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

Figure 1 Bicycle and skateboard ramps for advanced riders have a steeper incline than those designed for novices.

Bicycle and skateboard 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 tan θ = 5 3 . tan θ = 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,

If we let α = β = θ , α = β = θ , then we have

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

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

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

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

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

The second variation is:

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

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

tan ( α + β ) = tan α + tan β 1 − tan α tan β tan ( θ + θ ) = tan θ + tan θ 1 − tan θ tan θ tan ( 2 θ ) = 2 tan θ 1 − tan 2 θ tan ( α + β ) = tan α + tan β 1 − tan α tan β tan ( θ + θ ) = tan θ + tan θ 1 − tan θ tan θ tan ( 2 θ ) = 2 tan θ 1 − tan 2 θ

Double-Angle Formulas

The double-angle formulas are summarized as follows:

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

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

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

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

1. ⓐ sin ( 2 θ ) sin ( 2 θ )
2. ⓑ cos ( 2 θ ) cos ( 2 θ )
3. ⓒ tan ( 2 θ ) tan ( 2 θ )

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 tan θ = − 3 4 , tan θ = − 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.

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

cos ( 2 θ ) = ( − 4 5 ) 2 − ( 3 5 ) 2 = 16 25 − 9 25 = 7 25 cos ( 2 θ ) = ( − 4 5 ) 2 − ( 3 5 ) 2 = 16 25 − 9 25 = 7 25

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

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

Try It #1

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

Example 2

Using the Double-Angle Formula for Cosine without Exact Values

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

Solution

cos ( 6 x ) = cos ( 2 ( 3 x ) ) = 2 cos 2 ( 3 x ) − 1 cos ( 6 x ) = cos ( 2 ( 3 x ) ) = 2 cos 2 ( 3 x ) − 1

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

Using the Double-Angle Formulas to Verify an Identity

Verify the following identity using double-angle formulas:

Solution

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

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

Analysis

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

where a = sin θ a = sin θ and b = cos θ . b = cos θ . 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 #2

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

Example 4

Verifying a Double-Angle Identity for Tangent

Verify the identity:

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

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

Let’s work on the right side.

2 cot θ − tan θ = 2 1 tan θ − tan θ ( tan θ tan θ ) = 2 tan θ 1 tan θ ( tan θ ) − tan θ ( tan θ ) = 2 tan θ 1 − tan 2 θ 2 cot θ − tan θ = 2 1 tan θ − tan θ ( tan θ tan θ ) = 2 tan θ 1 tan θ ( tan θ ) − tan θ ( tan θ ) = 2 tan θ 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 #3

Verify the identity: cos ( 2 θ ) cos θ = cos 3 θ − cos θ sin 2 θ . cos ( 2 θ ) cos θ = cos 3 θ − cos θ 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 ( 2 θ ) = 1 − 2 sin 2 θ . cos ( 2 θ ) = 1 − 2 sin 2 θ . Solve for sin 2 θ : sin 2 θ :

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

Next, we use the formula cos ( 2 θ ) = 2 cos 2 θ − 1. cos ( 2 θ ) = 2 cos 2 θ − 1. Solve for cos 2 θ : cos 2 θ :

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

Reduction Formulas

The reduction formulas are summarized as follows: