12

π

. To this end, we have the following:

Step 1: Using the odd identity for sine and Exercise 1 yields

6 2sin sin

12 12 4

π π − − = − = −

Step 2: Using the even identity for cosine and the addition formula

cos( ) cos cos sin sinA B A B A B− = + yields

cos cos cos cos cos sin sin12 12 3 4 3 4 3 4

1 2 3 2 2 1 3 2 6

2 2 2 2 2 2 2 4

π π π π π π π π − = = − = +

+ = + = + =

Step 3: Using the definition of tangent and the computations in Steps 1 and 2 yields

6 2sin sin

4 6 212 12tan

12 6 2 6 2cos cos

12 12 4

6 2 6 2 6 2 2 6 2 8 4 32 3

6 2 46 2 6 2

π π

π

π π

− − − − − + − = = = = + + −

− + − − + − − += ⋅ = = = − +

−+ −

367

7. Using the addition formula sin( ) sin cos cos sinA B A B A B+ = + yields

( ) ( ) ( ) ( ) ( ) ( )sin 105 sin 60 45 sin 60 cos 45 cos 60 sin 45

3 2 1 2 2 6

2 2 2 2 4

= + = + =

+ = + =

� � � � � � �

9. First, we need to compute both ( )sin 105− � and ( )cos 105− � . Then, we use the

definition of tangent to compute ( )tan 105− � . To this end, we have the following:

Step 1: Using the odd identity for sine and Exercise 7 yields

( ) ( ) 2 6sin 105 sin 105

4

+− = − = −

� �


cos( ) cos cos sin sinA B A B A B+ = − yields

( ) ( ) ( ) ( ) ( ) ( ) ( )cos 105 cos 105 cos 60 45 cos 60 cos 45 sin 60 sin 45

1 2 3 2 2 6

2 2 2 2 4

− = = + = − =

− = − =

� � � � � � � �

Step 3: Using the definition of tangent and the computations in Steps 1 and 2 yields

( )( )( )

2 6

sin 105 4 6 2tan 105

cos 105 2 6 2 6

4

6 2 2 6 6 2 2 6 2 8 4 32 3

2 6 42 6 2 6

+−

− − − − = = =− − −

− − + − − − − −= ⋅ = = = +

− −− +

�

�

�

11. Using the odd identity of tangent, followed by Exercise 5, yields

1 1 1 1cot

12 2 3tan tantan12 1212

1 1 2 3 2 32 3

4 32 3 2 3 2 3

π

π ππ

= = = =

− − +− −− −

+ += = ⋅ = = +

−− − +

368

13. First, we need to compute 11

cos12

π −

. Then, we use the definition of secant to

compute11

sec12

π −

. To this end, we have the following:


cos( ) cos cos sin sinA B A B A B− = + yields

1 0

11 11 11cos cos cos

12 12 12

11 11cos cos sin sin cos

12 12 12

π π ππ

π π ππ π

=− =

− = = −

= + = −

(1)

Using the addition formula cos( ) cos cos sin sinA B A B A B− = + (again) now yields

cos cos cos cos sin sin12 3 4 3 4 3 4

1 2 3 2 2 1 3 2 6

2 2 2 2 2 2 2 4

π π π π π π π = − = +

+ = + = + =

Hence, using this in (1) yields 11 2 6

cos12 4

π + − = −

.

Step 2: Using the definition of secant and the computation in Step 1 now yields

( )

11 1 1 4sec

1112 2 62 6cos12 4

4 2 64 2 62 6

2 62 6 2 6

π

π

− − = = =

+ +− −

− −− −= ⋅ = = −

−+ −

15. ( )( ) ( ) ( ) ( ) ( )

( )

1 1csc 255

sin 45 300 sin 45 cos 300 cos 45 sin 300

1

2 1 2 3

2 2 2 2

4 2 66 2

2 62 1 3

− = =− −

=

− −

−= ⋅ = −

−+

�

� � � � � �

17. Using the addition formula cos( ) cos cos sin sinA B A B A B− = + with 2A x= and

3B x= , followed by the even identity for cosine at the very end, yields

sin 2 sin 3 cos 2 cos 3 cos(2 3 ) cos( ) cosx x x x x x x x+ = − = − = .

369

19. Using the addition formula sin( ) sin cos cos sinA B A B A B− = − with A x= and

2B x= , followed by the odd identity for sine at the very end, yields

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

21. Using the addition formula sin( ) sin cos cos sinA B A B A B+ = + with A xπ= − and

B x= yields sin cos( ) cos sin( ) sin( ) sin 0x x x x x xπ π π π− + − = + − = = .

