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

Module by: First Last. E-mail the author

Summary: In this section, you will:

• Use double-angle formulas to find exact values.
• Use double-angle formulas to verify identities.
• Use reduction formulas to simplify an expression.
• Use half-angle formulas to find exact values.

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Bicycle ramps made for competition (see Figure 1) must vary in height depending on the skill level of the competitors. For advanced competitors, the angle formed by the ramp and the ground should be θ θ such that tanθ= 5 3 . tanθ= 5 3 . The angle is divided in half for novices. What is the steepness of the ramp for novices? In this section, we will investigate three additional categories of identities that we can use to answer questions such as this one.

## Using Double-Angle Formulas to Find Exact Values

In the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The double-angle formulas are a special case of the sum formulas, where α=β. α=β. Deriving the double-angle formula for sine begins with the sum formula,

sin( α+β )=sinαcosβ+cosαsinβ sin( α+β )=sinαcosβ+cosαsinβ

If we let α=β=θ, α=β=θ, then we have

sin( θ+θ )=sinθcosθ+cosθsinθ     sin( 2θ )=2sinθcosθ sin( θ+θ )=sinθcosθ+cosθsinθ     sin( 2θ )=2sinθcosθ

Deriving the double-angle for cosine gives us three options. First, starting from the sum formula, cos( α+β )=cosαcosβsinαsinβ, cos( α+β )=cosαcosβsinαsinβ, and letting α=β=θ, α=β=θ, we have

cos(θ+θ)=cosθcosθsinθsinθ     cos(2θ)= cos 2 θ sin 2 θ cos(θ+θ)=cosθcosθsinθsinθ     cos(2θ)= cos 2 θ sin 2 θ

Using the Pythagorean properties, we can expand this double-angle formula for cosine and get two more interpretations. The first one is:

cos(2θ)= cos 2 θ sin 2 θ             =(1 sin 2 θ) sin 2 θ             =12 sin 2 θ cos(2θ)= cos 2 θ sin 2 θ             =(1 sin 2 θ) sin 2 θ             =12 sin 2 θ

The second interpretation is:

cos(2θ)= cos 2 θ sin 2 θ             = cos 2 θ(1 cos 2 θ)             =2 cos 2 θ1 cos(2θ)= cos 2 θ sin 2 θ             = cos 2 θ(1 cos 2 θ)             =2 cos 2 θ1

Similarly, to derive the double-angle formula for tangent, replacing α=β=θ α=β=θ in the sum formula gives

tan( α+β )= tanα+tanβ 1tanαtanβ tan( θ+θ )= tanθ+tanθ 1tanθtanθ tan( 2θ )= 2tanθ 1 tan 2 θ tan( α+β )= tanα+tanβ 1tanαtanβ tan( θ+θ )= tanθ+tanθ 1tanθtanθ tan( 2θ )= 2tanθ 1 tan 2 θ

### A General Note: Double-Angle Formulas:

The double-angle formulas are summarized as follows:

sin( 2θ )=2sinθcosθ sin( 2θ )=2sinθcosθ
(7)

cos(2θ)= cos 2 θ sin 2 θ             =12 sin 2 θ             =2 cos 2 θ1 cos(2θ)= cos 2 θ sin 2 θ             =12 sin 2 θ             =2 cos 2 θ1
(8)

tan( 2θ )= 2tanθ 1 tan 2 θ tan( 2θ )= 2tanθ 1 tan 2 θ
(9)

### How To:

Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value.

1. Draw a triangle to reflect the given information.
2. Determine the correct double-angle formula.
3. Substitute values into the formula based on the triangle.
4. Simplify.

### Example 1

#### Problem 1

##### Using a Double-Angle Formula to Find the Exact Value Involving Tangent

Given that tanθ= 3 4 tanθ= 3 4 and θ θ is in quadrant II, find the following:

1. sin( 2θ ) sin( 2θ )
2. cos( 2θ ) cos( 2θ )
3. tan( 2θ ) tan( 2θ )
##### Solution

If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given tanθ= 3 4 , tanθ= 3 4 , such that θ θ is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because θ θ is in the second quadrant, the adjacent side is on the x-axis and is negative. Use the Pythagorean Theorem to find the length of the hypotenuse:

(−4) 2 + (3) 2 = c 2 16+9= c 2 25= c 2 c=5  (−4) 2 + (3) 2 = c 2 16+9= c 2 25= c 2 c=5

Now we can draw a triangle similar to the one shown in Figure 2.

1. Let’s begin by writing the double-angle formula for sine.
sin(2θ)=2sinθcosθ sin(2θ)=2sinθcosθ

We see that we to need to find sinθ sinθ and cosθ. cosθ. Based on Figure 2, we see that the hypotenuse equals 5, so sinθ= 3 5 , sinθ= 3 5 , and cosθ= 4 5 . cosθ= 4 5 . Substitute these values into the equation, and simplify.

