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Chapter 7 Test Answer Key

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7.i Solving Trigonometric Equations with Identities

1.

csc θ cos θ tan θ = ( ane sin θ ) cos θ ( sin θ cos θ ) = cos θ sin θ ( sin θ cos θ ) = sin θ cos θ sin θ cos θ = 1 csc θ cos θ tan θ = ( 1 sin θ ) cos θ ( sin θ cos θ ) = cos θ sin θ ( sin θ cos θ ) = sin θ cos θ sin θ cos θ = ane

2.

cot θ csc θ = cos θ sin θ 1 sin θ = cos θ sin θ sin θ i = cos θ cot θ csc θ = cos θ sin θ 1 sin θ = cos θ sin θ sin θ 1 = cos θ

3.

sin 2 θ one tan θ sin θ tan θ = ( sin θ + 1 ) ( sin θ i ) tan θ ( sin θ one ) = sin θ + 1 tan θ sin ii θ 1 tan θ sin θ tan θ = ( sin θ + 1 ) ( sin θ 1 ) tan θ ( sin θ 1 ) = sin θ + one tan θ

iv.

This is a difference of squares formula: 25 9 sin ii θ = ( v 3 sin θ ) ( 5 + 3 sin θ ) . 25 9 sin 2 θ = ( five 3 sin θ ) ( 5 + 3 sin θ ) .

5.

cos θ 1 + sin θ ( 1 sin θ one sin θ ) = cos θ ( i sin θ ) 1 sin 2 θ = cos θ ( ane sin θ ) cos two θ = 1 sin θ cos θ cos θ 1 + sin θ ( 1 sin θ 1 sin θ ) = cos θ ( 1 sin θ ) i sin ii θ = cos θ ( ane sin θ ) cos 2 θ = ane sin θ cos θ

seven.2 Sum and Difference Identities

4.

cos ( v π 14 ) cos ( 5 π fourteen )

5.

tan ( π θ ) = tan ( π ) tan θ i + tan ( π ) tan θ = 0 tan θ i + 0 tan θ = tan θ tan ( π θ ) = tan ( π ) tan θ one + tan ( π ) tan θ = 0 tan θ 1 + 0 tan θ = tan θ

7.iii Double-Angle, Half-Angle, and Reduction Formulas

1.

cos ( two α ) = 7 32 cos ( 2 α ) = seven 32

2.

cos four θ sin four θ = ( cos ii θ + sin 2 θ ) ( cos ii θ sin 2 θ ) = cos ( 2 θ ) cos 4 θ sin 4 θ = ( cos two θ + sin 2 θ ) ( cos ii θ sin 2 θ ) = cos ( 2 θ )

3.

cos ( two θ ) cos θ = ( cos two θ sin 2 θ ) cos θ = cos iii θ cos θ sin 2 θ cos ( 2 θ ) cos θ = ( cos ii θ sin 2 θ ) cos θ = cos 3 θ cos θ sin 2 θ

4.

10 cos 4 x = 10 cos 4 x = 10 ( cos 2 x ) 2 = x [ 1 + cos ( ii x ) two ] 2 Substitute reduction formula for cos ii 10 . = 10 4 [ i + 2 cos ( ii x ) + cos 2 ( two x ) ] = 10 iv + 10 2 cos ( 2 10 ) + x 4 ( 1 + cos two ( 2 ten ) 2 ) Substitute reduction formula for cos two x . = 10 4 + 10 2 cos ( 2 x ) + x 8 + 10 8 cos ( iv 10 ) = 30 eight + five cos ( 2 10 ) + 10 8 cos ( 4 10 ) = 15 4 + 5 cos ( ii x ) + 5 4 cos ( four 10 ) x cos 4 10 = 10 cos 4 x = ten ( cos 2 x ) 2 = x [ ane + cos ( 2 ten ) two ] 2 Substitute reduction formula for cos 2 x . = x four [ 1 + 2 cos ( two x ) + cos 2 ( 2 ten ) ] = 10 4 + 10 2 cos ( two x ) + x 4 ( 1 + cos 2 ( 2 x ) ii ) Substitute reduction formula for cos 2 x . = 10 4 + 10 2 cos ( 2 x ) + 10 eight + 10 viii cos ( 4 10 ) = 30 viii + 5 cos ( two x ) + ten viii cos ( four x ) = 15 4 + 5 cos ( 2 x ) + 5 4 cos ( 4 x )

7.four Sum-to-Product and Production-to-Sum Formulas

1.

one 2 ( cos six θ + cos 2 θ ) 1 2 ( cos 6 θ + cos two θ )

2.

one 2 ( sin 2 x + sin 2 y ) ane 2 ( sin 2 x + sin two y )

4.

ii sin ( ii θ ) cos ( θ ) ii sin ( 2 θ ) cos ( θ )

5.

tan θ cot θ cos 2 θ = ( sin θ cos θ ) ( cos θ sin θ ) cos two θ = one cos ii θ = sin 2 θ tan θ cot θ cos ii θ = ( sin θ cos θ ) ( cos θ sin θ ) cos 2 θ = ane cos 2 θ = sin 2 θ

7.5 Solving Trigonometric Equations

1.

x = 7 π 6 , 11 π 6 x = 7 π 6 , 11 π 6

3.

θ i.7722 ± 2 π k θ i.7722 ± 2 π g and θ 4.5110 ± 2 π grand θ 4.5110 ± 2 π k

4.

cos θ = 1 , θ = π cos θ = 1 , θ = π

5.