23. Observe that

( ) ( )

( )

2 2

2 2 2 2

2 2

1

sin sin cos cos 2

sin 2sin sin sin cos 2cos cos cos 2

sin cos

A B A B

A A B B A A B B

A A

=

− + − − =

− + + − + − =

+��

( )2 2

1

sin cosB B

=

+ +��

[ ]2 sin sin cos cos 2A B A B− + −

[ ] [ ]2 sin sin cos cos 2 cos cos sin sin 2cos( )A B A B A B A B A B

=

− + = − + = − −

25. Observe that

( ) ( )

( ) ( )

( )

2 2

2 2 2 2

2 2

1

2 sin cos cos sin

2 sin 2sin cos cos cos 2cos sin sin

2 sin cos

A B A B

A A B B A A B B

A A

=

− + − + =

− + + − + +

− +��

( )2 2

1

sin cosB B

=

− +��

[ ]

[ ]

2 sin cos cos sin

2 sin cos cos sin 2sin( )

A B A B

A B A B A B

− + =

− + = − +

27. Using the addition formula tan tan

tan( )1 tan tan

A BA B

A B

−− =

+ with 49A = �

and 23B = �

yields tan 49 tan 23

tan(49 23 ) tan(26 )1 tan 49 tan 23

−= − =

+

� ��

� �

370

29. Observe that 1 1

cos( ) cos cos sin sin sin sin3 4

α β α β α β α β

+ = − = − − −

. We

need to compute both sinα and sin β . We proceed as follows:

Since the terminal side of α is in QIII, we

know that sin 0α < . As such, since we

are given that 1

cos3

α = − , the diagram is:

From the Pythagorean Theorem, we have

2 2 2( 1) 3z + − = , so that 2 2z = − .

Thus, 2 2sin

3α = − .

Since the terminal side of β is in QII, we

know that sin 0β > . As such, since we

are given that 1

cos4

β = − , the diagram is:


2 2 2( 1) 4z + − = , so that 15z = .

Thus, 15sin

4β = .

Hence,

1 1 1 1 2 2 15 1 2 30 1 2 30cos( ) sin sin

3 4 3 4 3 4 12 12 12α β α β

+ + = − − − = − − − − = + =

α β

371

31. Observe that 3 1

sin( ) sin cos cos sin cos cos5 5

α β α β α β β α

− = − = − −

. We

need to compute both cosα and cos β . We proceed as follows:


know that cos 0α < . As such, since we

are given that 3

sin5

α = − , the diagram is

as follows:


2 2 23 5z + = , so that 4z = − .

Thus, 4

cos5

α = − .

Since the terminal side of β is in QI, we

know that cos 0β > . As such, since we

are given that 1

sin5

β = , the diagram is as

follows:


2 2 21 5z + = , so that 24 2 6z = = .

Thus, 2 6

cos5

β = .

Hence,

3 1 3 2 6 4 1 6 6 4 6 6 4sin( ) cos cos

5 5 5 5 5 5 25 25 25α β β α

− − + − = − − = − − − = + =

αβ

372

33. Observe that sincetan tan

tan( )1 tan tan

α βα β

α β

++ =

−, we need to determine the values of

cosα and sin β so that we can then use this information in combination with the given

values of sinα and cos β in order to compute tan and tanα β . We proceed as

follows:


know that cos 0α < . As such, since we

are given that 3

sin5

α = − , the diagram is:


2 2 2( 3) 5z + − = , so that 4z = − .

As such, 4

cos5

α = − , and so,

3sin 35tan

4cos 45

αα

α

−= = =

−

Since the terminal side of β is in QII, we

know that sin 0β > . As such, since we

are given that 1cos

4β = − , the diagram is:


2 2 2( 1) 4z + − = , so that 15z = .

As such, 15

sin4

β = , and so

15sin 4tan 15

1cos4

ββ

β= = = −

−

Hence,

( )

3 3 4 1515

tan tan 4 4tan( )31 tan tan 4 3 15

1 154 4

3 4 15 4 3 15 12 16 15 9 15 12(15) 192 25 15

16 9(15) 1194 3 15 4 3 15

α βα β

α β

−−

++ = = =

− +− −

− − − − + −= ⋅ = =

− −+ −

35. Observe that

sin( ) sin( ) (sin cos sin cosA B A B A B B A+ + − = + ) (sin cos sin cosA B B A+ − )

2sin cosA B=


α

β

373

37. The equation sin cos2 2

x xπ π

− = +

is a conditional since it is false for 2

xπ

= , for

instance.

39. Observe that ( ) ( ) ( ) ( )2

4 4 4 2sin sin cos sin cos sin cosx x x x xπ π π+ = + = + . So, the

given equation is an identity.

41. Observe that

( ) ( )( )2 2

22 2

1 cos 1 cos1 cos cos sin sin1 cos 2

2 2 2

2 2cos1 cos sin

2

x xx x x xx

xx x

− − −− −−= =

−= = − =


43. Observe that

sin 2 sin( ) sin cos sin cos 2sin cosx x x x x x x x x= + = + = .


45. The equation sin( ) sin sinA B A B+ = + is a conditional since it is false for

2A B

π= = , for instance.