Thus,

sin(2θ)=2( 3 5 )( 4 5 )             = 24 25 sin(2θ)=2( 3 5 )( 4 5 )             = 24 25
2. Write the double-angle formula for cosine.
cos( 2θ )= cos 2 θ sin 2 θ cos( 2θ )= cos 2 θ sin 2 θ

Again, substitute the values of the sine and cosine into the equation, and simplify.

cos(2θ)= ( 4 5 ) 2 ( 3 5 ) 2             = 16 25 9 25             = 7 25 cos(2θ)= ( 4 5 ) 2 ( 3 5 ) 2             = 16 25 9 25             = 7 25
3. Write the double-angle formula for tangent.
tan(2θ)= 2tanθ 1 tan 2 θ tan(2θ)= 2tanθ 1 tan 2 θ

In this formula, we need the tangent, which we were given as tanθ= 3 4 . tanθ= 3 4 . Substitute this value into the equation, and simplify.

tan(2θ)= 2( 3 4 ) 1 ( 3 4 ) 2            = 3 2 1 9 16            = 3 2 ( 16 7 )            = 24 7 tan(2θ)= 2( 3 4 ) 1 ( 3 4 ) 2            = 3 2 1 9 16            = 3 2 ( 16 7 )            = 24 7

### Try It:

#### Exercise 1

Given sinα= 5 8 , sinα= 5 8 , with θ θ in quadrant I, find cos( 2α ). cos( 2α ).

##### Solution

cos( 2α )= 7 32 cos( 2α )= 7 32

### Example 2

#### Problem 1

##### Using the Double-Angle Formula for Cosine without Exact Values

Use the double-angle formula for cosine to write cos( 6x ) cos( 6x ) in terms of cos( 3x ). cos( 3x ).

##### Solution
cos(6x)=cos(3x+3x)             =cos3xcos3xsin3xsin3x             = cos 2 3x sin 2 3x cos(6x)=cos(3x+3x)             =cos3xcos3xsin3xsin3x             = cos 2 3x sin 2 3x
##### Analysis

This example illustrates that we can use the double-angle formula without having exact values. It emphasizes that the pattern is what we need to remember and that identities are true for all values in the domain of the trigonometric function.

## Using Double-Angle Formulas to Verify Identities

Establishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side.

### Example 3

#### Problem 1

##### Using the Double-Angle Formulas to Establish an Identity

Establish the following identity using double-angle formulas:

1+sin( 2θ )= ( sinθ+cosθ ) 2 1+sin( 2θ )= ( sinθ+cosθ ) 2
##### Solution

We will work on the right side of the equal sign and rewrite the expression until it matches the left side.

(sinθ+cosθ) 2 = sin 2 θ+2sinθcosθ+ cos 2 θ                        =( sin 2 θ+ cos 2 θ)+2sinθcosθ                        =1+2sinθcosθ                        =1+sin(2θ) (sinθ+cosθ) 2 = sin 2 θ+2sinθcosθ+ cos 2 θ                        =( sin 2 θ+ cos 2 θ)+2sinθcosθ                        =1+2sinθcosθ                        =1+sin(2θ)
##### Analysis

This process is not complicated, as long as we recall the perfect square formula from algebra:

( a±b ) 2 = a 2 ±2ab+ b 2 ( a±b ) 2 = a 2 ±2ab+ b 2

where a=sinθ a=sinθ and b=cosθ. b=cosθ. Part of being successful in mathematics is the ability to recognize patterns. While the terms or symbols may change, the algebra remains consistent.

### Try It:

#### Exercise 2

Establish the identity: cos 4 θ sin 4 θ=cos( 2θ ). cos 4 θ sin 4 θ=cos( 2θ ).

##### Solution

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

### Example 4

#### Problem 1

##### Verifying a Double-Angle Identity for Tangent

Verify the identity:

tan( 2θ )= 2 cotθtanθ tan( 2θ )= 2 cotθtanθ
##### Solution

In this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation.

tan( 2θ )= 2tanθ 1 tan 2 θ Double-angle formula            = 2tanθ( 1 tanθ ) ( 1 tan 2 θ )( 1 tanθ ) Multiply by a term that results in desired numerator.            = 2 1 tanθ tan 2 θ tanθ            = 2 cotθtanθ Use reciprocal identity for  1 tanθ . tan( 2θ )= 2tanθ 1 tan 2 θ Double-angle formula            = 2tanθ( 1 tanθ ) ( 1 tan 2 θ )( 1 tanθ ) Multiply by a term that results in desired numerator.            = 2 1 tanθ tan 2 θ tanθ            = 2 cotθtanθ Use reciprocal identity for  1 tanθ .
##### Analysis

Here is a case where the more complicated side of the initial equation appeared on the right, but we chose to work the left side. However, if we had chosen the left side to rewrite, we would have been working backwards to arrive at the equivalency. For example, suppose that we wanted to show

2tanθ 1 tan 2 θ = 2 cotθtanθ 2tanθ 1 tan 2 θ = 2 cotθtanθ

Let’s work on the right side.

