Double-Angle, Half-Angle, and Reduction Formulas

Learning Objectives

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.

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 θ𝜃 such that tanθ=53.tan𝜃=53. 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+cosαsinβsin(𝛼+𝛽)=sin𝛼cos𝛽+cos𝛼sin𝛽

If we let α=β=θ,𝛼=𝛽=𝜃, then we have

sin(θ+θ)sin(2θ)==sinθcosθ+cosθsinθ2sinθcosθsin(𝜃+𝜃)=sin𝜃cos𝜃+cos𝜃sin𝜃sin(2𝜃)=2sin𝜃cos𝜃

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

cos(θ+θ)cos(2θ)==cosθcosθ−sinθsinθcos2θ−sin2θcos(𝜃+𝜃)=cos𝜃cos𝜃−sin𝜃sin𝜃cos(2𝜃)=cos2𝜃−sin2𝜃

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

cos(2θ)===cos2θ−sin2θ(1−sin2θ)−sin2θ1−2sin2θcos(2𝜃)=cos2𝜃−sin2𝜃=(1−sin2𝜃)−sin2𝜃=1−2sin2𝜃

The second variation is:

cos(2θ)===cos2θ−sin2θcos2θ−(1−cos2θ)2cos2θ−1cos(2𝜃)=cos2𝜃−sin2𝜃=cos2𝜃−(1−cos2𝜃)=2cos2𝜃−1

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

tan(α+β)tan(θ+θ)tan(2θ)===tanα+tanβ1−tanαtanβtanθ+tanθ1−tanθtanθ2tanθ1−tan2θtan(𝛼+𝛽)=tan𝛼+tan𝛽1−tan𝛼tan𝛽tan(𝜃+𝜃)=tan𝜃+tan𝜃1−tan𝜃tan𝜃tan(2𝜃)=2tan𝜃1−tan2𝜃

DOUBLE-ANGLE FORMULAS

The double-angle formulas are summarized as follows:

sin(2θ)=2sinθcosθsin(2𝜃)=2sin𝜃cos𝜃

cos(2θ)===cos2θ−sin2θ1−2sin2θ2cos2θ−1cos(2𝜃)=cos2𝜃−sin2𝜃=1−2sin2𝜃=2cos2𝜃−1

tan(2θ)=2tanθ1−tan2θtan(2𝜃)=2tan𝜃1−tan2𝜃

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 tanθ=−34tan𝜃=−34 and θ𝜃 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=−34,tan𝜃=−34, such that θ𝜃 is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because θ𝜃 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)216+925c====c2c2c25(−4)2+(3)2=𝑐216+9=𝑐225=𝑐2𝑐=5

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

Figure 2

1. ⓐ Let’s begin by writing the double-angle formula for sine.sin(2θ)=2sinθcosθsin(2𝜃)=2sin𝜃cos𝜃We see that we to need to find sinθsin𝜃 and cosθ.cos𝜃. Based on Figure 2, we see that the hypotenuse equals 5, so sinθ=35,sin𝜃=35, and cosθ=−45.cos𝜃=−45. Substitute these values into the equation, and simplify.Thus,sin(2θ)==2(35)(−45)−2425sin(2𝜃)=2(35)(−45)=−2425
2. ⓑ Write the double-angle formula for cosine.cos(2θ)=cos2θ−sin2θcos(2𝜃)=cos2𝜃−sin2𝜃Again, substitute the values of the sine and cosine into the equation, and simplify.cos(2θ)===(−45)2−(35)21625−925725cos(2𝜃)=(−45)2−(35)2=1625−925=725
3. ⓒ Write the double-angle formula for tangent.tan(2θ)=2tanθ1−tan2θtan(2𝜃)=2tan𝜃1−tan2𝜃In this formula, we need the tangent, which we were given as tanθ=−34.tan𝜃=−34. Substitute this value into the equation, and simplify.tan(2θ)====2(−34)1−(−34)2−321−916−32(167)−247tan(2𝜃)=2(−34)1−(−34)2=−321−916=−32(167)=−247

TRY IT #1

Given sinα=58,sin𝛼=58, with α𝛼 in quadrant I, find cos(2α).cos(2𝛼).