47. Observe that

( )tan tan 0 tan

tan( ) tan1 tan tan 1 0 tan

π β βπ β β

π β β

+ ++ = = =

− −.


49. Observe that

1 1cot(3 )

tan(3 ) tanx

x xπ

π+ = =

+.


51. The given equation is a conditional since it is false when 4

x π= , for instance.

53. Using the addition formula sin( ) sin cos cos sinA B A B A B+ = + with A x= and

3B

π= yields cos sin sin cos sin

3 3 3y x x x

π π π = + = +

.

So, the graph is the graph of siny x= shifted to the left 3

π units, as seen below.

374

55. Using the addition formula cos( ) cos cos sin sinA B A B A B− = + with A x= and

4B

π= yields

sin sin cos cos cos cos sin sin cos4 4 4 4 4

y x x x x xπ π π π π

= + = + = −

.

So, the graph is the graph of cosy x= shifted to the right 4

π units, as seen below.

57. Using the addition formula sin( ) sin cos cos sinA B A B A B+ = + with A x= and

3B x= yields

[ ] ( )sin cos3 cos sin 3 sin cos3 cos sin 3 sin 3 sin 4y x x x x x x x x x x x= − − = − + = − + = −

So, the general shape of the graph is that of siny x= , but with period is 2

4 2

π π= and

then reflected over the x-axis, as seen below.

59. Observe that ( )4

1 tantan

1 tan

xx

x

π+= −

−.

So, the graph is as follows:

61. Observe that ( )6

1 3 tantan

3 tan

xx

x

π+= +

−.


63. ( ) ( ) ( )2 3 4 52 2

4 4 4 2 2 2! 3! 4! 5!cos cos cos sin sin cos sin 1 ...x x x xx x x x x xπ π π− = + = + = + − − + + − − + +

375

65. Observe that ( ) 2 1 2 12 1

2 1 2 1

tan tantan

1 tan tan 1

m m

m m

θ θθ θ

θ θ

− −− = =

+ +.

67. ( ) ( ) ( ) ( ) ( )cos cos cos sin sinE A kz ct A kz ct kz ct= − = +

If z

nλ

= , an integer, then z nλ= , so that 2

2kz kn kn nk

πλ π

= = =

. Hence,

sin( ) 0kz = . So, the formula for E simplifies to ( ) ( ) ( )cos cos cosE A kz ct A kz ct= − = .

69. tan( ) tan tanA B A B+ ≠ + . Should have used tan tan

tan( )1 tan tan

A BA B

A B

++ =

−.

71. False. Observe that

( ) ( ) ( ) ( ) ( ) ( )cos 15 cos 45 30 cos 45 cos 30 sin 45 sin 30

2 3 2 1 6 2

2 2 2 2 4

= − = +

+ = + =

� � � � � � �

while

( ) ( ) 2 3 2 3cos 45 cos 30

2 2 2

−− = − =

� � .

73. False. Observe that ( )4

1 tantan

1 tan

xx

x

π+= +

−.

75. Approaches may vary slightly here, but they will result in equivalent answers as

long as the identities are applied correctly. The final form of the simplification depends

on how you initially choose to group the input quantities. For instance, the following is

a legitimate simplification:

( ) ( )( )[ ] [ ]

sin sin sin( )cos sin cos( )

sin cos sin cos cos sin cos cos sin sin

sin cos cos sin cos cos cos cos sin sin sin sin

A B C A B C A B C C A B

A B B A C C A B A B

A B C B A C A B C A B C

+ + = + + = + + +

= + + −

= + + −

77. Observe that sin( ) sin cos sin cosA B A B B A− = − . We seek values of A and B such

that the right-side equals sin sinA B− . Certainly, if we choose values of A and B such

that cos cos 1B A= = , then this occurs. And this occurs precisely when A and B are

integer multiples of 2π (and they need not be the same multiple of 2π !). So, the

solutions are 2 , 2B m A nπ π= = , where n and m are integers.

376

79. a. The following is the graph of

1

sin1 1 cos1cos sin

1 1y x x

− = −

, along

with the graph of cosy x= (dotted

graph):

b. The following is the graph of

2

sin 0.1 1 cos 0.1cos sin

0.1 0.1y x x

− = −

,

along with the graph of cosy x= (dotted

graph):

c. The following is the graph of

3

sin 0.01 1 cos 0.01cos sin

0.01 0.01y x x

− = −

,

along with the graph of cosy x= (dotted

graph):

Refer back to Exercise 63 for a

development of this difference quotient

formula. Looking at the sequence of graphs

in parts a – c, it is clear that this sequence of

graphs better approximates the graph of

cosy x= as 0h → .

