unit 2 the trigonometric functions homework answers

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Question: Unit 2 Lab B: Trigonometric FunctionsMAT 172 (Section 5.4, 5.6, 5.7 & 5.8)Please print and complete this lab. Once you have completed the lab, please enter your answers in Canvas in the "Unit 2 Lab Answer Entry Sheet".For problems 1-4, determine whether the statement is true or false.The secant function is a continuous function.The sine function is a

5) y = 3 2 sin ⁡ x

Use the form a sin ⁡ ( b x − c ) + d to find the variables used to find the amplitude, period, phase shift, and vertical ...

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9.1 Verifying Trigonometric Identities and Using Trigonometric Identities to Simplify Trigonometric Expressions

sin 2 θ − 1 tan θ sin θ − tan θ = ( sin θ + 1 ) ( sin θ − 1 ) tan θ ( sin θ − 1 ) = sin θ + 1 tan θ sin 2 θ − 1 tan θ sin θ − tan θ = ( sin θ + 1 ) ( sin θ − 1 ) tan θ ( sin θ − 1 ) = sin θ + 1 tan θ

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

9.2 Sum and Difference Identities

2 + 6 4 2 + 6 4

2 − 6 4 2 − 6 4

1 − 3 1 + 3 1 − 3 1 + 3

cos ( 5 π 14 ) cos ( 5 π 14 )

9.3 Double-Angle, Half-Angle, and Reduction Formulas

cos ( 2 α ) = 7 32 cos ( 2 α ) = 7 32

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

cos ( 2 θ ) cos θ = ( cos 2 θ − sin 2 θ ) cos θ = cos 3 θ − cos θ sin 2 θ cos ( 2 θ ) cos θ = ( cos 2 θ − sin 2 θ ) cos θ = cos 3 θ − cos θ sin 2 θ

10 cos 4 x = 10 ( cos 2 x ) 2 = 10 [ 1 + cos ( 2 x ) 2 ] 2 Substitute reduction formula for cos 2 x . = 10 4 [ 1 + 2 cos ( 2 x ) + cos 2 ( 2 x ) ] = 10 4 + 10 2 cos ( 2 x ) + 10 4 ( 1 + cos 2 ( 2 x ) 2 ) Substitute reduction formula for cos 2 x . = 10 4 + 10 2 cos ( 2 x ) + 10 8 + 10 8 cos ( 4 x ) = 30 8 + 5 cos ( 2 x ) + 10 8 cos ( 4 x ) = 15 4 + 5 cos ( 2 x ) + 5 4 cos ( 4 x ) 10 cos 4 x = 10 ( cos 2 x ) 2 = 10 [ 1 + cos ( 2 x ) 2 ] 2 Substitute reduction formula for cos 2 x . = 10 4 [ 1 + 2 cos ( 2 x ) + cos 2 ( 2 x ) ] = 10 4 + 10 2 cos ( 2 x ) + 10 4 ( 1 + cos 2 ( 2 x ) 2 ) Substitute reduction formula for cos 2 x . = 10 4 + 10 2 cos ( 2 x ) + 10 8 + 10 8 cos ( 4 x ) = 30 8 + 5 cos ( 2 x ) + 10 8 cos ( 4 x ) = 15 4 + 5 cos ( 2 x ) + 5 4 cos ( 4 x )

− 2 5 − 2 5

9.4 Sum-to-Product and Product-to-Sum Formulas

1 2 ( cos 6 θ + cos 2 θ ) 1 2 ( cos 6 θ + cos 2 θ )

1 2 ( sin 2 x + sin 2 y ) 1 2 ( sin 2 x + sin 2 y )

− 2 − 3 4 − 2 − 3 4

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

tan θ cot θ − cos 2 θ = ( sin θ cos θ ) ( cos θ sin θ ) − cos 2 θ = 1 − cos 2 θ = sin 2 θ tan θ cot θ − cos 2 θ = ( sin θ cos θ ) ( cos θ sin θ ) − cos 2 θ = 1 − cos 2 θ = sin 2 θ