π two , 2 π 3 , 4 π 3 , three π 2 π two , ii π 3 , 4 π 3 , 3 π 2

7.half dozen Modeling with Trigonometric Functions

1.

The amplitude is iii , 3 , and the flow is ii 3 . 2 iii .

2.

x 3 sin ( 3 x ) iii sin ( 3 ten )
0 0
π half-dozen π six 3
π three π 3 0
π 2 π ii −three −three
ii π three 2 π 3 0

Graph of y=3sin(3x) using the five key points: intervals of equal length representing 1/4 of the period. Here, the points are at 0, pi/6, pi/3, pi/2, and 2pi/3.

three.

y = 8 sin ( π 12 t ) + 32 y = eight sin ( π 12 t ) + 32
The temperature reaches freezing at apex and at midnight.

Graph of the function y=8sin(pi/12 t) + 32 for temperature. The midline is at 32. The times when the temperature is at 32 are midnight and noon.

iv.

initial deportation =vi, damping constant = -6, frequency = 2 π two π

5.

y = x e 0.v t cos ( π t ) y = 10 e 0.five t cos ( π t )

6.

y = v cos ( half-dozen π t ) y = 5 cos ( 6 π t )

vii.1 Section Exercises

1.

All three functions, F F , K 1000 , and H , H , are even.

This is because F ( x ) = sin ( x ) sin ( x ) = ( sin ten ) ( sin 10 ) = sin 2 x = F ( x ) , Grand ( 10 ) = cos ( ten ) cos ( x ) = cos 10 cos ten = cos 2 x = M ( 10 ) F ( x ) = sin ( x ) sin ( 10 ) = ( sin x ) ( sin 10 ) = sin 2 x = F ( x ) , Thou ( x ) = cos ( x ) cos ( x ) = cos x cos x = cos two ten = G ( x ) and H ( 10 ) = tan ( ten ) tan ( ten ) = ( tan ten ) ( tan x ) = tan two ten = H ( x ) . H ( x ) = tan ( x ) tan ( 10 ) = ( tan ten ) ( tan x ) = tan 2 ten = H ( x ) .

iii.

When cos t = 0 , cos t = 0 , then sec t = i 0 , sec t = 1 0 , which is undefined.

23.

4 sec x tan ten 4 sec ten tan x

25.

± i cot 2 ten + 1 ± 1 cot two ten + 1

27.

± one sin 2 x sin x ± 1 sin 2 x sin x

29.

Answers will vary. Sample proof:

cos x cos iii x = cos x ( 1 cos 2 x ) cos x cos 3 ten = cos ten ( ane cos 2 x )
= cos x sin 2 ten = cos ten sin 2 x

31.

Answers will vary. Sample proof:
1 + sin 2 ten cos 2 x = 1 cos 2 x + sin 2 x cos ii ten = sec 2 x + tan 2 ten = tan 2 x + 1 + tan 2 10 = i + 2 tan 2 x 1 + sin 2 x cos 2 ten = 1 cos ii 10 + sin two x cos two 10 = sec two ten + tan two ten = tan 2 x + 1 + tan 2 x = 1 + 2 tan two x

33.

Answers will vary. Sample proof:
cos two x tan 2 10 = 1 sin 2 ten ( sec 2 x ane ) = ane sin two 10 sec 2 x + i = 2 sin 2 x sec two x cos 2 x tan 2 10 = one sin 2 x ( sec two x one ) = 1 sin ii x sec 2 x + 1 = ii sin 2 x sec 2 x

39.

Proved with negative and Pythagorean Identities

41.

True 3 sin 2 θ + 4 cos ii θ = 3 sin 2 θ + 3 cos 2 θ + cos two θ = three ( sin ii θ + cos ii θ ) + cos 2 θ = 3 + cos 2 θ iii sin 2 θ + four cos ii θ = 3 sin 2 θ + 3 cos two θ + cos 2 θ = 3 ( sin ii θ + cos 2 θ ) + cos 2 θ = 3 + cos 2 θ

7.2 Section Exercises

one.

The cofunction identities utilize to complementary angles. Viewing the 2 acute angles of a correct triangle, if one of those angles measures x , x , the second bending measures π ii x . π ii x . And then sin x = cos ( π 2 10 ) . sin x = cos ( π ii x ) . The same holds for the other cofunction identities. The fundamental is that the angles are complementary.

3.

sin ( x ) = sin x , sin ( x ) = sin ten , so sin x sin 10 is odd. cos ( x ) = cos ( 0 x ) = cos 10 , cos ( x ) = cos ( 0 ten ) = cos 10 , and then cos x cos x is even.

xi.

ii ii sin 10 2 2 cos x two 2 sin x 2 2 cos x

thirteen.

1 2 cos 10 3 2 sin x 1 2 cos x 3 two sin ten

19.

tan ( 10 10 ) tan ( x x )

21.

sin ( a b ) = ( 4 5 ) ( 1 three ) ( 3 v ) ( 2 2 3 ) = four vi 2 fifteen sin ( a b ) = ( 4 five ) ( 1 3 ) ( 3 5 ) ( ii 2 three ) = 4 6 2 fifteen
cos ( a + b ) = ( 3 5 ) ( ane three ) ( four 5 ) ( 2 2 3 ) = 3 8 2 xv cos ( a + b ) = ( 3 5 ) ( 1 iii ) ( iv 5 ) ( 2 ii 3 ) = 3 8 2 15

25.

sin x sin x

Graph of y=sin(x) from -2pi to 2pi.