2 cotθtanθ = 2 1 tanθ tanθ ( tanθ tanθ )                    = 2tanθ 1 tanθ ( tanθ )tanθ(tanθ)                    = 2tanθ 1 tan 2 θ 2 cotθtanθ = 2 1 tanθ tanθ ( tanθ tanθ )                    = 2tanθ 1 tanθ ( tanθ )tanθ(tanθ)                    = 2tanθ 1 tan 2 θ

When using the identities to simplify a trigonometric expression or solve a trigonometric equation, there are usually several paths to a desired result. There is no set rule as to what side should be manipulated. However, we should begin with the guidelines set forth earlier.

### Try It:

#### Exercise 3

Verify the identity: cos(2θ)cosθ= cos 3 θcosθ sin 2 θ. cos(2θ)cosθ= cos 3 θcosθ sin 2 θ.

##### Solution

cos( 2θ )cosθ=( cos 2 θ sin 2 θ )cosθ= cos 3 θcosθ sin 2 θ cos( 2θ )cosθ=( cos 2 θ sin 2 θ )cosθ= cos 3 θcosθ sin 2 θ

## Use Reduction Formulas to Simplify an Expression

The double-angle formulas can be used to derive the reduction formulas, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the double-angle formulas.

We can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let’s begin with cos( 2θ )=12 sin 2 θ. cos( 2θ )=12 sin 2 θ. Solve for sin 2 θ: sin 2 θ:

cos(2θ)=12 sin 2 θ  2 sin 2 θ=1cos(2θ)     sin 2 θ= 1cos(2θ) 2 cos(2θ)=12 sin 2 θ  2 sin 2 θ=1cos(2θ)     sin 2 θ= 1cos(2θ) 2

Next, we use the formula cos( 2θ )=2 cos 2 θ1. cos( 2θ )=2 cos 2 θ1. Solve for cos 2 θ: cos 2 θ:

cos(2θ)=2 cos 2 θ1  1+cos(2θ)=2 cos 2 θ 1+cos(2θ) 2 = cos 2 θ        cos(2θ)=2 cos 2 θ1  1+cos(2θ)=2 cos 2 θ 1+cos(2θ) 2 = cos 2 θ

The last reduction formula is derived by writing tangent in terms of sine and cosine:

tan 2 θ= sin 2 θ cos 2 θ          = 1cos(2θ) 2 1+cos(2θ) 2 Substitute the reduction formulas.          =( 1cos(2θ) 2 )( 2 1+cos(2θ) )          = 1cos(2θ) 1+cos(2θ) tan 2 θ= sin 2 θ cos 2 θ          = 1cos(2θ) 2 1+cos(2θ) 2 Substitute the reduction formulas.          =( 1cos(2θ) 2 )( 2 1+cos(2θ) )          = 1cos(2θ) 1+cos(2θ)

### A General Note: Reduction Formulas:

The reduction formulas are summarized as follows:

sin 2 θ= 1cos( 2θ ) 2 sin 2 θ= 1cos( 2θ ) 2
(28)
cos 2 θ= 1+cos( 2θ ) 2 cos 2 θ= 1+cos( 2θ ) 2
(29)
tan 2 θ= 1cos( 2θ ) 1+cos( 2θ ) tan 2 θ= 1cos( 2θ ) 1+cos( 2θ )
(30)

### Example 5

#### Problem 1

##### Writing an Equivalent Expression Not Containing Powers Greater Than 1

Write an equivalent expression for cos 4 x cos 4 x that does not involve any powers of sine or cosine greater than 1.

##### Solution

We will apply the reduction formula for cosine twice.

cos 4 x= ( cos 2 x) 2           = ( 1+cos(2x) 2 ) 2 Substitute reduction formula for cos 2 x.           = 1 4 ( 1+2cos(2x)+ cos 2 (2x) )           = 1 4 + 1 2 cos(2x)+ 1 4 ( 1+cos2(2x) 2 )  Substitute reduction formula for cos 2 x.           = 1 4 + 1 2 cos(2x)+ 1 8 + 1 8 cos(4x)           = 3 8 + 1 2 cos(2x)+ 1 8 cos(4x) cos 4 x= ( cos 2 x) 2           = ( 1+cos(2x) 2 ) 2 Substitute reduction formula for cos 2 x.           = 1 4 ( 1+2cos(2x)+ cos 2 (2x) )           = 1 4 + 1 2 cos(2x)+ 1 4 ( 1+cos2(2x) 2 )  Substitute reduction formula for cos 2 x.           = 1 4 + 1 2 cos(2x)+ 1 8 + 1 8 cos(4x)           = 3 8 + 1 2 cos(2x)+ 1 8 cos(4x)
##### Analysis

The solution is found by using the reduction formula twice, as noted, and the perfect square formula from algebra.