9.5 Solving Trigonometric Equations

x = 7 π 6 , 11 π 6 x = 7 π 6 , 11 π 6

π 3 ± π k π 3 ± π k

θ ≈ 1.7722 ± 2 π k θ ≈ 1.7722 ± 2 π k and θ ≈ 4.5110 ± 2 π k θ ≈ 4.5110 ± 2 π k

cos θ = − 1 , θ = π cos θ = − 1 , θ = π

π 2 , 2 π 3 , 4 π 3 , 3 π 2 π 2 , 2 π 3 , 4 π 3 , 3 π 2

9.1 Section Exercises

All three functions, F F , G G , and H H , are even.

This is because F ( − x ) = sin ( − x ) sin ( − x ) = ( − sin x ) ( − sin x ) = sin 2 x = F ( x ) F ( − x ) = sin ( − x ) sin ( − x ) = ( − sin x ) ( − sin x ) = sin 2 x = F ( x ) , G ( − x ) = cos ( − x ) cos ( − x ) = cos x cos x = cos 2 x = G ( x ) G ( − x ) = cos ( − x ) cos ( − x ) = cos x cos x = cos 2 x = G ( x ) and H ( − x ) = tan ( − x ) tan ( − x ) = ( − tan x ) ( − tan x ) = tan 2 x = H ( x ) . H ( − x ) = tan ( − x ) tan ( − x ) = ( − tan x ) ( − tan x ) = tan 2 x = H ( x ) .

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

sin x sin x

sec x sec x

csc t csc t

sec 2 x sec 2 x

sin 2 x + 1 sin 2 x + 1

1 sin x 1 sin x

1 cot x 1 cot x

tan x tan x

− 4 sec x tan x − 4 sec x tan x

± 1 cot 2 x + 1 ± 1 cot 2 x + 1

± 1 − sin 2 x sin x ± 1 − sin 2 x sin x

Answers will vary. Sample proof:

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

Answers will vary. Sample proof: 1 + sin 2 x cos 2 x = 1 cos 2 x + sin 2 x cos 2 x = sec 2 x + tan 2 x = tan 2 x + 1 + tan 2 x = 1 + 2 tan 2 x 1 + sin 2 x cos 2 x = 1 cos 2 x + sin 2 x cos 2 x = sec 2 x + tan 2 x = tan 2 x + 1 + tan 2 x = 1 + 2 tan 2 x

Answers will vary. Sample proof: cos 2 x − tan 2 x = 1 − sin 2 x − ( sec 2 x − 1 ) = 1 − sin 2 x − sec 2 x + 1 = 2 − sin 2 x − sec 2 x cos 2 x − tan 2 x = 1 − sin 2 x − ( sec 2 x − 1 ) = 1 − sin 2 x − sec 2 x + 1 = 2 − sin 2 x − sec 2 x

Proved with negative and Pythagorean identities

True 3 sin 2 θ + 4 cos 2 θ = 3 sin 2 θ + 3 cos 2 θ + cos 2 θ = 3 ( sin 2 θ + cos 2 θ ) + cos 2 θ = 3 + cos 2 θ 3 sin 2 θ + 4 cos 2 θ = 3 sin 2 θ + 3 cos 2 θ + cos 2 θ = 3 ( sin 2 θ + cos 2 θ ) + cos 2 θ = 3 + cos 2 θ

9.2 Section Exercises

The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures x , x , the second angle measures π 2 − x . π 2 − x . Then sin x = cos ( π 2 − x ) .   sin x = cos ( π 2 − x ) .   The same holds for the other cofunction identities. The key is that the angles are complementary.

sin ( − x ) = − sin x , sin ( − x ) = − sin x , so sin x sin x is odd. cos ( − x ) = cos ( 0 − x ) = cos x , cos ( − x ) = cos ( 0 − x ) = cos x , so cos x cos x is even.