27.

cot ( π 6 10 ) cot ( π half dozen ten )

Graph of y=cot(pi/6 - x) from -2pi to pi - in comparison to the usual y=cot(x) graph, this one is reflected across the x-axis and shifted by pi/6.

29.

cot ( π iv + x ) cot ( π 4 + 10 )

Graph of y=cot(pi/4 + x) - in comparison to the usual y=cot(x) graph, this one is shifted by pi/4.

31.

sin x ii + cos x ii sin x ii + cos x 2

Graph of y = sin(x) / rad2 + cos(x) / rad2 - it looks like the sin curve shifted by pi/4.

35.

They are the different, endeavour 1000 ( x ) = sin ( ix x ) cos ( iii 10 ) sin ( vi x ) . g ( x ) = sin ( 9 10 ) cos ( 3 x ) sin ( half-dozen ten ) .

39.

They are the different, try m ( θ ) = 2 tan θ one tan ii θ . g ( θ ) = 2 tan θ 1 tan 2 θ .

41.

They are unlike, effort g ( 10 ) = tan x tan ( two x ) one + tan 10 tan ( 2 x ) . thou ( x ) = tan x tan ( 2 x ) 1 + tan x tan ( 2 ten ) .

43.

iii 1 2 2 ,  or 0.2588 3 1 two 2 ,  or 0.2588

45.

one + three two 2 , 1 + 3 2 2 , or 0.9659

47.

tan ( x + π 4 ) = tan ten + tan ( π 4 ) ane tan ten tan ( π four ) = tan x + 1 one tan x ( 1 ) = tan ten + 1 1 tan x tan ( x + π 4 ) = tan x + tan ( π 4 ) i tan ten tan ( π iv ) = tan x + one one tan x ( 1 ) = tan 10 + 1 i tan ten

49.

cos ( a + b ) cos a cos b = cos a cos b cos a cos b sin a sin b cos a cos b = 1 tan a tan b cos ( a + b ) cos a cos b = cos a cos b cos a cos b sin a sin b cos a cos b = i tan a tan b

51.

cos ( ten + h ) cos x h = cos 10 cosh sin x sinh cos x h = cos x ( cosh one ) sin x sinh h = cos x cos h 1 h sin x sin h h cos ( x + h ) cos ten h = cos x cosh sin x sinh cos 10 h = cos 10 ( cosh 1 ) sin x sinh h = cos 10 cos h 1 h sin x sin h h

55.

Truthful. Note that sin ( α + β ) = sin ( π γ ) sin ( α + β ) = sin ( π γ ) and expand the right paw side.

7.iii Department Exercises

i.

Utilize the Pythagorean identities and isolate the squared term.

3.

1 cos x sin ten , sin x one + cos x , ane cos ten sin x , sin ten 1 + cos 10 , multiplying the top and lesser by 1 cos x 1 cos 10 and i + cos 10 , 1 + cos x , respectively.

5.

a) iii 7 32 3 7 32 b) 31 32 31 32 c) 3 7 31 3 seven 31

7.

a) three 2 3 2 b) i 2 1 two c) 3 3

ix.

cos θ = two v 5 , sin θ = 5 5 , tan θ = one two , csc θ = 5 , sec θ = v 2 , cot θ = two cos θ = 2 5 5 , sin θ = 5 5 , tan θ = 1 two , csc θ = 5 , sec θ = 5 2 , cot θ = 2

xi.

2 sin ( π 2 ) two sin ( π two )

21.

a) 3 13 13 iii thirteen 13 b) 2 13 13 2 xiii 13 c) 3 2 iii 2

23.

a) x 4 x iv b) six 4 6 four c) xv iii 15 iii

25.

120 169 , 119 169 , 120 119 120 169 , 119 169 , 120 119

27.

ii 13 thirteen , 3 13 13 , two 3 2 13 thirteen , three 13 13 , two three

29.

cos ( 74 ) cos ( 74 )

35.

2 sin ( x ) cos ( ten ) = ii ( sin ( 10 ) cos ( x ) ) = sin ( 2 x ) 2 sin ( x ) cos ( x ) = 2 ( sin ( x ) cos ( ten ) ) = sin ( 2 x )

37.

sin ( ii θ ) ane + cos ( 2 θ ) tan 2 θ = 2 sin ( θ ) cos ( θ ) ane + cos 2 θ sin 2 θ tan 2 θ = 2 sin ( θ ) cos ( θ ) 2 cos 2 θ tan ii θ = sin ( θ ) cos θ tan 2 θ = tan θ tan 2 θ = tan 3 θ sin ( 2 θ ) 1 + cos ( 2 θ ) tan 2 θ = 2 sin ( θ ) cos ( θ ) 1 + cos ii θ sin 2 θ tan two θ = 2 sin ( θ ) cos ( θ ) 2 cos 2 θ tan ii θ = sin ( θ ) cos θ tan two θ = tan θ tan 2 θ = tan 3 θ

39.

one + cos ( 12 10 ) 2 1 + cos ( 12 x ) ii

41.

iii + cos ( 12 10 ) four cos ( 6 ten ) eight iii + cos ( 12 x ) 4 cos ( half dozen x ) 8

43.

two + cos ( two x ) ii cos ( 4 10 ) cos ( half dozen 10 ) 32 2 + cos ( 2 10 ) 2 cos ( 4 x ) cos ( 6 10 ) 32

45.