377

81. Observe that

( )

( )

( )

sin 2( ) sin 2 sin 2 cos(2 ) sin(2 ) cos2 sin 2

sin 2 cos(2 ) 1 sin(2 ) cos 2

sin 2 cos(2 ) 1 sin(2 ) cos 2

sin(2 ) 1 cos(2 )cos 2 sin 2

x h x x h h x x

h h

x h h x

h

x h h x

h h

h hx x

h h

+ − + −=

− +=

−= +

− = −

a. The following is the graph of

1

sin(2) 1 cos(2)cos 2 sin 2

1 1y x x

− = −

,

along with the graph of 2cos 2y x=

(dotted graph):

b. The following is the graph of

2

sin(0.2) 1 cos(0.2)cos 2 sin 2

0.1 0.1y x x

− = −

,

along with the graph of 2cos 2y x= (dotted

graph):

c. The following is the graph of

3

sin(0.02) 1 cos(0.02)cos2 sin 2

0.01 0.01y x x

− = −

,

along with the graph of 2cos 2y x=

(dotted graph):

Looking at the sequence of graphs in parts a

– c, it is clear that this sequence of graphs

better approximates the graph of

2cos 2y x= as 0h → .

378

83. Observe that ( ) ( )

( )

sin cos( ) sin sin( )cos sin cos( ) 1 sin( )cossin( ) sin sin cos( ) sin( ) cos sin

sin cos( ) 1 sin( )cos cos( ) 1 sin( ) sin( ) 1 cos( )sin cos cos sin

x h x h x x h h xx h x x h h x x

h h h h

x h h x h h h hx x x x

h h h h h h

− + − ++ − + −= = =

− − − = + = + = −

85. sin( ) 0

sin cos sin cos 0

sin cos sin cos

sin sin

cos cos

tan tan

x y

x y y x

x y y x

x y

x y

x y

+ =

+ =

= −

= −

= −

87.

( )

tan( ) 2

tan tan2

1 tan tan

tan tan 2 2 tan tan

tan 1 2 tan 2 tan

2 tantan

1 2 tan

x y

x y

x y

x y x y

x y y

yx

y

+ =

+=

−

+ = −

+ = −

−=

+

Section 6.3 Solutions --------------------------------------------------------------------------------

1. Since 1

sin5

x = and cos 0x < , the

terminal side of x must be in QII.

The diagram is as follows:


( )2

2 21 5z + = , so that 2z = − .

As such, 2

cos5

x = − . Therefore,

1 2 4sin 2 2sin cos 2

55 5x x x

= = − = −

3. Since 5

cos13

x = and sin 0x < , the

terminal side of x must be in QIV.

The diagram is as follows:


2 2 25 13z + = , so that 12z = − .

As such, 12

sin13

x = − , so that

12

sin 1213tan5cos 5

13

xx

x

−= = = − .

Therefore,

22

12 2422 tan 1205 5tan 2

1191 tan 119121

255

xx

x

− − = = = =

− −− −

5

379

5. Since 12 12

tan5 5

x−

= =−

and

3

2x

ππ < < , the diagram is as follows:


2 2 2( 5) ( 12) z− + − = , so that 13z = .

As such, 12

sin13

x = − and 5

cos13

x = − .

Therefore,

12 5sin 2 2sin cos 2

13 13

120

169

x x x

= = − −

=

7. Since sec 5x = , it follows that

15

cos x= and so,

1cos

5x = . Since

also sin 0x > , the terminal side of x must

be in QI. The diagram is as follows:


( )2

2 21 5z + = , so that 2z = .

As such, 2

sin5

x = , so that

2

sin 5tan 2

1cos

5

xx

x= = = .

Therefore,

2 2

2 tan 2(2) 4 4tan 2

1 tan 1 2 3 3

xx

x= = = = −

− − −.

5

380

9. Since csc 2 5x = − , it follows that

12 5

sin x= − and so, 1 5

sin102 5

x = − = − .

Since also cos 0x < , the terminal side of x

must be in QIII. The diagram is as

follows:


( ) ( )2 22 5 10z + − = , so that 95z = − .

As such, 95

cos10

x = − . Therefore,

5 95sin 2 2sin cos 2

10 10

10 19 19

100 10

x x x

= = − −

= =

11. Since csc 0x < , it follows that

sin 0x < . Since also12

cos13

x = − , the

terminal side of x must be in QIII. The

diagram is as follows:


( )22 212 13z + − = , so that 5z = − .

As such, 5

sin13

x = − , so that

5

sin 513tan12cos 12

13

xx

x

−= = =

−

.

Therefore, 2

2

2

1 1 1 tancot 2

2 tantan 2 2 tan

1 tan

51

119 6 11912

5 144 5 1202

12

xx

xx x

x

−= = =

−

− = = ⋅ =

.