6 − 2 4 6 − 2 4

− 2 − 3 − 2 − 3

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

− 1 2 cos x − 3 2 sin x − 1 2 cos x − 3 2 sin x

csc θ csc θ

cot x cot x

tan ( x 10 ) tan ( x 10 )

sin ( a − b ) = ( 4 5 ) ( 1 3 ) − ( 3 5 ) ( 2 2 3 ) = 4 − 6 2 15 cos ( a + b ) = ( 3 5 ) ( 1 3 ) − ( 4 5 ) ( 2 2 3 ) = 3 − 8 2 15 sin ( a − b ) = ( 4 5 ) ( 1 3 ) − ( 3 5 ) ( 2 2 3 ) = 4 − 6 2 15 cos ( a + b ) = ( 3 5 ) ( 1 3 ) − ( 4 5 ) ( 2 2 3 ) = 3 − 8 2 15

cot ( π 6 − x ) cot ( π 6 − x )

cot ( π 4 + x ) cot ( π 4 + x )

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

They are the same.

They are the different, try g ( x ) = sin ( 9 x ) − cos ( 3 x ) sin ( 6 x ) . g ( x ) = sin ( 9 x ) − cos ( 3 x ) sin ( 6 x ) .

They are the different, try g ( θ ) = 2 tan θ 1 − tan 2 θ . g ( θ ) = 2 tan θ 1 − tan 2 θ .

They are different, try g ( x ) = tan x − tan ( 2 x ) 1 + tan x tan ( 2 x ) . g ( x ) = tan x − tan ( 2 x ) 1 + tan x tan ( 2 x ) .

− 3 − 1 2 2 , or  − 0.2588 − 3 − 1 2 2 , or  − 0.2588

1 + 3 2 2 , 1 + 3 2 2 , or 0.9659

tan ( x + π 4 ) = tan x + tan ( π 4 ) 1 − tan x tan ( π 4 ) = tan x + 1 1 − tan x ( 1 ) = tan x + 1 1 − tan x tan ( x + π 4 ) = tan x + tan ( π 4 ) 1 − tan x tan ( π 4 ) = tan x + 1 1 − tan x ( 1 ) = tan x + 1 1 − tan x

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 = 1 − tan a tan b

cos ( x + h ) − cos x h = cos x cosh − sin x sinh − cos x h = cos x ( cosh − 1 ) − sin x sinh h = cos x cos h − 1 h − sin x sin h h cos ( x + h ) − cos x h = cos x cosh − sin x sinh − cos x h = cos x ( cosh − 1 ) − sin x sinh h = cos x cos h − 1 h − sin x sin h h

True. Note that   sin ( α + β ) = sin ( π − γ )     sin ( α + β ) = sin ( π − γ )   and expand the right hand side.

9.3 Section Exercises

Use the Pythagorean identities and isolate the squared term.

1 − cos x sin x , sin x 1 + cos x , 1 − cos x sin x , sin x 1 + cos x , multiplying the top and bottom by 1 − cos x 1 − cos x and 1 + cos x , 1 + cos x , respectively.

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

a) 3 2 3 2 b) − 1 2 − 1 2 c) − 3 − 3

cos θ = − 2 5 5 , sin θ = 5 5 , tan θ = − 1 2 , csc θ = 5 , sec θ = − 5 2 , cot θ = − 2 cos θ = − 2 5 5 , sin θ = 5 5 , tan θ = − 1 2 , csc θ = 5 , sec θ = − 5 2 , cot θ = − 2

2 sin ( π 2 ) 2 sin ( π 2 )

2 − 2 2 2 − 2 2

2 − 3 2 2 − 3 2

2 + 3 2 + 3

− 1 − 2 − 1 − 2

a) 3 13 13 3 13 13 b) − 2 13 13 − 2 13 13 c) − 3 2 − 3 2

a) 10 4 10 4 b) 6 4 6 4 c) 15 3 15 3

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

2 13 13 , 3 13 13 , 2 3 2 13 13 , 3 13 13 , 2 3

cos ( 74° ) cos ( 74° )

cos ( 18 x ) cos ( 18 x )

3 sin ( 10 x ) 3 sin ( 10 x )