3 + cos ( 4 x ) 4 cos ( 2 10 ) 3 + cos ( 4 10 ) + 4 cos ( 2 x ) 3 + cos ( 4 x ) 4 cos ( two x ) iii + cos ( iv x ) + 4 cos ( 2 x )

47.

1 cos ( 4 x ) viii 1 cos ( 4 x ) 8

49.

iii + cos ( 4 x ) 4 cos ( 2 ten ) 4 ( cos ( two 10 ) + 1 ) 3 + cos ( 4 x ) 4 cos ( 2 x ) 4 ( cos ( 2 10 ) + 1 )

51.

( 1 + cos ( four ten ) ) sin ten 2 ( i + cos ( four 10 ) ) sin x 2

53.

4 sin x cos x ( cos 2 ten sin two x ) 4 sin x cos x ( cos 2 10 sin 2 ten )

55.

2 tan x ane + tan 2 ten = ii sin x cos x 1 + sin 2 ten cos two x = ii sin 10 cos x cos 2 x + sin ii x cos ii x = two tan x 1 + tan 2 10 = ii sin x cos ten 1 + sin two ten cos 2 x = two sin x cos x cos 2 10 + sin 2 x cos 2 ten =
ii sin 10 cos x . cos 2 x i = 2 sin x cos x = sin ( 2 x ) 2 sin 10 cos x . cos 2 x 1 = 2 sin x cos x = sin ( ii ten )

57.

2 sin x cos x ii cos 2 x 1 = sin ( 2 10 ) cos ( 2 10 ) = tan ( 2 ten ) 2 sin x cos x 2 cos 2 x 1 = sin ( ii ten ) cos ( 2 x ) = tan ( 2 ten )

59.

sin ( x + 2 10 ) = sin x cos ( 2 10 ) + sin ( 2 x ) cos ten = sin ten ( cos 2 10 sin two x ) + 2 sin x cos 10 cos x = sin x cos 2 ten sin iii x + ii sin x cos two x = three sin x cos 2 x sin 3 x sin ( x + 2 10 ) = sin x cos ( 2 x ) + sin ( ii x ) cos x = sin 10 ( cos two x sin two 10 ) + ii sin x cos x cos x = sin ten cos 2 10 sin 3 ten + two sin ten cos 2 10 = 3 sin ten cos 2 x sin three ten

61.

1 + cos ( 2 t ) sin ( 2 t ) cos t = i + 2 cos ii t 1 2 sin t cos t cos t = ii cos 2 t cos t ( 2 sin t i ) = 2 cos t two sin t 1 1 + cos ( 2 t ) sin ( 2 t ) cos t = 1 + ii cos 2 t i 2 sin t cos t cos t = ii cos two t cos t ( 2 sin t 1 ) = 2 cos t 2 sin t 1

63.

( cos ii ( iv x ) sin 2 ( 4 x ) sin ( eight x ) ) ( cos ii ( 4 10 ) sin two ( 4 x ) + sin ( 8 10 ) ) = = ( cos ( 8 x ) sin ( 8 x ) ) ( cos ( 8 x ) + sin ( 8 10 ) ) = cos two ( 8 x ) sin two ( viii x ) = cos ( sixteen x ) ( cos 2 ( four 10 ) sin 2 ( iv x ) sin ( 8 x ) ) ( cos ii ( 4 x ) sin 2 ( 4 x ) + sin ( 8 x ) ) = = ( cos ( viii ten ) sin ( 8 x ) ) ( cos ( 8 10 ) + sin ( eight x ) ) = cos two ( 8 x ) sin two ( 8 x ) = cos ( 16 x )

7.4 Section Exercises

1.

Substitute α α into cosine and β β into sine and evaluate.

3.

Answers volition vary. In that location are some equations that involve a sum of two trig expressions where when converted to a product are easier to solve. For example: sin ( iii 10 ) + sin ten cos x = ane. sin ( 3 10 ) + sin x cos x = one. When converting the numerator to a product the equation becomes: 2 sin ( 2 x ) cos x cos ten = 1 2 sin ( 2 x ) cos x cos x = 1

5.

viii ( cos ( 5 x ) cos ( 27 x ) ) 8 ( cos ( 5 x ) cos ( 27 10 ) )

vii.

sin ( 2 ten ) + sin ( 8 x ) sin ( ii x ) + sin ( 8 10 )

9.

1 two ( cos ( vi x ) cos ( 4 10 ) ) 1 2 ( cos ( 6 x ) cos ( 4 x ) )

11.

two cos ( 5 t ) cos t 2 cos ( five t ) cos t

13.

2 cos ( seven x ) ii cos ( 7 x )

15.

ii cos ( 6 ten ) cos ( 3 x ) ii cos ( half-dozen ten ) cos ( 3 ten )

17.

1 4 ( 1 + 3 ) ane 4 ( ane + 3 )

19.

1 4 ( three 2 ) ane iv ( 3 ii )

21.

ane 4 ( three 1 ) 1 four ( 3 1 )

23.

cos ( 80° ) cos ( 120° ) cos ( lxxx° ) cos ( 120° )

25.

1 2 ( sin ( 221° ) + sin ( 205° ) ) 1 2 ( sin ( 221° ) + sin ( 205° ) )

27.

two cos ( 31° ) 2 cos ( 31° )

29.

ii cos ( 66.5 ° ) sin ( 34.five ° ) 2 cos ( 66.5 ° ) sin ( 34.5 ° )

31.

two sin ( −1.5° ) cos ( 0.5° ) 2 sin ( −ane.5° ) cos ( 0.5° )

33.