### Example 6

#### Problem 1

##### Using the Power-Reducing Formulas to Prove an Identity

Use the power-reducing formulas to prove

sin 3 ( 2x )=[ 1 2 sin( 2x ) ][ 1cos( 4x ) ] sin 3 ( 2x )=[ 1 2 sin( 2x ) ][ 1cos( 4x ) ]
##### Solution

We will work on simplifying the left side of the equation:

sin 3 (2x)=[sin(2x)][ sin 2 (2x)]              =sin(2x)[ 1cos(4x) 2 ] Substitute the power-reduction formula.              =sin(2x)( 1 2 )[ 1cos(4x) ]              = 1 2 [sin(2x)][1cos(4x)] sin 3 (2x)=[sin(2x)][ sin 2 (2x)]              =sin(2x)[ 1cos(4x) 2 ] Substitute the power-reduction formula.              =sin(2x)( 1 2 )[ 1cos(4x) ]              = 1 2 [sin(2x)][1cos(4x)]
##### Analysis

Note that in this example, we substituted

1cos( 4x ) 2 1cos( 4x ) 2

for sin 2 ( 2x ). sin 2 ( 2x ). The formula states

sin 2 θ= 1cos( 2θ ) 2 sin 2 θ= 1cos( 2θ ) 2

We let θ=2x, θ=2x, so 2θ=4x. 2θ=4x.

### Try It:

#### Exercise 4

Use the power-reducing formulas to prove that 10 cos 4 x= 15 4 +5cos( 2x )+ 5 4 cos( 4x ). 10 cos 4 x= 15 4 +5cos( 2x )+ 5 4 cos( 4x ).

##### Solution

10 cos 4 x=10 cos 4 x=10 ( cos 2 x) 2             =10 [ 1+cos(2x) 2 ] 2 Substitute reduction formula for cos 2 x.             = 10 4 [1+2cos(2x)+ cos 2 (2x)]             = 10 4 + 10 2 cos(2x)+ 10 4 ( 1+cos2(2x) 2 ) Substitute reduction formula for cos 2 x.             = 10 4 + 10 2 cos(2x)+ 10 8 + 10 8 cos(4x)             = 30 8 +5cos(2x)+ 10 8 cos(4x)             = 15 4 +5cos(2x)+ 5 4 cos(4x) 10 cos 4 x=10 cos 4 x=10 ( cos 2 x) 2             =10 [ 1+cos(2x) 2 ] 2 Substitute reduction formula for cos 2 x.             = 10 4 [1+2cos(2x)+ cos 2 (2x)]             = 10 4 + 10 2 cos(2x)+ 10 4 ( 1+cos2(2x) 2 ) Substitute reduction formula for cos 2 x.             = 10 4 + 10 2 cos(2x)+ 10 8 + 10 8 cos(4x)             = 30 8 +5cos(2x)+ 10 8 cos(4x)             = 15 4 +5cos(2x)+ 5 4 cos(4x)

## Using Half-Angle Formulas to Find Exact Values

The next set of identities is the set of half-angle formulas, which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace θ θ with α 2 , α 2 , the half-angle formula for sine is found by simplifying the equation and solving for sin( α 2 ). sin( α 2 ). Note that the half-angle formulas are preceded by a ± ± sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which α 2 α 2 terminates.

The half-angle formula for sine is derived as follows:

sin 2 θ= 1cos(2θ) 2 sin 2 ( α 2 )= 1( cos2 α 2 ) 2            = 1cosα 2  sin( α 2 )=± 1cosα 2     sin 2 θ= 1cos(2θ) 2 sin 2 ( α 2 )= 1( cos2 α 2 ) 2            = 1cosα 2  sin( α 2 )=± 1cosα 2

To derive the half-angle formula for cosine, we have

cos 2 θ= 1+cos(2θ) 2 cos 2 ( α 2 )= 1+cos( 2 α 2 ) 2              = 1+cosα 2   cos( α 2 )=± 1+cosα 2     cos 2 θ= 1+cos(2θ) 2 cos 2 ( α 2 )= 1+cos( 2 α 2 ) 2              = 1+cosα 2   cos( α 2 )=± 1+cosα 2

For the tangent identity, we have

tan 2 θ= 1cos(2θ) 1+cos(2θ) tan 2 ( α 2 )= 1cos( 2 α 2 ) 1+cos( 2 α 2 )             = 1cosα 1+cosα   tan( α 2 )=± 1cosα 1+cosα     tan 2 θ= 1cos(2θ) 1+cos(2θ) tan 2 ( α 2 )= 1cos( 2 α 2 ) 1+cos( 2 α 2 )             = 1cosα 1+cosα   tan( α 2 )=± 1cosα 1+cosα

### A General Note: Half-Angle Formulas:

The half-angle formulas are as follows:

sin( α 2 )=± 1cosα 2 sin( α 2 )=± 1cosα 2
(39)
cos( α 2 )=± 1+cosα 2 cos( α 2 )=± 1+cosα 2
(40)
tan( α 2 )=± 1cosα 1+cosα = sinα 1+cosα = 1cosα sinα tan( α 2 )=± 1cosα 1+cosα = sinα 1+cosα = 1cosα sinα
(41)

### Example 7

#### Problem 1

##### Using a Half-Angle Formula to Find the Exact Value of a Sine Function

Find sin( 15 ) sin( 15 ) using a half-angle formula.