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

sin ( 2 θ ) 1 + cos ( 2 θ ) tan 2 θ = 2 sin ( θ ) cos ( θ ) 1 + cos 2 θ − sin 2 θ tan 2 θ = 2 sin ( θ ) cos ( θ ) 2 cos 2 θ tan 2 θ = sin ( θ ) cos θ tan 2 θ = cot ( θ ) tan 2 θ = tan 3 θ sin ( 2 θ ) 1 + cos ( 2 θ ) tan 2 θ = 2 sin ( θ ) cos ( θ ) 1 + cos 2 θ − sin 2 θ tan 2 θ = 2 sin ( θ ) cos ( θ ) 2 cos 2 θ tan 2 θ = sin ( θ ) cos θ tan 2 θ = cot ( θ ) tan 2 θ = tan 3 θ

1 + cos ( 12 x ) 2 1 + cos ( 12 x ) 2

3 + cos ( 12 x ) − 4 cos ( 6 x ) 8 3 + cos ( 12 x ) − 4 cos ( 6 x ) 8

2 + cos ( 2 x ) − 2 cos ( 4 x ) − cos ( 6 x ) 32 2 + cos ( 2 x ) − 2 cos ( 4 x ) − cos ( 6 x ) 32

3 + cos ( 4 x ) − 4 cos ( 2 x ) 3 + cos ( 4 x ) + 4 cos ( 2 x ) 3 + cos ( 4 x ) − 4 cos ( 2 x ) 3 + cos ( 4 x ) + 4 cos ( 2 x )

1 − cos ( 4 x ) 8 1 − cos ( 4 x ) 8

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

( 1 + cos ( 4 x ) ) sin x 2 ( 1 + cos ( 4 x ) ) sin x 2

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

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

2 sin x cos x 2 cos 2 x − 1 = sin ( 2 x ) cos ( 2 x ) = tan ( 2 x ) 2 sin x cos x 2 cos 2 x − 1 = sin ( 2 x ) cos ( 2 x ) = tan ( 2 x )

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

1 + cos ( 2 t ) sin ( 2 t ) − cos t = 1 + 2 cos 2 t − 1 2 sin t cos t − cos t = 2 cos 2 t cos t ( 2 sin t − 1 ) = 2 cos t 2 sin t − 1 1 + cos ( 2 t ) sin ( 2 t ) − cos t = 1 + 2 cos 2 t − 1 2 sin t cos t − cos t = 2 cos 2 t cos t ( 2 sin t − 1 ) = 2 cos t 2 sin t − 1

( cos 2 ( 4 x ) − sin 2 ( 4 x ) − sin ( 8 x ) ) ( cos 2 ( 4 x ) − sin 2 ( 4 x ) + sin ( 8 x ) ) = = ( cos ( 8 x ) − sin ( 8 x ) ) ( cos ( 8 x ) + sin ( 8 x ) ) = cos 2 ( 8 x ) − sin 2 ( 8 x ) = cos ( 16 x ) ( cos 2 ( 4 x ) − sin 2 ( 4 x ) − sin ( 8 x ) ) ( cos 2 ( 4 x ) − sin 2 ( 4 x ) + sin ( 8 x ) ) = = ( cos ( 8 x ) − sin ( 8 x ) ) ( cos ( 8 x ) + sin ( 8 x ) ) = cos 2 ( 8 x ) − sin 2 ( 8 x ) = cos ( 16 x )

9.4 Section Exercises

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

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

8 ( cos ( 5 x ) − cos ( 27 x ) ) 8 ( cos ( 5 x ) − cos ( 27 x ) )

sin ( 2 x ) + sin ( 8 x ) sin ( 2 x ) + sin ( 8 x )

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

2 cos ( 5 t ) cos t 2 cos ( 5 t ) cos t

2 cos ( 7 x ) 2 cos ( 7 x )

2 cos ( 6 x ) cos ( 3 x ) 2 cos ( 6 x ) cos ( 3 x )

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

1 4 ( 3 − 2 ) 1 4 ( 3 − 2 )