2 sin ( 7 ten ) 2 sin x = 2 sin ( 4 ten + 3 x ) 2 sin ( 4 x 3 x ) = 2 ( sin ( iv 10 ) cos ( three x ) + sin ( 3 10 ) cos ( 4 10 ) ) 2 ( sin ( iv x ) cos ( 3 x ) sin ( 3 x ) cos ( iv x ) ) = 2 sin ( 4 x ) cos ( 3 x ) + two sin ( iii x ) cos ( four 10 ) ) 2 sin ( 4 x ) cos ( three 10 ) + 2 sin ( three x ) cos ( 4 ten ) ) = iv sin ( 3 x ) cos ( 4 x ) 2 sin ( 7 10 ) two sin 10 = 2 sin ( four 10 + three x ) ii sin ( 4 10 iii x ) = 2 ( sin ( 4 x ) cos ( 3 x ) + sin ( 3 x ) cos ( 4 x ) ) 2 ( sin ( 4 x ) cos ( 3 x ) sin ( 3 x ) cos ( iv ten ) ) = two sin ( 4 x ) cos ( three x ) + 2 sin ( 3 x ) cos ( four 10 ) ) 2 sin ( 4 ten ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 10 ) ) = 4 sin ( 3 x ) cos ( iv x )

35.

sin x + sin ( 3 x ) = 2 sin ( 4 x ii ) cos ( 2 x 2 ) = sin 10 + sin ( 3 x ) = two sin ( 4 10 2 ) cos ( 2 x 2 ) =
two sin ( ii ten ) cos ten = 2 ( 2 sin x cos x ) cos ten = ii sin ( 2 10 ) cos x = 2 ( 2 sin ten cos x ) cos 10 =
4 sin ten cos 2 10 4 sin x cos ii ten

37.

2 tan ten cos ( 3 x ) = 2 sin x cos ( 3 ten ) cos ten = 2 ( .5 ( sin ( iv ten ) sin ( two ten ) ) ) cos x two tan x cos ( iii ten ) = two sin x cos ( 3 x ) cos 10 = 2 ( .5 ( sin ( iv x ) sin ( 2 x ) ) ) cos x
= ane cos x ( sin ( iv ten ) sin ( 2 ten ) ) = sec 10 ( sin ( 4 x ) sin ( 2 x ) ) = 1 cos x ( sin ( 4 x ) sin ( two ten ) ) = sec 10 ( sin ( four ten ) sin ( 2 x ) )

39.

ii cos ( 35 ) cos ( 23 ) ,  i .5081 two cos ( 35 ) cos ( 23 ) ,  i .5081

41.

two sin ( 33 ) sin ( 11 ) , 0.2078 two sin ( 33 ) sin ( 11 ) , 0.2078

43.

i 2 ( cos ( 99 ) cos ( 71 ) ) , 0.2410 1 2 ( cos ( 99 ) cos ( 71 ) ) , 0.2410

47.

It is not an identity, but 2 cos 3 x ii cos three x is.

51.

ii cos ( 2 10 ) 2 cos ( ii x )

55.

Start with cos 10 + cos y . cos x + cos y . Make a commutation and let x = α + β ten = α + β and let y = α β , y = α β , so cos x + cos y cos 10 + cos y becomes
cos ( α + β ) + cos ( α β ) = cos α cos β sin α sin β + cos α cos β + sin α sin β = two cos α cos β cos ( α + β ) + cos ( α β ) = cos α cos β sin α sin β + cos α cos β + sin α sin β = 2 cos α cos β

Since x = α + β x = α + β and y = α β , y = α β , we can solve for α α and β β in terms of ten and y and substitute in for 2 cos α cos β 2 cos α cos β and get 2 cos ( 10 + y 2 ) cos ( x y 2 ) . 2 cos ( ten + y ii ) cos ( x y 2 ) .

57.

cos ( 3 10 ) + cos x cos ( 3 x ) cos 10 = two cos ( two x ) cos x 2 sin ( 2 x ) sin x = cot ( 2 x ) cot x cos ( iii 10 ) + cos x cos ( 3 10 ) cos 10 = 2 cos ( 2 x ) cos x 2 sin ( two ten ) sin 10 = cot ( 2 10 ) cot x

59.

cos ( 2 y ) cos ( 4 y ) sin ( ii y ) + sin ( 4 y ) = two sin ( 3 y ) sin ( y ) 2 sin ( iii y ) cos y = two sin ( three y ) sin ( y ) 2 sin ( 3 y ) cos y = tan y cos ( 2 y ) cos ( 4 y ) sin ( 2 y ) + sin ( 4 y ) = 2 sin ( 3 y ) sin ( y ) 2 sin ( 3 y ) cos y = 2 sin ( three y ) sin ( y ) 2 sin ( 3 y ) cos y = tan y

61.

cos 10 cos ( 3 x ) = 2 sin ( two 10 ) sin ( x ) = 2 ( 2 sin x cos x ) sin ten = iv sin ii x cos x cos x cos ( 3 ten ) = 2 sin ( two x ) sin ( 10 ) = 2 ( 2 sin x cos ten ) sin x = four sin ii 10 cos x

63.

tan ( π 4 t ) = tan ( π iv ) tan t 1 + tan ( π 4 ) tan ( t ) = 1 tan t 1 + tan t tan ( π 4 t ) = tan ( π 4 ) tan t 1 + tan ( π 4 ) tan ( t ) = ane tan t 1 + tan t

7.v Section Exercises

i.

