5.7 solving trigonometric equations homework answers

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5.7 Solving trigonometric equations

5.7 Solving trigonometric equations (EMA3T)

In this section we will first look at finding unknown lengths in right-angled triangles and then we will look at finding unknown angles in right-angled triangles. Finally we will look at how to solve more general trigonometric equations.

Finding lengths (EMA3V)

From the definitions of the trigonometric ratios and what we have learnt about determining the values of these ratios for any angle we can now use this to help us find unknown lengths in right-angled triangles. The following worked examples will show you how.

Worked example 4: Finding lengths

Find the length of \(x\) in the following right-angled triangle using the appropriate trigonometric ratio (round your answer to two decimal places).

Identify the opposite and adjacent sides and the hypotenuse with reference to the given angle

Remember that the hypotenuse side is always opposite the right angle, it never changes position. The opposite side is opposite the angle we are interested in and the adjacent side is the remaining side.

Rearrange the equation to solve for \(x\)

Use your calculator to find the answer, worked example 5: finding lengths.

Worked example 6: Finding lengths

Find the length of \(x\) and \(y\) in the following right-angled triangle using the appropriate trigonometric ratio (round your answers to two decimal places).

Rearrange the equations to solve for \(x\) and \(y\)

Use your calculator to find the answers.

The following video shows an example of finding unknown lengths in a triangle using the trigonometric ratios.

Video: 2FQT

In each triangle find the length of the side marked with a letter. Give your answers correct to \(\text{2}\) decimal places.

Write down two ratios for each of the following in terms of the sides: \(AB; BC; BD; AD; DC \text{ and } AC\).

\(\sin \hat{B}\)

We note that triangles \(ABC\) and \(ABD\) both contain angle \(B\) so we can use these triangles to write down the ratios:

\(\cos \hat{D}\)

We note that triangles \(ACD\) and \(ABD\) both contain angle \(D\) so we can use these triangles to write down the ratios:

\(\tan \hat{B}\)

In \(\triangle MNP\), \(\hat{N} = 90°\), \(MP=20\) and \(\hat{P} = 40°\). Calculate \(NP\) and \(MN\) (correct to \(\text{2}\) decimal places).

Sketch the triangle:

To find \(MN\) we use the sine ratio:

To find \(NP\) we can use the cosine ratio:

Therefore \(MN = \text{12,86} \text{ and } NP = \text{15,32}\)

Calculate \(x\) and \(y\) in the following diagram.

To find \(x\) we use \(\triangle ABC\) and the tangent ratio. To find \(y\) we use \(\triangle ABD\) and the tangent ratio.

Therefore \(x = \text{29,82} \text{ and } y = \text{31,98}\).

Finding an angle (EMA3W)

If the length of two sides of a triangle are known, the angles can be calculated using trigonometric ratios. In this section, we are finding angles inside right-angled triangles using the ratios of the sides.

Worked example 7: Finding angles

Find the value of \(\theta\) in the following right-angled triangle using the appropriate trigonometric ratio.

Identify the opposite and adjacent sides with reference to the given angle and the hypotenuse

In this case you have the opposite side and the adjacent side for angle \(\theta\).

Use your calculator to solve for \(\theta\)

To solve for \(\theta\), you will need to use the inverse tangent function on your calculator. This works backwards by using the ratio of the sides to determine the angle which resulted in that ratio.

Press \(\boxed{\text{SHIFT}} \enspace \boxed{\text{tan}} \enspace \boxed{50} \enspace \boxed{\div} \enspace \boxed{100} \enspace \boxed{)} \enspace \boxed{=} \enspace \text{26,56505...} \approx \text{26,6}\)

Write the final answer

Determine \(\alpha\) in the following right-angled triangles:

We have now seen how to solve trigonometric equations in right-angled triangles. We can use the same techniques to help us solve trigonometric equations when the triangle is not shown.

Worked example 8: Solving trigonometric equations

Find the value of \(\theta\) if \(\cos \theta = \text{0,2}\).

To solve for \(\theta\), you will need to use the inverse cosine function on your calculator. This works backwards by using the ratio of the sides to determine the angle which resulted in that ratio.