##### Solution

Since 15 = 30 2 , 15 = 30 2 , we use the half-angle formula for sine:

sin 30 2 = 1cos 30 2            = 1 3 2 2            = 2 3 2 2            = 2 3 4            = 2 3 2 sin 30 2 = 1cos 30 2            = 1 3 2 2            = 2 3 2 2            = 2 3 4            = 2 3 2
##### Analysis

Notice that we used only the positive root because sin( 15 o ) sin( 15 o ) is positive.

### How To:

Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle.

1. Draw a triangle to represent the given information.
2. Determine the correct half-angle formula.
3. Substitute values into the formula based on the triangle.
4. Simplify.

### Example 8

#### Problem 1

##### Finding Exact Values Using Half-Angle Identities

Given that tanα= 8 15 tanα= 8 15 and α α lies in quadrant III, find the exact value of the following:

1. sin( α 2 ) sin( α 2 )
2. cos( α 2 ) cos( α 2 )
3. tan( α 2 ) tan( α 2 )
##### Solution

Using the given information, we can draw the triangle shown in Figure 3. Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate sinα= 8 17 sinα= 8 17 and cosα= 15 17 . cosα= 15 17 .

1. Before we start, we must remember that, if α α is in quadrant III, then 180°<α<270°, 180°<α<270°, so 180° 2 < α 2 < 270° 2 . 180° 2 < α 2 < 270° 2 . This means that the terminal side of α 2 α 2 is in quadrant II, since 90°< α 2 <135°. 90°< α 2 <135°.

To find sin α 2 , sin α 2 , we begin by writing the half-angle formula for sine. Then we substitute the value of the cosine we found from the triangle in Figure 3 and simplify.

sin α 2 =± 1cosα 2         =± 1( 15 17 ) 2         =± 32 17 2         =± 32 17 1 2         =± 16 17         =± 4 17         = 4 17 17 sin α 2 =± 1cosα 2         =± 1( 15 17 ) 2         =± 32 17 2         =± 32 17 1 2         =± 16 17         =± 4 17         = 4 17 17

We choose the positive value of sin α 2 sin α 2 because the angle terminates in quadrant II and sine is positive in quadrant II.

2. To find cos α 2 , cos α 2 , we will write the half-angle formula for cosine, substitute the value of the cosine we found from the triangle in Figure 3, and simplify.
cos α 2 =± 1+cosα 2         =± 1+( 15 17 ) 2         =± 2 17 2         =± 2 17 1 2         =± 1 17         = 17 17 cos α 2 =± 1+cosα 2         =± 1+( 15 17 ) 2         =± 2 17 2         =± 2 17 1 2         =± 1 17         = 17 17

We choose the negative value of cos α 2 cos α 2 because the angle is in quadrant II because cosine is negative in quadrant II.

3. To find tan α 2 , tan α 2 , we write the half-angle formula for tangent. Again, we substitute the value of the cosine we found from the triangle in Figure 3 and simplify.
tan α 2 =± 1cosα 1+cosα         =± 1( 15 17 ) 1+( 15 17 )         =± 32 17 2 17         =± 32 2         = 16         =4 tan α 2 =± 1cosα 1+cosα         =± 1( 15 17 ) 1+( 15 17 )         =± 32 17 2 17         =± 32 2         = 16         =4

We choose the negative value of tan α 2 tan α 2 because α 2 α 2 lies in quadrant II, and tangent is negative in quadrant II.

### Try It:

#### Exercise 5

Given that sinα= 4 5 sinα= 4 5 and α α lies in quadrant IV, find the exact value of cos( α 2 ). cos( α 2 ).

2 5 2 5

### Example 9

#### Problem 1

##### Finding the Measurement of a Half Angle

Now, we will return to the problem posed at the beginning of the section. A bicycle ramp is constructed for high-level competition with an angle of θ θ formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If tanθ= 5 3 tanθ= 5 3 for higher-level competition, what is the measurement of the angle for novice competition?

##### Solution

Since the angle for novice competition measures half the steepness of the angle for the high level competition, and tanθ= 5 3 tanθ= 5 3 for high competition, we can find cosθ cosθ from the right triangle and the Pythagorean theorem so that we can use the half-angle identities. See Figure 4.

