## Double‐Angle and Half‐Angle Identities

Special cases of the sum and difference formulas for sine and cosine yields what is known as the **double‐angle identities** and the **half‐angle identities**. First, using the sum identity for the sine,

sin 2α = sin (α + α)

sin 2α = sin α cos α + cos α sin α

sin 2α = 2 sin α cos α

Similarly for the cosine,

Using the Pythagorean identity, sin ^{2} α+cos ^{2}α=1, two additional cosine identities can be derived.

and

The half‐angle identities for the sine and cosine are derived from two of the cosine identities described earlier.

The sign of the two preceding functions depends on the quadrant in which the resulting angle is located.

**Example 1:** Find the exact value for sin 105° using the half‐angle identity.

In the following verification, remember that 105° is in the second quadrant, and sine functions in the second quadrant are positive. Also, 210° is in the third quadrant, and cosine functions in the third quadrant are negative. From Figure 1, the reference triangle of 210° in the third quadrant is a 30°–60°–90° triangle. Therefore, cos 210° = −cos 30°.

**Figure 1 ** Drawing for Example 1.

Using the half‐angle identity for sine,

**Example 2:** Find the exact value for cos 165° using the half‐angle identity.

In the following verification, remember that 165° is in the second quadrant, and cosine functions in the second quadrant are negative. Also, 330° is in the fourth quadrant, and cosine functions in the fourth quadrant are positive. From Figure 2 , the reference triangle of 330° in the fourth quadrant is a 30°–60°–90° triangle. Therefore, cos 330° = cos 30°.

**Figure 2** Drawing for Example 2.

Using the half‐angle identity for the cosine,

**Example 4:** Verify the identity 1 − cos 2 *x* = tan *x* sin 2 *x*.

## Trigonometry Half Angle Formulas

It is sometimes very crucial to determine the value of the trigonometric functions for half-angles. For instance, using some half angle formula we can convert an expression with exponents to one without exponents, and whose angles are multiples of the original angle. It is to note that we get half-angle formulas from double angle formulas. Both sin (2A) and cos (2A) are obtained from the double angle formula for the cosine. The formula for half angle identities is as below:

**Half – Angle Formula**

**Double‐Angle and Half‐Angle Identities**

Certain cases of the sums and differences formulas for sine and cosine generate what is called the double‐angle identities and the half‐angle identities. We will first start by incorporating the sum identity for the sine as given in the reference,

In the same manner, for the cosine,

Cos 2α = cos (α + α)

cos 2α = (cos α cos α – sin α sin α)

cos 2α = (cos^{2}α – sin^{2}α)

Applying the property of Pythagorean, sin 2 α + cos 2α =1, we can obtain two additional cosine identities i.e.

Cos 2α = cos^{2}α – sin^{2}α

Cos 2α = [-1 sin^{2}α] – sin^{2}α

Cos 2α = 1- 2 sin^{2}α

Cos 2α = cos^{2}α – sin^{2}α

cos 2α = cos^{2}α – [1 – sin^{2}α]

cos 2α = 2cos^{2}α – 1

The half‐angle identities for the sine and cosine are derived from two of the cosine identities explained above.

Cos 2α = 2cos^{2}α – 1

Cos 2 (β/2) – 2cos^{2} (β/2) – 1

Cosβ = 2cos^{2} (β/2) – 1

2cos^{2} (β/2) = 1 ± cosβ

Cos 2α = 1 – 2 sin^{2}α

Cos 2 (β/2) = 1 – 2 sin^{2} (β/2)

Cosβ = 1 – 2 sin^{2} (β/2)

2 sin^{2} (β/2) = 1 Cosβ

Cos^{2} (β/2) = 1 + cosβ/2

Cos^{ }(β/2) = ± √1+*c**o**s*(*β*/2)

Similarly,

sin^{2} (β/2) = 1 – cosβ/2

sin (β/2) = √1−*c**o**s*(*β*/2)

The symbol of the two preceding functions is dependent on the quadrant in which the resulting angle is positioned.

For computing the tangent of the half-angle, tan (2A), we need to combine the identities for sine and cosine:

tan^{2} (A) = 1 – cos (2A)/2 / 1 + cos (2A)/2 = 1–*c**o**s*(2*A*)/ 1+*c**o**s*(2*A*)

Again replacing A by (1/2)A, we obtain

**Trigonometry Angles and Its Notations**

These are actually the use of Greek letters such as alpha (α), beta (β), theta (θ) and gamma (γ) to denote angles. It represents the signs of trigonometric functions in each quadrant. A couple of different units of angle measure are extensively used, including degree, radian, and gradian (gons): As an example,

1 complete circle (turn) = 360° = 2π radian = 400 gon.

However it is not particularly annotated by (°) for degree or (^{g}) for gradian, all values for angles are assumed to be given in radian.

The graphical representation shown below depicts some common angles and their values and the conversions of the basic trigonometric functions:

**Solved Examples**

**Example1:**

Find the absolute value for cosine 165° using the half angle identities.

**Solution1:**

In the given verification, note that 165 degrees is in the 2^{nd} quadrant, and cosine functions in the 2^{nd} quadrant are negative. Also, 330 degrees is in the 4^{th} quadrant, and cosine functions in the 4^{th} quadrant are positive.

That said, the reference triangle of 330° in the 4^{th} quadrant is a 30°–60°–90° triangle. Thus, cos 330° = cos 30°.

Now, applying the half angle identity for the cosine:

We get,

**Key Facts**

- In the half-angle formula problems for sine and cosine, observe that a plus/minus sign occurs in front of each square root (radical).
- Whether your answer is negative or positive depends on which quadrant the new half angle is in.
- The tan half angle formula doesn’t have a ± sign in front, so the above doesn’t apply to tangent.

## FAQs (Frequently Asked Questions)

**Q1. What are Trigonometric Identities?**

Ans. In mathematical terms, trigonometric identities demonstrate the equalities that involve trigonometric functions and are true for every value of the variables taking place where both sides of the equality are described. However geometrically, these are identities taking into account only specified functions of one or more angles. They are quite different from triangle identities, which are identities likely to involve angles in addition to involving side lengths or other lengths of a triangle.

These identities are of help whenever expressions involving trigonometric functions require to be simplified. An important application of using half angle identities is the integration of non-trigonometric functions: a general method entails first using the substitution law with a trigonometric function, and afterwards simplifying the resulting integral using a trigonometric identity.

**Q2. What is the Use of Half Angle Formulas?**

Ans. There are many applications of trigonometry half angle formulas to science and engineering with respect to light and sound. Many of these processes need equations involving the sine and cosine of x, 2x, 3x, 4x and more. Neither, doubling the sin x will provide you the value of sin 2x, nor will taking half of sin x, provide you sin (x/2). Moreover, we can create the double angle formulas just by using the addition formulas for sine, cosine and tangent.

## Half Angle Formulas

We study **half angle formulas** (or half-angle identities) in Trigonometry. Half angle formulas can be derived using the double angle formulas. As we know, the double angle formulas can be derived using the angle sum and difference formulas of trigonometry. Half-angles in half angle formulas are usually denoted by θ/2, x/2, A/2, etc and the half-angle is a sub-multiple angle. The half angle formulas are used to find the exact values of the trigonometric ratios of the angles like 22.5° (which is half of the standard angle 45°), 15° (which is half of the standard angle 30°), etc.

Let us explore the half angle formulas along with their proofs and with a few solved examples here.

## What are Half Angle Formulas?

In this section, we will see the half angle formulas of sin, cos, and tan. We know the values of the trigonometric functions (sin, cos , tan, cot, sec, cosec) for the angles like 0°, 30°, 45°, 60°, and 90° from the trigonometric table. But to know the exact values of sin 22.5°, tan 15°, etc, the half angle formulas are extremely useful. Also, they are helpful in proving several trigonometric identities. We have half angle formulas that are derived from the double angle formulas and they are expressed in terms of half angles like θ/2, x/2, A/2, etc. Here is the list of important half angle formulas:

## Half Angle Identities

Here are the popular half angle identities that we use in solving many trigonometry problems are as follows:

**Half angle formula of sin:**sin A/2 = ±√[(1 – cos A) / 2]**Half angle formula of cos:**cos A/2 = ±√[(1 + cos A) / 2]**Half angle formula of tan:**tan A/2 = ±√[1 – cos A] / [1 + cos A] (or) sin A / (1 + cos A) (or) (1 – cos A) / sin A

## Half Angle Formulas Derivation Using Double Angle Formulas

To derive the above formulas, first, let us derive the following half angle formulas. The double angle formulas are in terms of the double angles like 2θ, 2A, 2x, etc. We know that the double angle formulas of sin, cos, and tan are

- sin 2x = 2 sin x cos x
- cos 2x = cos
^{2}x – sin^{2}x (or)

= 1 – 2 sin^{2}x (or)

= 2 cos^{2}x – 1 - tan 2x = 2 tan x / (1 – tan
^{2}x)

If we replace x with A/2 on both sides of every equation of double angle formulas, we get half angle identities (as 2x = 2(A/2) = A).