Press \(\boxed{\text{SHIFT}} \enspace \boxed{\text{cos}} \enspace \boxed{0} \enspace \boxed{.} \enspace \boxed{2} \enspace \boxed{)} \enspace \boxed{=} \enspace \text{78,46304} \approx \text{78,46}\)

Worked example 9: Solving trigonometric equations

Find the value of \(\theta\) if \(3\sin \theta = \text{2,4}\).

Rearrange the equation

We need to rearrange the equation so that \(\sin \theta\) is on one side of the equation.

To solve for \(\theta\), you will need to use the inverse sine function on your calculator. This works backwards by using the ratio of the sides to determine the angle which resulted in that ratio.

Press \(\boxed{\text{SHIFT}} \enspace \boxed{\text{sin}} \enspace \boxed{(} \enspace \boxed{2} \enspace \boxed{.} \enspace \boxed{4} \enspace \boxed{\div} \enspace \boxed{3} \enspace \boxed{)} \enspace \boxed{=} \enspace \text{53,1301...} \approx \text{53,13}\).

When you are solving trigonometric equations you might find that you get an error when you try to calculate \(\sin\) or \(\cos\) (remember that both the sine and cosine functions have a maximum value of 1). For these cases there is no solution to the equation.

If learners get a math error on their calculator encourage them to think about what might have happened. It is also important to ensure that they know they must write down no solution rather than math error when this happens.

Worked example 10: Solving trigonometric equations

Solve for \(\alpha\): \(3 \sec \alpha = \text{1,4}\).

Convert \(\sec\) to \(\cos\)

There is no “\(\sec\)” button on the calculator and so we need to convert \(\sec\) to \(\cos\) so we can find \(\alpha\).

We need to rearrange the equation so that we have \(\cos \alpha\) on one side of the equation.

Use your calculator to solve for \(\alpha\)

To solve for \(\alpha\), you will need to use the inverse cosine function on your calculator. This works backwards by using the ratio of the sides to determine the angle which resulted in that ratio.

Press \(\boxed{\text{SHIFT}} \enspace \boxed{\text{cos}} \enspace \boxed{3} \enspace \boxed{\div} \enspace \boxed{1} \enspace \boxed{.} \enspace \boxed{4} \enspace \boxed{)} \enspace \boxed{=}\) math error

In this case we get an error when we try to do the calculation. This is because \(\dfrac{3}{\text{1,4}}\) is greater than 1 and the maximum value of the cosine function is 1. Therefore there is no solution. It is important in this case to write no solution and not math error.

There is no solution.

Determine the angle (correct to \(\text{1}\) decimal place):

\(\tan \theta = \text{1,7}\)

\(\sin \theta = \text{0,8}\)

\(\cos \alpha = \text{0,32}\)

\(\tan \beta = \text{4,2}\)

\(\tan \theta = 5\frac{3}{4}\)

\(\sin \theta = \frac{2}{3}\)

\(\cos \beta = \text{1,2}\)

\(4\cos \theta = 3\)

\(\cos 4\theta = \text{0,3}\)

\(\sin \beta + 2 = \text{2,65}\)

\(2\sin \theta + 5 = \text{0,8}\)

\(3 \tan \beta = 1\)

\(\sin 3 \alpha = \text{1,2}\)

\(\tan \frac{\theta}{3} = \sin 48°\)

\(\frac{1}{2}\cos 2\beta = \text{0,3}\)

\(2 \sin 3\theta + 1 = \text{2,6}\)

If \(x = 16°\) and \(y = 36°\), use your calculator to evaluate each of the following, correct to 3 decimal places.

\(\sin(x - y)\)

\(3\sin x\)

\(\tan x - \tan y\)

\(\cos x + \cos y\)

\(\frac{1}{3} \tan y\)

\(\text{cosec }(x - y)\)

\(2\cos x + \cos 3y\)

\(\tan(2x - 5y)\)

In each of the following find the value of \(x\) correct to two decimal places.

\(\sin x = \text{0,814}\)

\(\sin x = \tan \text{45}°\)

\(\tan 2x = \text{3,123}\)

\(\tan x = 3 \sin \text{41}°\)

\(\sin(2x + 45) = \text{0,123}\)

\(\sin(x - 10°) = \cos\text{57}°\)

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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Chapter 5: Equations and Identities

Exercises: Chapter 5 Review Problems

Suggested problems.