EXAMPLE 2

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(2(3x))2cos2(3x)−1cos(6𝑥)=cos(2(3𝑥))=2cos2(3𝑥)−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:

1+sin(2θ)=(sinθ+cosθ)21+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====sin2θ+2sinθcosθ+cos2θ(sin2θ+cos2θ)+2sinθcosθ1+2sinθcosθ1+sin(2θ)(sin𝜃+cos𝜃)2=sin2𝜃+2sin𝜃cos𝜃+cos2𝜃=(sin2𝜃+cos2𝜃)+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=a2±2ab+b2(𝑎±𝑏)2=𝑎2±2𝑎𝑏+𝑏2

where a=sinθ𝑎=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 #2

Verify the identity: cos4θ−sin4θ=cos(2θ).cos4𝜃−sin4𝜃=cos(2𝜃).

EXAMPLE 4

Verifying a Double-Angle Identity for Tangent

Verify the identity:

tan(2θ)=2cotθ−tanθtan(2𝜃)=2cot𝜃−tan𝜃

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)====2tanθ1−tan2θ2tanθ(1tanθ)(1−tan2θ)(1tanθ)21tanθ−tan2θtanθ2cotθ−tanθDouble-angle formulaMultiply by a term that results in desired numerator.Use reciprocal identity for  1tanθ.tan(2𝜃)=2tan𝜃1−tan2𝜃Double-angle formula=2tan𝜃(1tan𝜃)(1−tan2𝜃)(1tan𝜃)Multiply by a term that results in desired numerator.=21tan𝜃−tan2𝜃tan𝜃=2cot𝜃−tan𝜃Use reciprocal identity for  1tan𝜃.

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

2tanθ1−tan2θ=2cotθ−tanθ2tan𝜃1−tan2𝜃=2cot𝜃−tan𝜃

Let’s work on the right side.

2cotθ−tanθ===21tanθ−tanθ(tanθtanθ)2tanθ1tanθ(tanθ)−tanθ(tanθ)2tanθ1−tan2θ2cot𝜃−tan𝜃=21tan𝜃−tan𝜃(tan𝜃tan𝜃)=2tan𝜃1tan𝜃(tan𝜃)−tan𝜃(tan𝜃)=2tan𝜃1−tan2𝜃

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(2θ)cosθ=cos3θ−cosθsin2θ.cos(2𝜃)cos𝜃=cos3𝜃−cos𝜃sin2𝜃.

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)=1−2sin2θ.cos(2𝜃)=1−2sin2𝜃. Solve for sin2θ:sin2𝜃:

cos(2θ)2sin2θsin2θ===1−2sin2θ1−cos(2θ)1−cos(2θ)2cos(2𝜃)=1−2sin2𝜃2sin2𝜃=1−cos(2𝜃)sin2𝜃=1−cos(2𝜃)2

Next, we use the formula cos(2θ)=2cos2θ−1.cos(2𝜃)=2cos2𝜃−1. Solve for cos2θ:cos2𝜃:

cos(2θ)1+cos(2θ)1+cos(2θ)2=== 2cos2θ−12cos2θcos2θcos(2𝜃)= 2cos2𝜃−11+cos(2𝜃)=2cos2𝜃1+cos(2𝜃)2=cos2𝜃

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

tan2θ====sin2θcos2θ1−cos(2θ)21+cos(2θ)2(1−cos(2θ)2)(21+cos(2θ))1−cos(2θ)1+cos(2θ)Substitute the reduction formulas.tan2𝜃=sin2𝜃cos2𝜃=1−cos(2𝜃)21+cos(2𝜃)2Substitute the reduction formulas.=(1−cos(2𝜃)2)(21+cos(2𝜃))=1−cos(2𝜃)1+cos(2𝜃)

REDUCTION FORMULAS

The reduction formulas are summarized as follows:

sin2θ=1−cos(2θ)2sin2𝜃=1−cos(2𝜃)2

cos2θ=1+cos(2θ)2cos2𝜃=1+cos(2𝜃)2

tan2θ=1−cos(2θ)1+cos(2θ)tan2𝜃=1−cos(2𝜃)1+cos(2𝜃)

EXAMPLE 5

Writing an Equivalent Expression Not Containing Powers Greater Than 1

Write an equivalent expression for cos4xcos4𝑥 that does not involve any powers of sine or cosine greater than 1.