In that location will not ever be solutions to trigonometric function equations. For a bones example, cos ( x ) = −v. cos ( x ) = −five.

3.

If the sine or cosine part has a coefficient of one, isolate the term on one side of the equals sign. If the number information technology is set equal to has an absolute value less than or equal to i, the equation has solutions, otherwise it does not. If the sine or cosine does not have a coefficient equal to 1, still isolate the term just then dissever both sides of the equation by the leading coefficient. Then, if the number it is set equal to has an absolute value greater than one, the equation has no solution.

seven.

3 π 4 , five π iv 3 π 4 , 5 π 4

eleven.

π 4 , 3 π 4 , 5 π four , 7 π 4 π 4 , iii π iv , v π 4 , 7 π 4

thirteen.

π 4 , seven π 4 π four , 7 π 4

15.

seven π 6 , 11 π 6 7 π half dozen , 11 π six

17.

π xviii π 18 , v π eighteen v π eighteen , 13 π xviii xiii π eighteen , 17 π 18 17 π eighteen , 25 π xviii 25 π 18 , 29 π eighteen 29 π 18

19.

3 π 12 , 5 π 12 , xi π 12 , xiii π 12 , nineteen π 12 , 21 π 12 3 π 12 , 5 π 12 , xi π 12 , 13 π 12 , 19 π 12 , 21 π 12

21.

1 6 , five half-dozen , xiii 6 , 17 6 , 25 six , 29 6 , 37 six 1 6 , 5 vi , 13 six , 17 6 , 25 6 , 29 half-dozen , 37 6

23.

0 , π iii , π , v π 3 0 , π 3 , π , 5 π three

25.

π three , π , 5 π 3 π three , π , 5 π 3

27.

π 3 , iii π 2 , v π 3 π 3 , 3 π ii , 5 π iii

31.

π sin 1 ( 1 4 ) , 7 π 6 , 11 π 6 , ii π + sin ane ( i 4 ) π sin 1 ( 1 4 ) , seven π six , xi π half dozen , two π + sin 1 ( 1 four )

33.

1 3 ( sin ane ( 9 10 ) ) , π iii one 3 ( sin one ( 9 10 ) ) , two π 3 + one iii ( sin 1 ( 9 ten ) ) , π 1 3 ( sin ane ( 9 10 ) ) , four π 3 + 1 three ( sin 1 ( 9 10 ) ) , five π 3 ane three ( sin 1 ( 9 10 ) ) 1 3 ( sin one ( ix x ) ) , π 3 1 3 ( sin 1 ( ix 10 ) ) , two π iii + 1 3 ( sin i ( 9 10 ) ) , π 1 3 ( sin one ( 9 10 ) ) , four π 3 + 1 3 ( sin 1 ( nine 10 ) ) , 5 π three one iii ( sin 1 ( nine 10 ) )

37.

θ = sin - one 2 3 , π - sin - ane ii 3 , π + sin - 1 2 three , 2 π - sin - 1 2 three θ = sin - one two three , π - sin - 1 two 3 , π + sin - one ii 3 , 2 π - sin - 1 2 three

39.

three π 2 , π vi , 5 π 6 iii π 2 , π half-dozen , v π half dozen

41.

0 , π 3 , π , 4 π 3 0 , π 3 , π , four π 3

43.

There are no solutions.

45.

cos i ( 1 3 ( 1 7 ) ) , 2 π cos 1 ( 1 iii ( ane vii ) ) cos i ( 1 3 ( one vii ) ) , ii π cos 1 ( ane iii ( 1 seven ) )

47.

tan 1 ( ane 2 ( 29 5 ) ) , π + tan 1 ( 1 two ( 29 5 ) ) , π + tan 1 ( 1 ii ( 29 5 ) ) , 2 π + tan 1 ( 1 2 ( 29 5 ) ) tan 1 ( 1 2 ( 29 5 ) ) , π + tan ane ( 1 2 ( 29 5 ) ) , π + tan 1 ( 1 2 ( 29 5 ) ) , ii π + tan i ( one 2 ( 29 5 ) )

49.

There are no solutions.

51.

At that place are no solutions.

53.

0 , 2 π 3 , iv π 3 0 , two π 3 , four π 3

55.

π 4 , 3 π 4 , v π 4 , 7 π 4 π 4 , 3 π iv , 5 π four , 7 π four

57.

sin i ( 3 5 ) , π 2 , π sin 1 ( 3 5 ) , 3 π two sin 1 ( 3 5 ) , π 2 , π sin ane ( 3 five ) , 3 π 2

59.

cos 1 ( 1 4 ) cos 1 ( 1 4 ) , 2 π cos ane ( ane 4 ) 2 π cos ane ( i 4 )

61.

π three , cos i ( three four ) , 2 π cos 1 ( three four ) , 5 π three π 3 , cos 1 ( 3 4 ) , 2 π cos 1 ( 3 4 ) , 5 π iii

63.

cos ane ( 3 4 ) , cos one ( 2 3 ) , 2 π cos one ( 2 3 ) cos 1 ( 3 4 ) , cos i ( ii three ) , 2 π cos i ( 2 three ) , two π cos ane ( 3 4 ) two π cos 1 ( three 4 )

65.

0 , π 2 , π , iii π 2 0 , π 2 , π , three π 2

67.

π three , cos −1 ( 1 four ) , 2 π cos −1 ( ane 4 ) , 5 π iii π 3 , cos −1 ( 1 4 ) , 2 π cos −1 ( i 4 ) , 5 π 3

69.

There are no solutions.

71.

π + tan −1 ( −ii ) π + tan −i ( −ii ) , π + tan −1 ( 3 2 ) , ii π + tan −1 ( −ii ) , ii π + tan −1 ( iii 2 ) π + tan −one ( 3 2 ) , 2 π + tan −1 ( −ii ) , 2 π + tan −1 ( 3 two )

73.

2 π one thousand + 0.2734 , 2 π 1000 + ii.8682 2 π k + 0.2734 , 2 π k + two.8682

77.

0.6694 , 1.8287 , iii.8110 , 4.9703 0.6694 , 1.8287 , 3.8110 , 4.9703

79.

1.0472 , iii.1416 , five.2360 i.0472 , 3.1416 , 5.2360

81.

0.5326 , 1.7648 , 3.6742 , 4.9064 0.5326 , 1.7648 , iii.6742 , four.9064

83.

sin 1 ( i 4 ) , π sin 1 ( 1 4 ) , 3 π 2 sin 1 ( 1 iv ) , π sin 1 ( 1 4 ) , 3 π ii

85.

π 2 , 3 π two π 2 , 3 π 2

87.

There are no solutions.

89.

0 , π two , π , iii π 2 0 , π two , π , 3 π 2

91.

There are no solutions.

7.6 Section Exercises

ane.

Concrete behavior should be periodic, or cyclical.

3.

Since cumulative rainfall is always increasing, a sinusoidal part would non be ideal hither.

v.

y = 3 cos ( π vi x ) 1 y = 3 cos ( π 6 x ) one

viii.

y = 4 - 6 cos ( x π ii ) y = iv - half dozen cos ( x π 2 )

ten.

y = tan ( x π 8 ) y = tan ( x π viii )

12.

tan ( x π 12 ) tan ( 10 π 12 )

23.

From June xv through November 16

25.

From mean solar day 31 through twenty-four hour period 58

27.

Floods: Apr 16 to July 15. Drought: Oct 16 to Jan 15.

29.

Aamplitude: 8, period: 1 iii , ane iii , frequency: iii Hz

31.

Amplitude: four, period: four , 4 , frequency: 1 4 one 4 Hz

33.

P ( t ) = 19 cos ( π 6 t ) + 800 + 160 12 t P ( t ) = 19 cos ( π vi t ) + 800 + forty 3 t P ( t ) = 19 cos ( π vi t ) + 800 + 160 12 t P ( t ) = 19 cos ( π six t ) + 800 + 40 three t

35.

P ( t ) = 33 cos ( π vi t ) + 900 + ( i.07 ) t P ( t ) = 33 cos ( π 6 t ) + 900 + ( 1.07 ) t

37.

D ( t ) = ten ( 0.85 ) t cos ( 36 π t ) D ( t ) = x ( 0.85 ) t cos ( 36 π t )

39.

D ( t ) = 17 ( 0.9145 ) t cos ( 28 π t ) D ( t ) = 17 ( 0.9145 ) t cos ( 28 π t )

45.

Jump 2 comes to remainder commencement after 7.3 seconds.

47.

234.3 miles, at 72.2°

49.

y = half dozen ( 4 ) ten + v sin ( π ii x ) y = vi ( 4 ) x + 5 sin ( π 2 x )

51.

y = 4 ( 2 ) 10 + eight sin ( π 2 x ) y = 4 ( two ) x + 8 sin ( π ii x )

53.

y = iii ( 2 ) x cos ( π 2 x ) + 1 y = three ( ii ) 10 cos ( π 2 x ) + i

Review Exercises

one.

sin 1 ( 3 3 ) , π sin 1 ( 3 3 ) , π + sin ane ( three three ) , two π sin 1 ( 3 3 ) sin one ( three 3 ) , π sin 1 ( 3 three ) , π + sin one ( iii three ) , ii π sin one ( iii 3 )

3.

seven π 6 , 11 π vi seven π 6 , eleven π 6

5.

sin i ( i iv ) , π sin ane ( 1 4 ) sin 1 ( one 4 ) , π sin 1 ( one 4 )

15.