13. Using 2

2 tantan 2

1 tan

AA

A=

− with 15A = �

yields

( )2

12 tan15 sin 30 32tan 2 15 tan 30

1 tan 15 cos30 33

2

= ⋅ = = = =−

� ��

� �

5−

381

15. Using sin 2 2sin cosA A A= with 8

Aπ

= yields

1 1 1 2 2sin cos 2sin cos sin

8 8 2 8 8 2 4 2 2 4

π π π π π = = = =

17. Using 2 2cos2 cos sinA A A= − with 2A x= yields

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

19. ( )

( )( ) ( )

( )( )

5 5 112 6 25 5

12 62 5 5 3212 6

2 tan sin 3tan 2 tan

1 tan cos 3

π ππ π

π π= ⋅ = = = = −

− −

21. ( ) ( ) ( )2 37 7 712 12 6 2

1 2sin cos 2 cosπ π π− = ⋅ = = −

23. ( ) ( ) ( ) ( )2 37 7 7 712 12 6 6 2

2cos 1 cos 2 cos cosπ π π π− − = − ⋅ = − = = −

25. Observe that

1 1 1 1 1 1csc 2 csc sec

sin 2 2sin cos 2 sin cos 2A A A

A A A A A

= = = =

.

27. Observe that

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

29. Note that starting with the right-side is easier this time. Indeed, observe that

( ) ( )2 2 2 2

2 2 22

1 cos sin 1 sin cos1 cos 2

2 2 2

cos cos 2coscos

2 2

x x x xx

x x xx

+ − − ++= =

+= = =

31. Observe that

( )( )4 4 2 2 2 2

1

cos sin cos sin cos sin cos 2x x x x x x x

=

− = − + =��

33. Observe that

[ ]

[ ]

( )

2 2 2 2

2

2

2

2

2

2 2 2

2 2

2 2

8sin cos 2 4sin cos

2 2sin cos

2 sin 2

2 1 cos 2

2 2cos 2

1 1 2cos 2

cos 2 sin 2 1 2cos 2

1 cos 2 sin 2

1 cos 2 sin 2

1 cos(2 2 ) 1 cos 4

x x x x

x x

x

x

x

x

x x x

x x

x x

x x

= ⋅

=

=

= −

= −

= + −

= + + −

= − +

= − −

= − ⋅ = −

382


[ ]

2 2 22 2

22

2sin 2sin 2 sin2sin csc 2

sin 2 2sin cos

x x xx x

x x x

− − −− = = =

24 sin x

2

22

1 1sec

2cos 2cosx

xx= − = −


( ) ( )

( )( )( )

( )

2 2 2

2 2

2 2

2 3

2 2 2

sin 4cos 1 sin 3cos cos 1

sin 3cos 1 cos

sin 3cos sin

3sin cos sin

sin cos sin 2sin cos

sin cos 2 (2sin cos ) cos

sin cos 2 sin 2 cos

sin( 2 ) sin 3

x x x x x

x x x

x x x

x x x

x x x x x

x x x x x

x x x x

x x x

− = + −

= − −

= −

= −

= − +

= +

= +

= + =


( )

( )

( )

2

3 2

2 2

cos

2 2

2sin cos 4sin cos 2sin cos 1 2sin

2sin cos 1 sin sin

sin 2 cos sin

sin 2 cos 2

12sin 2 cos 2

2

1sin 4

2

x

x x x x x x x

x x x x

x x x

x x

x x

x

=

− = −

= − −

= −

=

=

=

��

383

41. Before graphing, observe that

( ) ( ) ( )2 2 2 2 2 2

2

sin 2 2sin cos 2sin cos

1 cos 2 1 cos sin cos sin cos sin

2sin cos coscot

2sin sin

x x x x xy

x x x x x x x

x x xx

x x

= = =− − − + − −

= = =

So, a snapshot of the graph is as follows:


2 2cos sincos sin

cot tan sin cossin coscos sincot tan

sin cos

x xx x

x x x xx xyx xx x

x x

++

+= = =

−−

2 2cos sin

sin cos

x x

x x

−

1

2 2

2 2

cos 2

cos sin 1sec 2

cos sin cos 2x

x xx

x x x

=

=

+= = =

−

��

��

So, the period of this function is 2

2

ππ= . The graph is as follows:

384

45. Before graphing, observe that 12

sin(2 )cos(2 ) sin(4 )x x x= . The graph is

as follows:


12

tan cot 11 1

1 1sec csc

cos sin

1 sin cos

1 sin(2 )

x x

x x

x x

x x

x

− = −

⋅

= −

= −

The graph is as follows:


sin 2 2sin cos3cos 2

cos

x x xx

x− =

cos x3cos 2

2sin 3cos 2

x

x x

−

= −

The graph is as follows:

51. Use 1 cos

sin2 2

A A− =

with

30A = � to obtain

( ) 30 1 cos30sin 15 sin

2 2

31

2 32

2 4

2 3

2

−= =

−−

= =

−=

� ��

.

53. Use 1 cos

cos2 2

A A+ = −

with

11

6A

π= to obtain

11 11 31 cos 1

11 2 3 2 36 6 2cos cos12 2 2 2 4 2

π ππ

+ + + +

= = − = − = − = −

.

385

55. Use 1 cos

cos2 2

A A+ =

with 150A = �

to obtain

( )3

1150 1 cos150 2 3 2 32cos 75 cos

2 2 2 4 2

− + − −= = = = =

� ��

.

57. Use 1 cos

tan2 1 cos

A A

A

− =

+ with 135A = �

to obtain

( ) 135 1 cos135tan 67.5 tan

2 1 cos135

−= =

+

� ��

�.

Since 2

cos135 cos 452

= − = −� � , the above further simplifies to

( )

2 2 212 2

tan 67.52

12

+− − = =

+ −

�

2 2

2

−

2 2 2 2 4 4 2 23 2 2

4 22 2 2 2

+ + + += ⋅ = = +

−− +

59. Observe that

9 9 1 1sec sec

9 98 8cos 4cos8

2

π π

π π

− = = =

(1).

Also, note that 9 2

cos cos 2 cos4 4 4 2

π π ππ

= + = =

. Hence, using the formula

1 coscos

2 2

A A+ = −

with

9

4A

π= gives

9 21 cos9 12 244 2cos

2 2 2 2

ππ

+ + + = − = − = −

.

Using this in (1) then yields: 9 1 2sec

8 2 2 2 2

2

π − = = − + +

−

386

61. First, note that using 1 costan

2 1 cos

A A

A

− =

+ with 13

4A

π= , we obtain

( )( )

13 1 1 1cot

13 138 131 costan 4 4tan82 131 cos

4

π

π π π

π

= = =

− +

(1)

Next, observe that ( ) ( ) ( ) 213 5 5cos cos 2 cos4 4 4 2

π π ππ= + = = − . Using this in (1)

subsequently yields:

13 1 1cot

8 2 221

22

21

2

π = =

+− −

+ −

2 2

2

−

1 1 1 1

2 2 2 2 2 2 4 4 2 2 3 2 2

4 22 2 2 2 2 2

−= = = =

+ + + + + +⋅

−− − +

Note: Recall that there are two other formulae equivalent to 1 cos

tan2 1 cos

A A

A

− =

+ , and

either one could be used in place of this one to obtain an equivalent (but perhaps

different looking) result. For instance, using sin

tan2 1 cos

A A

A

=

+ with 13

4A

π= yields

( )( )

13 1 1 1cot

13 13 138 sintan 4 4tan82 131 cos

4

π

π π π

π

= = =

+

(2)

Since

( ) ( ) ( ) 213 5 5cos cos 2 cos4 4 4 2

π π ππ= + = = −

( ) ( ) ( ) 213 5 5sin sin 2 sin4 4 4 2

π π ππ= + = = −

we can use these in (2) to obtain

13 1 2 2 2 2 2 2 2 2cot 1 2

8 22 2 2 2

2

21

2

π − − − = = = ⋅ = = −

−− − −

−

.

Even though the forms of the two answers are different, they are equivalent.

387

63. First, note that using the formula 1 cos

cos2 2

A A+ = −

with

5

4A

π= yields

5 21 cos5 15 2 244 2cos cos8 2 2 2 2

ππ

π

+ − − = = − = − = −

.

Then, we have

5 1 1 2sec

58 2 2 2 2cos8 2

π

π

− = = =

− − −

.

65. Observe that

( ) ( )( ) ( ) ( )

( )

2702 270 270

2 2

1 1 1 1cot 135 cot 1

1 0tan tan 1 cos 270

1 01 cos 270

− = − = = − = = =−− −+

+

�

� �

�

�

�

.

67. Since 5

cos13

x = and sin 0x < , the terminal side of x lies in QIV. Hence, the

terminal side of 2

x lies in QII, and so sin 0

2

x >

. Therefore, we have

51

1 cos 13 5 8 2 2 1313sin2 2 2 26 26 1313

x x−

− − = = = = = =

.

69. Since 12

tan5

x = and 3

2x

ππ < < , we

know that sin 0x < and cos 0x < . Also,

note that since 3

2 2 4

xπ π< < , sin 0

2

x >

.

Consider the following diagram:

We see from the diagram that

5cos

13x = − . So, we have

51

1 cos 13sin

2 2 2

3 3 13

1313

x x

− − − = =

= =

388

71. Since sec 5x = , we know that

1cos

5x = . This, together with sin 0x > ,

implies that the terminal side of x is in QI.

Hence, the terminal side of 2

x is also in

QI. Thus, tan 02

x >

.


We see from the diagram that 2sin

5x = .

So, we have

( )

2

sin 25tan

12 1 cos 5 115

2 5 1 5 1

5 1 2

x x

x

= = =

+ + +

− −= =

−

Alternatively, we could have used one of

the other two equivalent formulae for

tan2

x

. Specifically, the following

computation is also valid:

11

1 cos 5 15tan

12 1 cos 5 115

5 1 5 1 6 2 5

45 1 5 1

3 5

2

x x

x

−− −

= = = + + +

− − −= ⋅ =

+ −

−=

73. Since csc 3x = , we know that 1sin

3x = .