3 2 + 5 2 =34           c= 34 3 2 + 5 2 =34           c= 34

We see that cosθ= 3 34 = 3 34 34 . cosθ= 3 34 = 3 34 34 . We can use the half-angle formula for tangent: tan θ 2 = 1cosθ 1+cosθ . tan θ 2 = 1cosθ 1+cosθ . Since tanθ tanθ is in the first quadrant, so is tan θ 2 . tan θ 2 . Thus,

tan θ 2 = 1 3 34 34 1+ 3 34 34         = 343 34 34 34+3 34 34         = 343 34 34+3 34         0.57 tan θ 2 = 1 3 34 34 1+ 3 34 34         = 343 34 34 34+3 34 34         = 343 34 34+3 34         0.57

We can take the inverse tangent to find the angle: tan 1 ( 0.57 ) 29.7 . tan 1 ( 0.57 ) 29.7 . So the angle of the ramp for novice competition is 29.7 . 29.7 .

### Media:

Access these online resources for additional instruction and practice with double-angle, half-angle, and reduction formulas.

## Key Equations

 Double-angle formulas sin(2θ)=2sin θ cos θ cos(2θ)= cos 2 θ− sin 2 θ            =1−2 sin 2 θ            =2 cos 2 θ−1 tan(2θ)= 2tan θ 1− tan 2 θ sin(2θ)=2sin θ cos θ cos(2θ)= cos 2 θ− sin 2 θ            =1−2 sin 2 θ            =2 cos 2 θ−1 tan(2θ)= 2tan θ 1− tan 2 θ Reduction formulas sin 2 θ= 1−cos( 2θ ) 2 cos 2 θ= 1+cos( 2θ ) 2 tan 2 θ= 1−cos( 2θ ) 1+cos( 2θ ) sin 2 θ= 1−cos( 2θ ) 2 cos 2 θ= 1+cos( 2θ ) 2 tan 2 θ= 1−cos( 2θ ) 1+cos( 2θ ) Half-angle formulas sin  α 2 =± 1−cos α 2 cos  α 2 =± 1+cos α 2 tan  α 2 =± 1−cos α 1+cos α         = sin α 1+cos α         = 1−cos α sin α sin  α 2 =± 1−cos α 2 cos  α 2 =± 1+cos α 2 tan  α 2 =± 1−cos α 1+cos α         = sin α 1+cos α         = 1−cos α sin α

## Key Concepts

• Double-angle identities are derived from the sum formulas of the fundamental trigonometric functions: sine, cosine, and tangent. See Example 1, Example 2, Example 3, and Example 4.
• Reduction formulas are especially useful in calculus, as they allow us to reduce the power of the trigonometric term. See Example 5 and Example 6.
• Half-angle formulas allow us to find the value of trigonometric functions involving half-angles, whether the original angle is known or not. See Example 7, Example 8, and Example 9.

## Section Exercises

### Verbal

#### Exercise 6

Explain how to determine the reduction identities from the double-angle identity cos( 2x )= cos 2 x sin 2 x. cos( 2x )= cos 2 x sin 2 x.

##### Solution

Use the Pythagorean identities and isolate the squared term.

#### Exercise 7

Explain how to determine the double-angle formula for tan(2x) tan(2x) using the double-angle formulas for cos(2x) cos(2x) and sin(2x). sin(2x).

#### Exercise 8

We can determine the half-angle formula for tan( x 2 )= 1cosx 1+cosx tan( x 2 )= 1cosx 1+cosx by dividing the formula for sin( x 2 ) sin( x 2 ) by cos( x 2 ). cos( x 2 ). Explain how to determine two formulas for tan( x 2 ) tan( x 2 ) that do not involve any square roots.

##### Solution

1cosx sinx , sinx 1+cosx , 1cosx sinx , sinx 1+cosx , multiplying the top and bottom by 1cosx 1cosx and 1+cosx , 1+cosx , respectively.

#### Exercise 9

For the half-angle formula given in the previous exercise for tan( x 2 ), tan( x 2 ), explain why dividing by 0 is not a concern. (Hint: examine the values of cosx cosx necessary for the denominator to be 0.)

### Algebraic

For the following exercises, find the exact values of a) sin( 2x ), sin( 2x ), b) cos( 2x ), cos( 2x ), and c) tan( 2x ) tan( 2x ) without solving for x. x.

#### Exercise 10

If sinx= 1 8 , sinx= 1 8 , and x x is in quadrant I.

##### Solution

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

#### Exercise 11

If cosx= 2 3 , cosx= 2 3 , and x x is in quadrant I.

#### Exercise 12

If cosx= 1 2 , cosx= 1 2 , and x x is in quadrant III.

##### Solution

a) 3 2 3 2 b) 1 2 1 2 c) 3 3

#### Exercise 13

If tanx=8, tanx=8, and x x is in quadrant IV.

For the following exercises, find the values of the six trigonometric functions if the conditions provided hold.