1 4 ( 3 − 1 ) 1 4 ( 3 − 1 )

cos ( 80° ) − cos ( 120° ) cos ( 80° ) − cos ( 120° )

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

2 cos ( 31° ) 2 cos ( 31° )

2 cos ( 66.5° ) sin ( 34.5° ) 2 cos ( 66.5° ) sin ( 34.5° )

2 sin ( −1.5° ) cos ( 0.5° ) 2 sin ( −1.5° ) cos ( 0.5° )

2 sin ( 7 x ) − 2 sin x = 2 sin ( 4 x + 3 x ) − 2 sin ( 4 x − 3 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 ( 4 x ) ) = 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) − 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) = 4 sin ( 3 x ) cos ( 4 x ) 2 sin ( 7 x ) − 2 sin x = 2 sin ( 4 x + 3 x ) − 2 sin ( 4 x − 3 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 ( 4 x ) ) = 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) − 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) = 4 sin ( 3 x ) cos ( 4 x )

sin x + sin ( 3 x ) = 2 sin ( 4 x 2 ) cos ( − 2 x 2 ) = 2 sin ( 2 x ) cos x = 2 ( 2 sin x cos x ) cos x = 4 sin x cos 2 x sin x + sin ( 3 x ) = 2 sin ( 4 x 2 ) cos ( − 2 x 2 ) = 2 sin ( 2 x ) cos x = 2 ( 2 sin x cos x ) cos x = 4 sin x cos 2 x

2 tan x cos ( 3 x ) = 2 sin x cos ( 3 x ) cos x = 2 ( .5 ( sin ( 4 x ) − sin ( 2 x ) ) ) cos x = 1 cos x ( sin ( 4 x ) − sin ( 2 x ) ) = sec x ( sin ( 4 x ) − sin ( 2 x ) ) 2 tan x cos ( 3 x ) = 2 sin x cos ( 3 x ) cos x = 2 ( .5 ( sin ( 4 x ) − sin ( 2 x ) ) ) cos x = 1 cos x ( sin ( 4 x ) − sin ( 2 x ) ) = sec x ( sin ( 4 x ) − sin ( 2 x ) )

2 cos ( 35° ) cos ( 23° ) , 1.5081 2 cos ( 35° ) cos ( 23° ) , 1.5081

− 2 sin ( 33° ) sin ( 11° ) , − 0.2078 − 2 sin ( 33° ) sin ( 11° ) , − 0.2078

1 2 ( cos ( 99° ) − cos ( 71° ) ) , −0.2410 1 2 ( cos ( 99° ) − cos ( 71° ) ) , −0.2410

It is an identity.

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

tan ( 3 t ) tan ( 3 t )

2 cos ( 2 x ) 2 cos ( 2 x )

− sin ( 14 x ) − sin ( 14 x )

Start with cos x + cos y . cos x + cos y . Make a substitution and let x = α + β x = α + β and let y = α − β , y = α − β , so cos x + cos y cos x + 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 = 2 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 x and y and substitute in for 2 cos α cos β 2 cos α cos β and get 2 cos ( x + y 2 ) cos ( x − y 2 ) . 2 cos ( x + y 2 ) cos ( x − y 2 ) .

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

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 ( 3 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 ( 3 y ) sin ( y ) 2 sin ( 3 y ) cos y = tan y

cos x − cos ( 3 x ) = − 2 sin ( 2 x ) sin ( − x ) = 2 ( 2 sin x cos x ) sin x = 4 sin 2 x cos x cos x − cos ( 3 x ) = − 2 sin ( 2 x ) sin ( − x ) = 2 ( 2 sin x cos x ) sin x = 4 sin 2 x cos x

tan ( π 4 − t ) = tan ( π 4 ) − 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 ) = 1 − tan t 1 + tan t

9.5 Section Exercises

There will not always be solutions to trigonometric function equations. For a basic example, cos ( x ) = −5. cos ( x ) = −5.