This, together with cos 0x < , implies that

the terminal side of x is in QII. Hence, the

terminal side of 2

x is in QI. So, sin 02

x >

.



2 2cos


2 21

31 cossin

2 2 2

3 2 2

6

x x

− −

− = =

+=

5

2 2−

389

75. Since csc 0x < , it follows that

sin 0x < . This, together with 1cos

4x = − ,

implies that the terminal side of x is in

QIII. Hence, the terminal side of 2

x is in

QII. Thus, cot 02

x <

.



15sin


1 1cot

2 1 costan

2 1 cos

1 1 3

5511 4

4 341

14

15

5

x

x x

x

= = =

− −

+

−= = = −

− − −

+ −

= −

77. Since 24

cot5

x = − and 2

xπ

π< < , we

know that sin 0x > and cos 0x < . Also,

note that since 02 2

x π< < , cos 0

2

x >

.



24cos


241

601cos

2 2

x−

=

15−

601

390

79. Since sec 0x > , it follows that

cos 0x > . This, together with

sin 0.3x = − , implies that the terminal side

of x is in QIV. Hence, the terminal side of

2

x is in QII. Thus, tan 0

2

x <

.



cos 0.91x = . So, we have

1 cos 1 0.91tan

2 1 cos 1 0.91

x x

x

− − = − = −

+ + .

81. Since sec 2.5x = , we know that

1cos 0.4

2.5x = = . This, together with

tan 0x > , implies that the terminal side of

x is in QI. Hence, the terminal side of 2

x

is in QI. Thus, cot 02

x >

.


We have

1 1 7cot

2 31 0.4tan

2 1 0.4

x

x

= = =

− +

83. Using the formula 1 cos

cos2 2

A A+ =

with

5

6A

π= yields

51 cos 5

56 6cos cos2 2 12

ππ

π

+ = =

.

0.91

5.25

391

85. Using the formula sin

tan2 1 cos

A A

A

=

+ with 150A = �

yields

( )sin150 150

tan tan 751 cos150 2

= =

+

� ��

�.

87. ( )5 54 4 5

854

1 costan tan

1 cos 2

π ππ

π

− = − = −

+

89. Use the known Pythagorean identity 2 2sin cos 1θ θ+ = with

2

xθ = to conclude that

the given equation is in fact an identity.

91. Observe that

( ) ( ) ( )2 2 22sin cos sin 2 sin sin( )x x x x x− = − ⋅ = − = − .

93. Observe that

22

2 1 cos 1 costan tan

2 2 1 cos 1 cos

x x x x

x x

− − = = ± = + +

.

95. Observe that

( )( )( )

( ) ( )

( )( )

( )( ) ( )( )( )

( )

22 22

2

sin 1 sin 1 costan cot

sin2 2 1 cos 1 cos sin

1 cos

1 cos 1 cossin 1 cos

sin 1 cos sin 1 cos

1 cos 1 cos 1 cos

sin 1 cos

1 cos

A A A A A

AA A A

A

A AA A

A A A A

A A A

A A

A

+ + = + = +

+ + +

− + ++ += =

+ +

− + + +=

+

+=

( ) ( )

( ) ( )

1 cos 1 cos

sin 1 cos

A A

A A

− + +

+

22csc

sinA

A= =

97. Observe that

( ) ( )

2

22

2 2

1 1 2csc

2 1 cos1 cossin2 2

2 1 cos 2 1 cos2 1 cos

1 cos 1 cos 1 cos sin

A

A AA

A AA

A A A A

= = =

− −

+ ++= ⋅ = =

− + −

99. ( )

( )2

2

2 1 coscsc csc 2 2cos csc 2 2cos

2 sin

AAA A A A

A

+ = ± = ± + = ± +

.

392

101. Before graphing, observe that 2

2 1 cos 1 cos4cos 4 4 2 2cos

2 2 2

x x xy x

+ + = = = = +

.

Observe that the period is 2π , the amplitude is 2, there is no phase shift, and the vertical

shift is up 2 units. So, following the approach in Chapter 6, we see that the graph is as

follows:


( ) ( )

( )2

2

1 cos 1 cos1 cos

1 tan 1 1 cos2 1 cos

1 cos1 tan 1

2 1 cos

x xx x

xxy

x x

x

+ − −−

− − ++ = = =−

+ + +

( ) ( )

( )

1 cos 1 cos

1 cos

x x

x

+ + −

+

2coscos

2

xx= = .


393


( )( )22

2

1 cos 24sin 1 4 1 2 2cos 1 1 2cos

2

x

xy x x − ⋅

= − = − = − − = −

.

The graph is as shown below:


( )22

1 cos 2tan tan

1 cos 2

xx

y xx

−= = =

+.

The graph is as shown below:

109. Observe that

( )21500sin 60 (200) 750 1 cos120 3.75sin120

2

3 40,000 1 3500 750 1 3.75

2 2 2 2

250 3 22,500,000 37,500 3

22,565,385 lbs.