cos ( 4 x ) cos ( 3 ten ) cos x = cos ( 2 x + two x ) cos ( ten + two x ) cos ten = cos ( 2 ten ) cos ( ii ten ) sin ( 2 10 ) sin ( two 10 ) cos x cos ( two x ) cos x + sin x sin ( 2 10 ) cos x = ( cos 2 x sin 2 10 ) ii iv cos 2 ten sin ii x cos two ten ( cos ii x sin two x ) + sin 10 ( ii ) sin x cos ten cos x = ( cos ii 10 sin ii 10 ) 2 4 cos two x sin 2 ten cos 2 ten ( cos two ten sin 2 x ) + ii sin 2 ten cos 2 x = cos four x 2 cos 2 x sin 2 x + sin 4 ten four cos 2 x sin 2 x cos 4 10 + cos 2 x sin ii x + 2 sin 2 10 cos 2 x = sin 4 ten iv cos 2 x sin ii 10 + cos 2 x sin 2 10 = sin 2 x ( sin ii ten + cos 2 10 ) iv cos 2 ten sin ii x = sin two x 4 cos two ten sin 2 x cos ( 4 x ) cos ( 3 10 ) cos x = cos ( 2 10 + 2 x ) cos ( ten + ii x ) cos x = cos ( 2 10 ) cos ( 2 x ) sin ( 2 x ) sin ( 2 ten ) cos ten cos ( two ten ) cos x + sin 10 sin ( 2 x ) cos x = ( cos two 10 sin 2 x ) ii 4 cos two ten sin 2 x cos 2 x ( cos 2 x sin 2 x ) + sin x ( two ) sin x cos x cos 10 = ( cos 2 x sin ii 10 ) ii iv cos ii x sin two x cos two 10 ( cos 2 10 sin ii ten ) + ii sin 2 10 cos 2 x = cos four x 2 cos ii 10 sin 2 x + sin 4 ten 4 cos 2 x sin 2 ten cos 4 x + cos 2 x sin 2 x + 2 sin 2 ten cos 2 x = sin 4 x four cos 2 x sin two 10 + cos 2 x sin two x = sin 2 10 ( sin 2 x + cos 2 x ) 4 cos 2 x sin 2 x = sin 2 x 4 cos 2 x sin 2 x

17.

tan ( 5 8 x ) tan ( v eight x )

21.

24 25 , 7 25 , 24 vii 24 25 , 7 25 , 24 7

25.

two 10 , 7 2 10 , one seven , 3 5 , iv 5 , 3 4 2 10 , 7 2 10 , 1 7 , 3 v , 4 5 , 3 4

27.

cot 10 cos ( 2 x ) = cot x ( 1 ii sin 2 x ) = cot x cos x sin x ( 2 ) sin 2 ten = ii sin 10 cos 10 + cot 10 = sin ( two x ) + cot x cot 10 cos ( 2 x ) = cot x ( 1 two sin 2 10 ) = cot 10 cos x sin x ( ii ) sin 2 ten = 2 sin ten cos 10 + cot x = sin ( 2 x ) + cot x

29.

10 sin x 5 sin ( 3 x ) + sin ( 5 x ) 8 ( cos ( 2 x ) + i ) ten sin x 5 sin ( 3 x ) + sin ( 5 x ) viii ( cos ( 2 10 ) + one )

35.

1 2 ( sin ( 6 10 ) + sin ( 12 x ) ) 1 ii ( sin ( half-dozen 10 ) + sin ( 12 x ) )

37.

2 sin ( 13 2 x ) cos ( nine 2 x ) two sin ( thirteen 2 x ) cos ( 9 two ten )

39.

iii π 4 , seven π four 3 π 4 , 7 π 4

41.

0 , π 6 , 5 π 6 , π 0 , π 6 , 5 π 6 , π

47.

0.2527 , 2.8889 , 4.7124 0.2527 , 2.8889 , iv.7124

49.

ane.3694 1.3694 , ane.9106 i.9106 , 4.3726 iv.3726 , iv.9137 4.9137

51.

3 sin ( ten π 2 ) ii 3 sin ( x π 2 ) 2

55.

P ( t ) = 950 450 sin ( π 6 t ) P ( t ) = 950 450 sin ( π 6 t )

57.

Aamplitude: 3, menstruation: 2, frequency: one 2 1 2 Hz

59.

C ( t ) = twenty sin ( 2 π t ) + 100 ( one.4427 ) t C ( t ) = twenty sin ( 2 π t ) + 100 ( 1.4427 ) t

Practice Exam

11.

2 cos ( three ten ) cos ( 5 x ) 2 cos ( 3 10 ) cos ( 5 x )

13.

ten = cos –ane ( 1 5 ) ten = cos –1 ( i 5 )

15.

iii 5 , 4 v , 3 four three 5 , 4 5 , 3 4

17.

tan 3 x tan x sec 2 ten = tan 10 ( tan 2 x sec ii x ) = tan ten ( tan 2 x ( 1 + tan ii 10 ) ) = tan x ( tan two ten 1 tan 2 x ) = tan x = tan ( ten ) = tan 10 ) tan iii x tan x sec two 10 = tan x ( tan 2 x sec two x ) = tan x ( tan 2 x ( 1 + tan 2 ten ) ) = tan 10 ( tan 2 10 1 tan two x ) = tan ten = tan ( 10 ) = tan 10 )

19.

sin ( ii x ) sin 10 cos ( 2 x ) cos ten = 2 sin 10 cos x sin 10 ii cos ii x ane cos x = two cos x 2 cos ten + ane cos x = 1 cos ten = sec 10 = sec x sin ( 2 x ) sin ten cos ( ii x ) cos 10 = ii sin 10 cos 10 sin x two cos 2 x 1 cos x = two cos 10 2 cos x + 1 cos x = 1 cos x = sec ten = sec x

21.

Amplitude: 1 four ane 4 , period i 60 1 60 , frequency: sixty Hz

23.

Aamplitude: 8 8 , fast period: 1 500 1 500 , fast frequency: 500 Hz, boring flow: 1 10 one 10 , wearisome frequency: ten Hz

25.

D ( t ) = 20 ( 0.9086 ) t cos ( 4 π t ) D ( t ) = twenty ( 0.9086 ) t cos ( four π t ) , 31 seconds

Chapter 7 Test Answer Key,

Source: https://openstax.org/books/precalculus/pages/chapter-7

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