Exercises for Chapter 5 Review

Exercise group.

For Problems 1–4, evaluate the expressions for [latex]x = 120°,~ y = 225°,[/latex] and [latex]z = 90°{.}[/latex] Give exact values for your answers.

[latex]\sin^2 x \cos y[/latex]

[latex]\sin z - \dfrac{1}{2} \sin y[/latex]

[latex]\tan (z - x) \cos (y - z)[/latex]

[latex]\dfrac{\tan^2 x}{2 \cos y}[/latex]

For Problems 5–8, evaluate the expressions using a calculator. Are they equal?

• [latex]\sin (20° + 40°)[/latex]
• [latex]\sin 20° + \sin 40°[/latex]
• [latex]\cos^2 70° - \sin^2 70°[/latex]
• [latex]\cos (2\cdot 70°)[/latex]
• [latex]\dfrac{\sin 55°}{\cos 55°}[/latex]
• [latex]\tan 55°[/latex]
• [latex]\tan 80° - \tan 10°[/latex]
• [latex]\tan (80° - 10°)[/latex]

For Problems 9–12, simplify the expression.

[latex]3\sin x - 2\sin x \cos y + 2\sin x - \cos y[/latex]

[latex]\cos t + 3\cos 3t - 3\cos t - 2\cos 3t[/latex]

[latex]6 \tan^2 \theta + 2\tan \theta - (4\tan \theta )^2[/latex]

[latex]\sin \theta (2\cos \theta - 2) + \sin \theta (1 - \sin \theta)[/latex]

For Problems 13–16, decide whether or not the expressions are equivalent. Explain.

[latex]\cos \theta + \cos 2\theta;~~\cos 3\theta[/latex]

[latex]1 + \sin^2 x;~~(1 + \sin x)^2[/latex]

[latex]3\tan^2 t - \tan^2 t;~~2\tan^2 t[/latex]

[latex]\cos 4\theta;~~2\cos 2\theta[/latex]

For Problems 17–20, multiply or expand.

[latex](\cos \alpha + 2)(2\cos \alpha - 3)[/latex]

[latex](1 - 3\tan \beta)^2[/latex]

[latex](\tan \phi - \cos \phi)^2 = 0[/latex]

[latex](\sin \rho - 2\cos \rho)(\sin \rho + \cos \rho)[/latex]

For Problems 21–24, factor the expression.

[latex]12\sin 3x - 6\sin 2x[/latex]

[latex]2\cos^2 \beta + \cos \beta[/latex]

[latex]1 - 9\tan^2 \theta[/latex]

[latex]\sin^2 \phi - \sin \phi \tan \phi - 2\tan^2 \phi[/latex]

For Problems 25–30, reduce the fraction.

[latex]\dfrac{\cos^2 \alpha - \sin^2 \alpha}{\cos \alpha - \sin \alpha}[/latex]

[latex]\dfrac{1 - \tan^2 \theta}{1 - \tan \theta}[/latex]

[latex]\dfrac{3\cos x + 9}{2\cos x + 6}[/latex]

[latex]\dfrac{5\sin \theta - 10}{\sin^2 \theta - 4}[/latex]

[latex]\dfrac{3\tan^2 C - 12}{\tan^2 C - 4\tan C + 4}[/latex]

[latex]\dfrac{\tan^2 \beta - \tan \beta - 6}{\tan \beta - 3}[/latex]

For Problems 31–32, use a graph to solve the equation for [latex]0° \le x \lt 360°{.}[/latex] Check your solutions by substitution.

[latex]8\cos x - 3 = 2[/latex]

[latex]6\tan x - 2 = 8[/latex]

For Problems 33–40, find all solutions between [latex]0°[/latex] and [latex]360°{.}[/latex] Give exact answers.

[latex]2\cos^2 \theta + \cos \theta = 0[/latex]

[latex]\sin^2 \alpha - \sin \alpha = 0[/latex]

[latex]2\sin^2 x - \sin x - 1 = 0[/latex]

[latex]\cos^2 B + 2\cos B + 1 = 0[/latex]

[latex]\tan^2 x = \dfrac{1}{3}[/latex]

[latex]\tan^2 t - \tan t = 0[/latex]

[latex]6\cos^2 \alpha - 3\cos \alpha - 3 = 0[/latex]

[latex]2\sin^2 \theta + 4\sin \theta + 2 = 0[/latex]

For Problems 41–44, solve the equation for [latex]0° \le x \lt 360°{.}[/latex] Round your answers to two decimal places.

[latex]2 - 5\tan \theta = -6[/latex]

[latex]3 + 5\cos \theta = 4[/latex]

[latex]3\cos^2 x + 7\cos x = 0[/latex]

[latex]8 - 9\sin^2 x = 0[/latex]

A light ray passes from glass to water, with a [latex]37°[/latex] angle of incidence. What is the angle of refraction? The index of refraction from water to glass is 1.1.

A light ray passes from glass to water, with a [latex]76°[/latex] angle of incidence. What is the angle of refraction? The index of refraction from water to glass is 1.1.

For Problems 47–50, decide which of the following equations are identities. Explain your reasoning.

[latex]\cos x\tan x = \sin x[/latex]

[latex]\sin \theta = 1 - \cos \theta[/latex]

[latex]\tan \phi + \tan \phi = \tan 2\phi[/latex]

[latex]\tan^2 x = \dfrac{\sin^2 x}{1 - \sin^2 x}[/latex]

For Problems 51–54, use graphs to decide which of the following equations are identities.

[latex]\cos 2\theta = 2 \cos \theta[/latex]

[latex]\cos (x - 90°) = \sin x[/latex]

[latex]\sin 2x = 2\sin x \cos x[/latex]

[latex]\cos (\theta + 90°) = \cos \theta - 1[/latex]

For Problems 55–58, show that the equation is an identity by transforming the left side into the right side.

[latex]\dfrac{1 - \cos^2 \alpha}{\tan \alpha} = \sin \alpha \cos \alpha[/latex]

[latex]\cos^2 \beta \tan^2 \beta + \cos^2 \beta = 1[/latex]

[latex]\dfrac{\tan \theta - \sin \theta \cos \theta}{\sin \theta \cos \theta} = \sin \theta[/latex]

[latex]\tan \phi - \dfrac{\sin^2 \phi}{\tan \phi} = \tan \phi \sin^2 \phi[/latex]

For Problems 59–62, simplify, using identities as necessary.

[latex]\tan \theta + \dfrac{\cos \theta}{\sin \theta}[/latex]

[latex]\dfrac{1 - 2\cos^2 \beta}{\sin \beta \cos \beta} + \dfrac{\cos \beta}{\sin \beta}[/latex]

[latex]\dfrac{1}{1 - \sin^2 v} - \tan^2 v[/latex]

[latex]\cos u + (\sin u)(\tan u)[/latex]

For Problems 63–66, evaluate the expressions without using a calculator.

[latex]\sin 137° - \tan 137° \cdot \cos 137°[/latex]

[latex]\cos^2 8° + \cos 8° \cdot \tan 8° \cdot \sin 8°[/latex]

[latex]\dfrac{1}{\cos^2 54°} - \tan^2 54°[/latex]

[latex]\dfrac{2}{\cos^2 7°} - 2\tan^2 7°[/latex]

For Problems 67–70, use identities to rewrite each expression.

Write [latex]\tan^2 \beta + 1[/latex] in terms of [latex]\cos^2 \beta{.}[/latex]

Write [latex]2\sin^2 t + \cos t[/latex] in terms of [latex]\cos t{.}[/latex]

Write [latex]\dfrac{\cos x}{\tan x}[/latex] in terms of [latex]\sin x{.}[/latex]

For Problems 71–74, find the values of the three trigonometric functions.