If the sine or cosine function has a coefficient of one, isolate the term on one side of the equals sign. If the number it is set equal to has an absolute value less than or equal to one, the equation has solutions, otherwise it does not. If the sine or cosine does not have a coefficient equal to one, still isolate the term but then divide 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.

π 3 , 2 π 3 π 3 , 2 π 3

3 π 4 , 5 π 4 3 π 4 , 5 π 4

π 4 , 5 π 4 π 4 , 5 π 4

π 4 , 3 π 4 , 5 π 4 , 7 π 4 π 4 , 3 π 4 , 5 π 4 , 7 π 4

π 4 , 7 π 4 π 4 , 7 π 4

7 π 6 , 11 π 6 7 π 6 , 11 π 6

π 18 , 5 π 18 , 13 π 18 , 17 π 18 , 25 π 18 , 29 π 18 π 18 , 5 π 18 , 13 π 18 , 17 π 18 , 25 π 18 , 29 π 18

3 π 12 , 5 π 12 , 11 π 12 , 13 π 12 , 19 π 12 , 21 π 12 3 π 12 , 5 π 12 , 11 π 12 , 13 π 12 , 19 π 12 , 21 π 12

1 6 , 5 6 , 13 6 , 17 6 , 25 6 , 29 6 , 37 6 1 6 , 5 6 , 13 6 , 17 6 , 25 6 , 29 6 , 37 6

0 , π 3 , π , 5 π 3 0 , π 3 , π , 5 π 3

π 3 , π , 5 π 3 π 3 , π , 5 π 3

π 3 , 3 π 2 , 5 π 3 π 3 , 3 π 2 , 5 π 3

0 , π 0 , π

π − sin − 1 ( − 1 4 ) , 7 π 6 , 11 π 6 , 2 π + sin − 1 ( − 1 4 ) π − sin − 1 ( − 1 4 ) , 7 π 6 , 11 π 6 , 2 π + sin − 1 ( − 1 4 )

1 3 ( sin − 1 ( 9 10 ) ) 1 3 ( sin − 1 ( 9 10 ) ) , π 3 − 1 3 ( sin − 1 ( 9 10 ) ) π 3 − 1 3 ( sin − 1 ( 9 10 ) ) , 2 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) 2 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) , π − 1 3 ( sin − 1 ( 9 10 ) ) π − 1 3 ( sin − 1 ( 9 10 ) ) , 4 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) 4 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) , 5 π 3 − 1 3 ( sin − 1 ( 9 10 ) ) 5 π 3 − 1 3 ( sin − 1 ( 9 10 ) )

π 6 , 5 π 6 , 7 π 6 , 11 π 6 π 6 , 5 π 6 , 7 π 6 , 11 π 6

3 π 2 , π 6 , 5 π 6 3 π 2 , π 6 , 5 π 6

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

There are no solutions.

cos − 1 ( 1 3 ( 1 − 7 ) ) cos − 1 ( 1 3 ( 1 − 7 ) ) , 2 π − cos − 1 ( 1 3 ( 1 − 7 ) ) 2 π − cos − 1 ( 1 3 ( 1 − 7 ) )

tan − 1 ( 1 2 ( 29 − 5 ) ) tan − 1 ( 1 2 ( 29 − 5 ) ) , π + tan − 1 ( 1 2 ( − 29 − 5 ) ) π + tan − 1 ( 1 2 ( − 29 − 5 ) ) , π + tan − 1 ( 1 2 ( 29 − 5 ) ) π + tan − 1 ( 1 2 ( 29 − 5 ) ) , 2 π + tan − 1 ( 1 2 ( − 29 − 5 ) ) 2 π + tan − 1 ( 1 2 ( − 29 − 5 ) )

0 , 2 π 3 , 4 π 3 0 , 2 π 3 , 4 π 3

sin − 1 ( 3 5 ) , π 2 , π − sin − 1 ( 3 5 ) , 3 π 2 sin − 1 ( 3 5 ) , π 2 , π − sin − 1 ( 3 5 ) , 3 π 2

cos − 1 ( − 1 4 ) , 2 π − cos − 1 ( − 1 4 ) cos − 1 ( − 1 4 ) , 2 π − cos − 1 ( − 1 4 )