#### Exercise 14

cos(2θ)= 3 5 cos(2θ)= 3 5 and 90 θ 180 90 θ 180

##### Solution

cosθ= 2 5 5 ,sinθ= 5 5 ,tanθ= 1 2 ,cscθ= 5 ,secθ= 5 2 ,cotθ=2 cosθ= 2 5 5 ,sinθ= 5 5 ,tanθ= 1 2 ,cscθ= 5 ,secθ= 5 2 ,cotθ=2

#### Exercise 15

cos(2θ)= 1 2 cos(2θ)= 1 2 and 180 θ 270 180 θ 270

For the following exercises, simplify to one trigonometric expression.

#### Exercise 16

2sin( π 4 )2cos( π 4 ) 2sin( π 4 )2cos( π 4 )

##### Solution

2sin( π 2 ) 2sin( π 2 )

#### Exercise 17

4sin( π 8 )cos( π 8 ) 4sin( π 8 )cos( π 8 )

For the following exercises, find the exact value using half-angle formulas.

#### Exercise 18

sin( π 8 ) sin( π 8 )

2 2 2 2 2 2

#### Exercise 19

cos( 11π 12 ) cos( 11π 12 )

#### Exercise 20

sin( 11π 12 ) sin( 11π 12 )

2 3 2 2 3 2

#### Exercise 21

cos( 7π 8 ) cos( 7π 8 )

#### Exercise 22

tan( 5π 12 ) tan( 5π 12 )

2+ 3 2+ 3

#### Exercise 23

tan( 3π 12 ) tan( 3π 12 )

#### Exercise 24

tan( 3π 8 ) tan( 3π 8 )

##### Solution

1 2 1 2

For the following exercises, find the exact values of a) sin( x 2 ), sin( x 2 ), b) cos( x 2 ), cos( x 2 ), and c) tan( x 2 ) tan( x 2 ) without solving for x, x, when 0 x 360 0 x 360

#### Exercise 25

If tanx= 4 3 , tanx= 4 3 , and x x is in quadrant IV.

#### Exercise 26

If sinx= 12 13 , sinx= 12 13 , and x x is in quadrant III.

##### Solution

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

#### Exercise 27

If cscx=7, cscx=7, and x x is in quadrant II.

#### Exercise 28

If secx=4, secx=4, and x x is in quadrant II.

##### Solution

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

For the following exercises, use Figure 5 to find the requested half and double angles.

#### Exercise 29

Find sin( 2θ ),cos(2θ), sin( 2θ ),cos(2θ), and tan(2θ). tan(2θ).

#### Exercise 30

Find sin(2α),cos(2α), sin(2α),cos(2α), and tan(2α). tan(2α).

##### Solution

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

#### Exercise 31

Find sin( θ 2 ),cos( θ 2 ), sin( θ 2 ),cos( θ 2 ), and tan( θ 2 ). tan( θ 2 ).

#### Exercise 32

Find sin( α 2 ),cos( α 2 ), sin( α 2 ),cos( α 2 ), and tan( α 2 ). tan( α 2 ).

##### Solution

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

For the following exercises, simplify each expression. Do not evaluate.

#### Exercise 33

cos 2 ( 28 ) sin 2 ( 28 ) cos 2 ( 28 ) sin 2 ( 28 )

#### Exercise 34

2 cos 2 ( 37 )1 2 cos 2 ( 37 )1

##### Solution

cos( 74 ) cos( 74 )

#### Exercise 35

12 sin 2 ( 17 ) 12 sin 2 ( 17 )

#### Exercise 36

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

##### Solution

cos(18x) cos(18x)

#### Exercise 37

4sin(8x)cos(8x) 4sin(8x)cos(8x)

#### Exercise 38

6sin(5x)cos(5x) 6sin(5x)cos(5x)

##### Solution

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

For the following exercises, prove the identity given.

#### Exercise 39

( sintcost ) 2 =1sin( 2t ) ( sintcost ) 2 =1sin( 2t )

#### Exercise 40

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

##### Solution

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

#### Exercise 41

cotxtanx=2cot( 2x ) cotxtanx=2cot( 2x )

#### Exercise 42

sin( 2θ ) 1+cos( 2θ ) tan 2 θ=tanθ sin( 2θ ) 1+cos( 2θ ) tan 2 θ=tanθ

##### Solution

sin( 2θ ) 1+cos( 2θ ) tan 2 θ= 2sin( θ )cos( θ ) 1+ cos 2 θ sin 2 θ tan 2 θ= 2sin( θ )cos( θ ) 2 cos 2 θ tan 2 θ= sin( θ ) cosθ tan 2 θ= cot( θ ) tan 2 θ=tanθ sin( 2θ ) 1+cos( 2θ ) tan 2 θ= 2sin( θ )cos( θ ) 1+ cos 2 θ sin 2 θ tan 2 θ= 2sin( θ )cos( θ ) 2 cos 2 θ tan 2 θ= sin( θ ) cosθ tan 2 θ= cot( θ ) tan 2 θ=tanθ

For the following exercises, rewrite the expression with an exponent no higher than 1.