[latex]7\tan \beta - 4 = 2, ~~ 180° \lt \beta \lt 270°[/latex]

[latex]3\tan C + 5 = 3, ~~-90° \lt C \lt 0°[/latex]

[latex]5\cos \alpha + 3 = 1, ~~ 90° \lt \alpha \lt 180°[/latex]

[latex]3\sin \theta + 2 = 4, ~~ 90° \lt \beta \lt 180°[/latex]

For Problems 75–82, solve the equation for [latex]0° \le x \lt 360°{.}[/latex] Round angles to three decimal places if necessary.

[latex]\sin w + 1 = \cos^2 w[/latex]

[latex]\cos^2 \phi - \cos \phi - \sin^2 \phi = 0[/latex]

[latex]\cos x + \sin x = 0[/latex]

[latex]3\sin \theta = \sqrt{3} \cos \theta[/latex]

[latex]2\sin \beta - \tan \beta = 0[/latex]

[latex]6\tan \theta \cos \theta + 6 = 0[/latex]

[latex]\cos^2 t - \sin^2 t = 1[/latex]

[latex]5\cos^2 \beta - 5\sin^2 \beta = -5[/latex]

Trigonometry Copyright © 2024 by Bimal Kunwor; Donna Densmore; Jared Eusea; and Yi Zhen. All Rights Reserved.

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Solving Simple Trigonometric Equations

A trigonometric equation is an equation that uses at least one variable inside a trigonometric function.

Here are some examples of trigonometric equations. Follow the links to see the solutions.

• Example 1:   A single trigonometric function with input $\,x\,,$ no other appearance of the variable: $$\cssId{s6}{\sqrt 2\sin x - 1 = 0}$$
• Example 2:   A single trigonometric function with input $\,kx\,,$ no other appearance of the variable: $$\cssId{s9}{2\cos(3x) = 1}$$
• Example 3 (next section):   A pseudo-quadratic (fake quadratic) trigonometric equation: $$\cssId{s12}{2\sin^2 x - \sin x - 1 = 0}$$
• Example 4:   Use graphical methods to approximate solutions to trigonometric equations:   find all solutions of the equation ‘$\,2\cos x = -3\sin x\,$’ in the interval $\,[0,2\pi]$
• Example 5:   Use WolframAlpha to solve trigonometric equations: $$\cssId{s18}{\tan(5x-3)\sec(x^2 - 7) = 4}$$

Note that an equation like ‘$\,5 - x\sin 3 = 7\,$’ is not a trigonometric equation, because it does not have a variable inside a trigonometric function . Here, $\,\sin 3\,$ is just a constant. This is a linear equation in one variable with unique solution $\,x = \frac{-2}{\sin 3}\,.$

A Single Trigonometric Function With Input $\,x\,,$ No Other Appearance of the Variable

The equation   ‘$\,\sqrt 2\sin x - 1 = 0\,$’   is a trigonometric equation because of the ‘$\,x\,$’ inside the sine function. Note that there is no other appearance of $\,x\,$ in this equation.

In cases such as this, we start by isolating the trigonometric function containing the variable. (Recall that isolate means to get all by itself on one side of the equation .)

Isolate the trigonometric function with variable input:

• Add $\,1\,$ to both sides (using the addition property of equality )
• Divide both sides by $\,\sqrt 2\,$ (using the multiplication property of equality)

At this point, several ‘checks’ should run through your head:

Is the equation ever true? Can an output from the sine function ever equal $\,\frac{1}{\sqrt 2}\,$?

The range of the sine function is $\,[-1,1]\,.$ Since $\,\sqrt 2 \approx 1.4 \gt 1\,,$ we have $\,\frac{1}{\sqrt 2} \approx 0.7 \lt 1 \,.$ So, the number $\,\frac{1}{\sqrt 2}\,$ is in the range of the sine function. Thus, there is at least one value for which the equation is true.

How many solutions does the equation have?

A quick sketch (below) shows that there are infinitely many solutions. This is typical of trigonometric equations—they often have infinitely many solutions, because of the periodic nature of trigonometric functions.