Solution

We will apply the reduction formula for cosine twice.

cos4x======(cos2x)2(1+cos(2x)2)214(1+2cos(2x)+cos2(2x))14+12cos(2x)+14(1+cos2(2x)2)14+12cos(2x)+18+18cos(4x)38+12cos(2x)+18cos(4x)Substitute reduction formula for cos2x.Substitute reduction formula for cos2x.cos4𝑥=(cos2𝑥)2=(1+cos(2𝑥)2)2Substitute reduction formula for cos2𝑥.=14(1+2cos(2𝑥)+cos2(2𝑥))=14+12cos(2𝑥)+14(1+cos2(2𝑥)2)Substitute reduction formula for cos2𝑥.=14+12cos(2𝑥)+18+18cos(4𝑥)=38+12cos(2𝑥)+18cos(4𝑥)

Analysis

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

EXAMPLE 6

Using the Power-Reducing Formulas to Prove an Identity

Use the power-reducing formulas to prove

sin3(2x)=[12sin(2x)][1−cos(4x)]sin3(2𝑥)=[12sin(2𝑥)][1−cos(4𝑥)]

Solution

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

sin3(2x)====[sin(2x)][sin2(2x)]sin(2x)[1−cos(4x)2]sin(2x)(12)[1−cos(4x)]12[sin(2x)][1−cos(4x)]Substitute the power-reduction formula.sin3(2𝑥)=[sin(2𝑥)][sin2(2𝑥)]=sin(2𝑥)[1−cos(4𝑥)2]Substitute the power-reduction formula.=sin(2𝑥)(12)[1−cos(4𝑥)]=12[sin(2𝑥)][1−cos(4𝑥)]

Analysis

Note that in this example, we substituted

1−cos(4x)21−cos(4𝑥)2

for sin2(2x).sin2(2𝑥). The formula states

sin2θ=1−cos(2θ)2sin2𝜃=1−cos(2𝜃)2

We let θ=2x,𝜃=2𝑥, so 2θ=4x.2𝜃=4𝑥.

TRY IT #4

Use the power-reducing formulas to prove that 10cos4x=154+5cos(2x)+54cos(4x).10cos4𝑥=154+5cos(2𝑥)+54cos(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 θ𝜃 with α2,𝛼2, the half-angle formula for sine is found by simplifying the equation and solving for sin(α2).sin(𝛼2). 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 α2𝛼2 terminates.

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

sin2θsin2(α2)sin(α2)====1−cos(2θ)21−(cos2⋅α2)21−cosα2±1−cosα2−−−−−−√sin2𝜃=1−cos(2𝜃)2sin2(𝛼2)=1−(cos2⋅𝛼2)2=1−cos𝛼2sin(𝛼2)=±1−cos𝛼2

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

cos2θcos2(α2)cos(α2)====1+cos(2θ)21+cos(2⋅α2)21+cosα2±1+cosα2−−−−−−√cos2𝜃=1+cos(2𝜃)2cos2(𝛼2)=1+cos(2⋅𝛼2)2=1+cos𝛼2cos(𝛼2)=±1+cos𝛼2

For the tangent identity, we have

tan2θtan2(α2)tan(α2)====1−cos(2θ)1+cos(2θ)1−cos(2⋅α2)1+cos(2⋅α2)1−cosα1+cosα±1−cosα1+cosα−−−−−−√tan2𝜃=1−cos(2𝜃)1+cos(2𝜃)tan2(𝛼2)=1−cos(2⋅𝛼2)1+cos(2⋅𝛼2)=1−cos𝛼1+cos𝛼tan(𝛼2)=±1−cos𝛼1+cos𝛼