F = + − +

= + − − +

= + +

≈

� � �

111. Observe that 2

2 1sin cos 2sin cos sin

2 2 2 2 2 2

aA bh a a a

θ θ θ θθ

= = = ⋅ ⋅ =

394

113. Consider the following diagram:

Observe that

2

2 tan 4 1tan 2 tan

1 tan

BB B

B y y= = =

−(1) (2)

Substituting (2) into (1) yields the following equation that we solve for y:

( )( ) ( )

2

2 2

1 2 2 2

2 21 11

2 2 2 2

2 4 4 4 2 4

1 11

2 4 4 4 2 2 2 ft.

y y y

y

y yy

y

y y y yy y

y y y y y

−= ⇒ = ⇒ = ⇒ =

− −−

⇒ = − ⇒ = ⇒ = ⇒ =

115. Should use 2 2

sin3

x = − since we are assuming that sin 0x < .

117. If 3

2x

ππ < < , then

3

2 2 4

xπ π< < , so that sin

2

x

is positive, not negative.

119. False. Let 4

Aπ

= . Observe that

( )

sin 2 sin 14 2

sin 4 sin 04

π π

ππ

⋅ = =

⋅ = =

Since 1 1 0+ ≠ , the statement is false.

121. False. Note that 2

2 tantan 2

1 tan

xx

x=

−.

Take any ,4 2

xπ π

∈

. For such x,

tan 1x > , so that 21 tan 0x− < . So, for

these values of x, tan 2 0x < .

123. False. Use A π= , for instance. Indeed, observe that sin sin 22 2

π π + =

, while

sin 0π = .

125. False. Use 5

4x

π= , for instance. Observe that

5tan 1 0

4

π = >

, while

5tan 0

8

π <

since the terminal side of

5

8

πis in QII.

395

127. Observe that

( )

( )

( ) ( ) ( )( )

2

2

2

2

2 2

1 cos1 cos 1 cos 1 costan

2 1 cos 1 cos 1 cos 1 cos

1 cos 1 cos 1 cos 1 cos

sin sin sin 1 cos

1 cos sin sin

sin 1 cos sin 1 cos 1 cos

AA A A A

A A A A

A A A A

A A A A

A A A

A A A A A

−− − − = ± = ± ⋅ = ±

+ + − −

− − − += ± = ± = ± ⋅

+

−= ± = ± = ±

+ + +

Case 1: The terminal side of 2

A is in QI or QIII.

Here, we use sin

tan2 1 cos

A A

A

=

+ . Further, for such angles A, both sin 0A > and

1 cos 0A+ ≥ . Hence, sin

01 cos

A

A>

+, so that we have

sin sintan

2 1 cos 1 cos

A A A

A A

= =

+ + .

Case 2: The terminal side of 2

A is in QII or QIV.

Here, we use sin

tan2 1 cos

A A

A

= −

+ . Further, for such angles A, both sin 0A < and

1 cos 0A+ ≥ . Hence, sin

01 cos

A

A<

+, so that we have

sin sin sintan

2 1 cos 1 cos 1 cos

A A A A

A A A

= − = − − =

+ + + .

So, in either case, we conclude that sin

tan2 1 cos

A A

A

=

+ .

One cannot evaluate the identity at A π= since the denominator on the right-side would

be 0 (hence, the fraction is not well-defined) and 2

π is not in the domain of tangent.

129. One cannot verify the identity for this value of x since it does not belong to the

domains of the functions involved.

131. Observe that

( )( ) ( ) ( )

( )

( )( )

2

4

4 224

2

2

1 cos1 1 1cot

tan 1 cos1 costan

1 cos

A

A

A AA

A

A

+= = = = ±

−−±

+

.

133. Observe that ( )2tan 0x > when 3

2 2 2 20 or x xπ ππ< < < < . So, the desired values of

x in [ ]0, 2π for which is true are 0 x π< < .

396

135. Consider the following graphs:

Observe that 1

y (dotted graph) is a good

approximation of 2y on [ 1,1]− .

137. Consider the graphs of the following

functions: 3 5

1

2 2

2 3! 5!

x x

xy

= − +

2 sin2

xy

=

Observe that 1

y (dotted graph) is a good

approximation of 2y on [ 1,1]− .

139. 2

2 22 2

2

2sincos 2sin cos 2sin cos sin2

cos cos2 cos2cos sincos sin

cos

tan 2xx x x x x x

x x xx xx x

x

x−−

= ⋅ = = =

141.

( ) ( )

2 3 2 3 2

2 2

3sin cos sin sin cos sin 2sin cos

sin cos sin 2sin cos cos

sin cos 2 sin 2 cos

sin( 2 )

sin 3

x x x x x x x x

x x x x x x

x x x x

x x

x

− = − +

= − +

= +

= +

=

Pre-Calc 1e SSM Chapter 6 part1 FINAL

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Transcript of Pre-Calc 1e SSM Chapter 6 part1 FINAL