Punchline: All the Solutions for Example 1

Reporting the solutions in degrees:

• $x = 45^\circ + {360k\,}^\circ\,$ for all integers $\,k\,,$ or
• $x = 135^\circ + {360k\,}^\circ\,$ for all integers $\,k$

Reporting the solutions in radians:

• $\displaystyle x = \frac{\pi}{4} + 2\pi k\,$ for all integers $\,k\,,$ or
• $\displaystyle x = \frac{3\pi}{4} + 2\pi k\,$ for all integers $\,k$

Here's Where These Solutions Come From

Precisely what angles have sine equal to $\,\frac{1}{\sqrt 2}\,$? From the special triangle below, we see that $\,\sin 45^\circ = \frac{1}{\sqrt 2}\,$; so, $\,45^\circ\,$ (or $\,\frac{\pi}{4}\,$ radians) is one of the solutions of the equation.

Recall that the sine function gives the $y$-values of points on the unit circle. From the unit circle above, we can now get all the angles with $y$-value equal to $\,\frac{1}{\sqrt 2}\,,$ as follows:

Recall that, when laying off angles:

Counterclockwise is the positive direction and clockwise is the negative direction

Also recall that the integers are:

Start at the terminal point for $\,45^\circ\,,$ in the first quadrant:

• $45^\circ+ 360^\circ$
• $45^\circ+ 2\cdot360^\circ$
• $45^\circ+ 3\cdot360^\circ\,,$ and so on
• $45^\circ - 360^\circ$
• $45^\circ - 2\cdot360^\circ$
• $45^\circ - 3\cdot360^\circ\,,$ and so on

See the pattern? Together, these give all the solutions in the first quadrant :

Next, start at the terminal point for $\,180^\circ - 45^\circ = 135^\circ\,,$ in the second quadrant:

• $135^\circ+ 360^\circ$
• $135^\circ+ 2\cdot360^\circ$
• $135^\circ+ 3\cdot360^\circ\,,$ and so on
• $135^\circ - 360^\circ$
• $135^\circ - 2\cdot360^\circ$
• $135^\circ - 3\cdot360^\circ\,,$ and so on

See the same pattern? Together, these give all the solutions in the second quadrant :

A Single Trigonometric Function With Input $\,kx\,,$ No Other Appearance of the Variable

The equation ‘$\,2\cos(3x) = 1\,$’ is a trigonometric equation because of the ‘$\,x\,$’ inside the cosine function. Note that there is no other appearance of $\,x\,$ in this equation.

Again, we start by isolating the trigonometric function containing the variable.

Isolate the trigonometric function with variable input. The cosine function takes on the value $\,\frac 12\,$ infinitely many times, so again there are infinitely many solutions.

Punchline: All the ‘Pre’-Solutions for Example 2

To get $\,x\,$ in degrees:

• $3x = 60^\circ + {360k\,}^\circ\,$ for all integers $\,k\,,$ or
• $3x = -60^\circ + {360k\,}^\circ\,$ for all integers $\,k$

To get $\,x\,$ in radians:

• $\displaystyle 3x = \frac{\pi}{3} + 2\pi k\,$ for all integers $\,k\,,$ or
• $\displaystyle 3x = -\frac{\pi}{3} + 2\pi k\,$ for all integers $\,k$

Here's Where These ‘Pre’-Solutions Come From

For the moment, ignore the ‘$\,3x\,$’ inside the cosine function. (We'll use it in a minute.)

Ask:   What angles have cosine equal to $\,\frac{1}{2}\,$?

From the special triangle below, we see that $\,\cos 60^\circ = \frac{1}{2}\,.$

Recall that the cosine function gives the $x$-values of points on the unit circle.

As in Example 1 , get all the angles with $x$-value equal to $\,\frac{1}{2}\,$:

$60^\circ + {360k\,}^\circ\,$ for all integers $\,k\,$ (in quadrant I)   (*)

$-60^\circ + {360k\,}^\circ\,$ for all integers $\,k\,$ (in quadrant IV)   (**)

These are all possible angles with cosine equal to $\,\frac 12\,.$

Now , we'll use the ‘$\,3x\,$’.

Note that if $\,x\,$ has units of degrees, then $\,3x\,$ also has units of degrees.