#### Exercise 43

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

#### Exercise 44

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

##### Solution

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

#### Exercise 45

sin 4 (8x) sin 4 (8x)

#### Exercise 46

sin 4 (3x) sin 4 (3x)

##### Solution

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

#### Exercise 47

cos 2 x sin 4 x cos 2 x sin 4 x

#### Exercise 48

cos 4 x sin 2 x cos 4 x sin 2 x

##### Solution

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

#### Exercise 49

tan 2 x sin 2 x tan 2 x sin 2 x

### Technology

For the following exercises, reduce the equations to powers of one, and then check the answer graphically.

#### Exercise 50

tan 4 x tan 4 x

##### Solution

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

#### Exercise 51

sin 2 (2x) sin 2 (2x)

#### Exercise 52

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

##### Solution

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

#### Exercise 53

tan 2 xsinx tan 2 xsinx

#### Exercise 54

tan 4 x cos 2 x tan 4 x cos 2 x

##### Solution

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

#### Exercise 55

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

#### Exercise 56

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

##### Solution

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

#### Exercise 57

tan 2 ( x 2 )sinx tan 2 ( x 2 )sinx

For the following exercises, algebraically find an equivalent function, only in terms of sinx sinx and/or cosx, cosx, and then check the answer by graphing both equations.

#### Exercise 58

sin(4x) sin(4x)

##### Solution

4sinxcosx( cos 2 x sin 2 x ) 4sinxcosx( cos 2 x sin 2 x )

cos(4x) cos(4x)

### Extensions

For the following exercises, prove the identities.

#### Exercise 60

sin( 2x )= 2tanx 1+ tan 2 x sin( 2x )= 2tanx 1+ tan 2 x

##### Solution

2tanx 1+ tan 2 x = 2sinx cosx 1+ sin 2 x cos 2 x = 2sinx cosx cos 2 x+ sin 2 x cos 2 x = 2tanx 1+ tan 2 x = 2sinx cosx 1+ sin 2 x cos 2 x = 2sinx cosx cos 2 x+ sin 2 x cos 2 x =
2sinx cosx . cos 2 x 1 =2sinxcosx=sin(2x) 2sinx cosx . cos 2 x 1 =2sinxcosx=sin(2x)

#### Exercise 61

cos(2α)= 1 tan 2 α 1+ tan 2 α cos(2α)= 1 tan 2 α 1+ tan 2 α

#### Exercise 62

tan(2x)= 2sinxcosx 2 cos 2 x1 tan(2x)= 2sinxcosx 2 cos 2 x1

##### Solution

2sinxcosx 2 cos 2 x1 = sin(2x) cos(2x) =tan(2x) 2sinxcosx 2 cos 2 x1 = sin(2x) cos(2x) =tan(2x)

#### Exercise 63

( sin 2 x1 ) 2 =cos( 2x )+ sin 4 x ( sin 2 x1 ) 2 =cos( 2x )+ sin 4 x

#### Exercise 64

sin( 3x )=3sinx cos 2 x sin 3 x sin( 3x )=3sinx cos 2 x sin 3 x

##### Solution

sin(x+2x)=sinxcos(2x)+sin(2x)cosx =sinx( cos 2 x sin 2 x)+2sinxcosxcosx =sinx cos 2 x sin 3 x+2sinx cos 2 x =3sinx cos 2 x sin 3 x sin(x+2x)=sinxcos(2x)+sin(2x)cosx =sinx( cos 2 x sin 2 x)+2sinxcosxcosx =sinx cos 2 x sin 3 x+2sinx cos 2 x =3sinx cos 2 x sin 3 x

#### Exercise 65

cos( 3x )= cos 3 x3 sin 2 xcosx cos( 3x )= cos 3 x3 sin 2 xcosx

#### Exercise 66

1+cos( 2t ) sin( 2t )cost = 2cost 2sint1 1+cos( 2t ) sin( 2t )cost = 2cost 2sint1

##### Solution

1+cos(2t) sin(2t)cost = 1+2 cos 2 t1 2sintcostcost = 2 cos 2 t cost(2sint1) = 2cost 2sint1 1+cos(2t) sin(2t)cost = 1+2 cos 2 t1 2sintcostcost = 2 cos 2 t cost(2sint1) = 2cost 2sint1

#### Exercise 67

sin( 16x )=16sinxcosxcos( 2x )cos( 4x )cos( 8x ) sin( 16x )=16sinxcosxcos( 2x )cos( 4x )cos( 8x )

#### Exercise 68

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

##### Solution

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

## Glossary

double-angle formulas:
identities derived from the sum formulas for sine, cosine, and tangent in which the angles are equal
half-angle formulas:
identities derived from the reduction formulas and used to determine half-angle values of trigonometric functions
reduction formulas:
identities derived from the double-angle formulas and used to reduce the power of a trigonometric function

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