HALF-ANGLE FORMULAS

The half-angle formulas are as follows:

sin(α2)=±1−cosα2−−−−−−−−√sin(𝛼2)=±1−cos𝛼2

cos(α2)=±1+cosα2−−−−−−−−√cos(𝛼2)=±1+cos𝛼2

tan(α2)=±1−cosα1+cosα−−−−−−√=sinα1+cosα=1−cosαsinαtan(𝛼2)=±1−cos𝛼1+cos𝛼=sin𝛼1+cos𝛼=1−cos𝛼sin𝛼

EXAMPLE 7

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:

sin30°2=====1−cos30°2−−−−−−−√1−3√22−−−−−√2−3√22−−−−√2−3√4−−−−√2−3√√2sin30°2=1−cos30°2=1−322=2−322=2−34=2−32

Remember that we can check the answer with a graphing calculator.

Analysis

Notice that we used only the positive root because sin(15°)sin(15°) 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

Finding Exact Values Using Half-Angle Identities

Given that tanα=815tan𝛼=815 and α𝛼 lies in quadrant III, find the exact value of the following:

1. ⓐ sin(α2)sin(𝛼2)
2. ⓑ cos(α2)cos(𝛼2)
3. ⓒ tan(α2)tan(𝛼2)

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=−817sin𝛼=−817 and cosα=−1517.cos𝛼=−1517.

Figure 3

1. ⓐ Before we start, we must remember that if α𝛼 is in quadrant III, then 180°<α<270°,180°<𝛼<270°, so 180°2<α2<270°2.180°2<𝛼2<270°2. This means that the terminal side of α2𝛼2 is in quadrant II, since 90°<α2<135°.90°<𝛼2<135°.To find sinα2,sin𝛼2, 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.sinα2=======±1−cosα2−−−−−−√±1−(−1517)2−−−−−−−√±32172−−−√±3217⋅12−−−−−√±1617−−√±417√417√17sin𝛼2=±1−cos𝛼2=±1−(−1517)2=±32172=±3217⋅12=±1617=±417=41717We choose the positive value of sinα2sin𝛼2 because the angle terminates in quadrant II and sine is positive in quadrant II.
2. ⓑ To find cosα2,cos𝛼2, 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.cosα2======±1+cosα2−−−−−−√±1+(−1517)2−−−−−−−√±2172−−−√±217⋅12−−−−−√±117−−√−17√17cos𝛼2=±1+cos𝛼2=±1+(−1517)2=±2172=±217⋅12=±117=−1717We choose the negative value of cosα2cos𝛼2 because the angle is in quadrant II because cosine is negative in quadrant II.
3. ⓒ To find tanα2,tan𝛼2, 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.tanα2======±1−cosα1+cosα−−−−−−√±1−(−1517)1+(−1517)−−−−−−√±3217217−−−√±322−−√−16−−√−4tan𝛼2=±1−cos𝛼1+cos𝛼=±1−(−1517)1+(−1517)=±3217217=±322=−16=−4We choose the negative value of tanα2tan𝛼2 because α2𝛼2 lies in quadrant II, and tangent is negative in quadrant II.

TRY IT #5

Given that sinα=−45sin𝛼=−45 and α𝛼 lies in quadrant IV, find the exact value of cos(α2).cos(𝛼2).

EXAMPLE 9

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 θ𝜃 formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If tanθ=53tan𝜃=53 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=53tan𝜃=53 for high competition, we can find cosθcos𝜃 from the right triangle and the Pythagorean theorem so that we can use the half-angle identities. See Figure 4.

32+52c==3434−−√32+52=34𝑐=34

Figure 4

We see that cosθ=334√=334√34.cos𝜃=334=33434. We can use the half-angle formula for tangent: tanθ2=1−cosθ1+cosθ−−−−−−√.tan𝜃2=1−cos𝜃1+cos𝜃. Since tanθtan𝜃 is in the first quadrant, so is tanθ2.tan𝜃2.

tanθ2===≈1−334√341+334√34−−−−−√34−334√3434+334√34−−−−−√34−334√34+334√−−−−−−√0.57tan𝜃2=1−334341+33434=34−3343434+33434=34−33434+334≈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°.