The angle inside the cosine function (which in this example is $\,3x\,$) must equal one of the angles in (*) or (**), in order to have cosine equal to $\,\frac 12\,.$

Thus, for all integers $\,k\,,$ and for $\,x\,$ in degrees:

Solve for $\,x\,$ (by dividing both sides of the equations by $\,3\,$):

Degree measure:

$\displaystyle x = \frac{60^\circ + {360k\ }^\circ}{3} = 20^\circ + {120k\,}^\circ\,$ for all integers $\,k$

$\displaystyle x = \frac{-60^\circ + {360k\ }^\circ}{3} = -20^\circ + {120k\,}^\circ\,$ for all integers $\,k$

Radian measure:

$\displaystyle x = \frac{\frac\pi 3 + 2\pi k}{3} = \frac{\pi}{9} + \frac{2\pi k}{3}\,$ for all integers $\,k$

$\displaystyle x = \frac{-\frac\pi 3 + 2\pi k}{3} = -\frac{\pi}{9} + \frac{2\pi k}{3}\,$ for all integers $\,k\,$

  Done!

Example 3 (next section)

Use graphical methods to approximate solutions to trigonometric equations.

Graphical methods for solving equations (and inequalities) were studied thoroughly in earlier sections:

• Graphical Interpretations of Sentences like $\,f(x) = 0\,$ and $\,f(x) \gt 0$
• Graphical Interpretations of Sentences like $\,f(x) = g(x)\,$ and $\,f(x) \gt g(x)$

Even if you're working with a simple equation (as in Examples 1 and 2 ), graphical methods are great for giving yourself confidence in your answers and/or catching mistakes. Graphical methods typically provide approximate solutions, not exact solutions.

The equation ‘$\,2\cos x = -3\sin x\,$’ is a trigonometric equation because there is at least one variable inside a trigonometric function.

In this equation, there are two trigonometric functions with variable inputs. Isolating either one would still cause a variable expression (not a constant) to appear on the other side of the equation. Thus, the technique illustrated in Example 1 and Example 2 doesn't work.

Both sides of the equation ‘$\,2\cos x = -3\sin x\,$’ are easy to graph—it is well-suited to graphical methods. Graph the left-hand side; graph the right-hand side; approximate the intersection point(s) in a desired interval.

original equation; find solutions in the interval $\,[0,2\pi]$

Punchline: Solutions for Example 4

$x\approx 2.6\,$  or  $\,x\approx 5.7$

Graph, on the interval $\,[0,2\pi]\,$:

• The left-hand side of the equation (in red): cosine curve, vertical stretch by a factor of $\,2$
• The right-hand side of the equation (in black): sine curve, vertical stretch by a factor of $\,3\,,$ reflect about the $x$-axis

It is clear there are exactly two solutions in the interval $\,[0,2\pi]\,$; these correspond to the intersection points of the two graphs:

• $x_1\,$ is a bit less than $\,\pi\,$:   $\color{red}{2\cos x_1} \approx -3\sin x_1$
• $x_2\,$ is a bit less than $\,2\pi\,$:   $\color{red}{2\cos x_2} \approx -3\sin x_2$

A graphing calculator or WolframAlpha (see Example 5 ) can be used to get decimal approximations for $\,x_1\,$ and $\,x_2\,.$

Use WolframAlpha to Solve Trigonometric Equations

WolframAlpha (opens in a new window) is a powerful, reliable, and free resource for mathematical information. WolframAlpha understands conventional math syntax and abbreviations. You can also type in a reasonable guess, and WolframAlpha will try to make sense of it.

For example, if you type in  ‘ sine of x ’  and press Enter, the  ‘ Input ’  area will display ‘$\,\sin(x)\,$’. This lets you know that WolframAlpha has understood you correctly. Always check the  ‘ Input ’  area, to make sure that you and WolframAlpha are on the same page!

Try solving each of the examples from this page (or the exercises) at WolframAlpha . WolframAlpha is a good resource for checking answers, gaining additional insight, or when you just need a quick reliable answer.

You can cut-and-paste each of the following. Have fun!

tan(5x-3)sec(x^2 - 7) = 4 sin(x) = 1/sqrt(2) 2cos(3x) = 1 2sin^2 x - sin x - 1 = 0 2cos(x) = -3sin(x), 0 <= x <= 2pi

WolframAlpha gives exact solutions when possible; click ‘Approximate forms’ if desired.

Concept Practice

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