π 3 π 3 , cos − 1 ( − 3 4 ) cos − 1 ( − 3 4 ) , 2 π − cos − 1 ( − 3 4 ) 2 π − cos − 1 ( − 3 4 ) , 5 π 3 5 π 3

cos − 1 ( 3 4 ) cos − 1 ( 3 4 ) , cos − 1 ( − 2 3 ) cos − 1 ( − 2 3 ) , 2 π − cos − 1 ( − 2 3 ) 2 π − cos − 1 ( − 2 3 ) , 2 π − cos − 1 ( 3 4 ) 2 π − cos − 1 ( 3 4 )

0 , π 2 , π , 3 π 2 0 , π 2 , π , 3 π 2

π 3 π 3 , cos −1 ( − 1 4 ) cos −1 ( − 1 4 ) , 2 π − cos −1 ( − 1 4 ) 2 π − cos −1 ( − 1 4 ) , 5 π 3 5 π 3

π + tan −1 ( −2 ) π + tan −1 ( −2 ) , π + tan −1 ( − 3 2 ) π + tan −1 ( − 3 2 ) , 2 π + tan −1 ( −2 ) 2 π + tan −1 ( −2 ) , 2 π + tan −1 ( − 3 2 ) 2 π + tan −1 ( − 3 2 )

2 π k + 0.2734 , 2 π k + 2.8682 2 π k + 0.2734 , 2 π k + 2.8682

π k − 0.3277 π k − 0.3277

0.6694 , 1.8287 , 3.8110 , 4.9703 0.6694 , 1.8287 , 3.8110 , 4.9703

1.0472 , 3.1416 , 5.2360 1.0472 , 3.1416 , 5.2360

0.5326 , 1.7648 , 3.6742 , 4.9064 0.5326 , 1.7648 , 3.6742 , 4.9064

sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 ) , 3 π 2 sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 ) , 3 π 2

π 2 , 3 π 2 π 2 , 3 π 2

7.2 ∘ 7.2 ∘

5.7 ∘ 5.7 ∘

82.4 ∘ 82.4 ∘

31.0 ∘ 31.0 ∘

88.7 ∘ 88.7 ∘

59.0 ∘ 59.0 ∘

36.9 ∘ 36.9 ∘

Review Exercises

sin − 1 ( 3 3 ) sin − 1 ( 3 3 ) , π − sin − 1 ( 3 3 ) π − sin − 1 ( 3 3 ) , π + sin − 1 ( 3 3 ) π + sin − 1 ( 3 3 ) , 2 π − sin − 1 ( 3 3 ) 2 π − sin − 1 ( 3 3 )

sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 ) sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 )

cos ( 4 x ) − cos ( 3 x ) cos x = cos ( 2 x + 2 x ) − cos ( x + 2 x ) cos x = cos ( 2 x ) cos ( 2 x ) − sin ( 2 x ) sin ( 2 x ) − cos x cos ( 2 x ) cos x + sin x sin ( 2 x ) cos x = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + sin x ( 2 ) sin x cos x cos x = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + 2 sin 2 x cos 2 x = cos 4 x − 2 cos 2 x sin 2 x + sin 4 x − 4 cos 2 x sin 2 x − cos 4 x + cos 2 x sin 2 x + 2 sin 2 x cos 2 x = sin 4 x − 4 cos 2 x sin 2 x + cos 2 x sin 2 x = sin 2 x ( sin 2 x + cos 2 x ) − 4 cos 2 x sin 2 x = sin 2 x − 4 cos 2 x sin 2 x cos ( 4 x ) − cos ( 3 x ) cos x = cos ( 2 x + 2 x ) − cos ( x + 2 x ) cos x = cos ( 2 x ) cos ( 2 x ) − sin ( 2 x ) sin ( 2 x ) − cos x cos ( 2 x ) cos x + sin x sin ( 2 x ) cos x = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + sin x ( 2 ) sin x cos x cos x = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + 2 sin 2 x cos 2 x = cos 4 x − 2 cos 2 x sin 2 x + sin 4 x − 4 cos 2 x sin 2 x − cos 4 x + cos 2 x sin 2 x + 2 sin 2 x cos 2 x = sin 4 x − 4 cos 2 x sin 2 x + cos 2 x sin 2 x = sin 2 x ( sin 2 x + cos 2 x ) − 4 cos 2 x sin 2 x = sin 2 x − 4 cos 2 x sin 2 x

tan ( 5 8 x ) tan ( 5 8 x )

− 24 25 , − 7 25 , 24 7 − 24 25 , − 7 25 , 24 7

2 ( 2 + 2 ) 2 ( 2 + 2 )

2 10 , 7 2 10 , 1 7 , 3 5 , 4 5 , 3 4 2 10 , 7 2 10 , 1 7 , 3 5 , 4 5 , 3 4

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

10 sin x − 5 sin ( 3 x ) + sin ( 5 x ) 8 ( cos ( 2 x ) + 1 ) 10 sin x − 5 sin ( 3 x ) + sin ( 5 x ) 8 ( cos ( 2 x ) + 1 )

− 2 2 − 2 2

1 2 ( sin ( 6 x ) + sin ( 12 x ) ) 1 2 ( sin ( 6 x ) + sin ( 12 x ) )

2 sin ( 13 2 x ) cos ( 9 2 x ) 2 sin ( 13 2 x ) cos ( 9 2 x )

3 π 4 , 7 π 4 3 π 4 , 7 π 4

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

3 π 2 3 π 2

No solution

0.2527 , 2.8889 , 4.7124 0.2527 , 2.8889 , 4.7124

1.3694 , 1.9106 , 4.3726 , 4.9137 1.3694 , 1.9106 , 4.3726 , 4.9137

Practice Test

sec ( θ ) sec ( θ )

− 1 2 cos θ + 3 2 sin θ − 1 2 cos θ + 3 2 sin θ

1 − cos ( 64 ∘ ) 2 1 − cos ( 64 ∘ ) 2

2 cos ( 3 x ) cos ( 5 x ) 2 cos ( 3 x ) cos ( 5 x )

4 sin ( 2 θ ) cos ( 6 θ ) 4 sin ( 2 θ ) cos ( 6 θ )

x = cos –1 ( 1 5 ) x = cos –1 ( 1 5 )

3 5 , – 4 5 , – 3 4 3 5 , – 4 5 , – 3 4

tan 3 x – tan x sec 2 x = tan x ( tan 2 x – sec 2 x ) = tan x ( tan 2 x – ( 1 + tan 2 x ) ) = tan x ( tan 2 x – 1 – tan 2 x ) = – tan x = tan ( – x ) = tan ( – x ) tan 3 x – tan x sec 2 x = tan x ( tan 2 x – sec 2 x ) = tan x ( tan 2 x – ( 1 + tan 2 x ) ) = tan x ( tan 2 x – 1 – tan 2 x ) = – tan x = tan ( – x ) = tan ( – x )

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

Amplitude: 1 4 1 4 , period: 1 60 1 60 , frequency: 60 Hz

Amplitude: 8, fast period: 1 500 1 500 , fast frequency: 500 Hz, slow period: 1 10 1 10 , slow frequency: 10 Hz

D ( t ) = 20 ( 0.9086 ) t cos ( 4 π t ) D ( t ) = 20 ( 0.9086 ) t cos ( 4 π t ) , 31 second

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Access for free at https://openstax.org/books/algebra-and-trigonometry/pages/1-introduction-to-prerequisites

• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: Algebra and Trigonometry
• Publication date: Feb 13, 2015
• Location: Houston, Texas
• Book URL: https://openstax.org/books/algebra-and-trigonometry/pages/1-introduction-to-prerequisites
• Section URL: https://openstax.org/books/algebra-and-trigonometry/pages/chapter-9

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