- sin A = 2 sin(A/2) cos(A/2)
- cos A = cos
^{2}(A/2) – sin^{2}(A/2) (or)

= 1 – 2 sin^{2}(A/2) (or)

= 2 cos^{2}(A/2) – 1 - tan A = 2 tan (A/2) / (1 – tan
^{2}(A/2))

We can also derive one half angle formula using another half angle formula. For example, just from the formula of cos A, we can derive 3 important half angle identities for sin, cos, and tan which are mentioned in the first section. Here is the half angle formulas proof.

## Half Angle Formula of Sin Proof

Now, we will prove the half angle formula for the sine function. Using one of the above formulas of cos A, we have

cos A = 1 – 2 sin^{2} (A/2)

From this,

2 sin^{2} (A/2) = 1 – cos A

sin^{2} (A/2) = (1 – cos A) / 2

sin (A/2) = ±√[(1 – cos A) / 2]

## Half Angle Formula of Cos Derivation

Now, we will prove the half angle formula for the cosine function. Using one of the above formulas of cos A,

cos A = 2 cos^{2}(A/2) – 1

From this,

2 cos^{2}(A/2) = 1 + cos A

cos^{2} (A/2) = (1 + cos A) / 2

cos (A/2) = ±√[(1 + cos A) / 2]

## Half Angle Formula of Tan Derivation

We know that tan (A/2) = [sin (A/2)] / [cos (A/2)]

From the half angle formulas of sin and cos,

tan (A/2) = [±√(1 – cos A)/2] / [±√(1 + cos A)/2]

= **±√[(1 – cos A) / (1 + cos A)]**

This is one of the formulas of tan (A/2). Let us derive the other two formulas by rationalizing the denominator here.

tan (A/2) = ±√[(1 – cos A) / (1 + cos A)] × √[(1 – cos A) / (1 – cos A)]

= √[(1 – cos A)^{2} / (1 – cos^{2}A)]

= √[(1 – cos A)^{2}/ sin^{2}A]

= **(1 – cos A) / sin A**

This is the second formula of tan (A/2). To derive another formula, let us multiply and divide the above formula by (1 + cos A). Then we get

tan (A/2) = [(1 – cos A) / sin A] × [(1 + cos A) / (1 + cos A)]

= (1 – cos^{2}A) / [sin A (1 + cos A)]

= sin^{2}A / [sin A (1 + cos A)]

= **sin A / (1 + cos A)**

Thus, tan (A/2) = ±√[(1 – cos A) / (1 + cos A)] = (1 – cos A) / sin A = sin A / (1 + cos A).

## Half Angle Formula Using Semiperimeter

In this section, we will see the half angle formulas using the semi perimeter. i.e., these are the half angle formulas in terms of sides of a triangle. Let us consider a triangle ABC where AB = c, BC = a, and CA = b.

Let us derive one of these formulas here. We know that the semi-perimeter of the triangle is s = (a + b + c)/2. From this, we have 2s = a + b + c. From one of the above formulas,

cos A = 2 cos²(A/2) – 1 (or)

2 cos²(A/2) = 1 + cos A

Now using the law of cosines,

2 cos^{2}(A/2) = 1 + [ (b^{2} + c^{2} – a^{2}) / (2bc) ]

2 cos^{2}(A/2) = [2bc + b² + c² – a²] / [2bc]

2 cos^{2}(A/2) = [ (b + c)² – a²] / [2bc] [Using (a+b)² formula]

2 cos^{2}(A/2) = [ (b + c + a) (b + c – a) ] / [2bc] [Using a² – b² formula]

2 cos^{2}(A/2) = [ 2s (2s – 2a) ] / [2bc] [As 2s = a + b + c]

2 cos^{2}(A/2) = [ 2s (s – a) ] / [bc]

cos^{2}(A/2) = [ s(s – a) ] / [bc]

cos (A/2) = √[ s (s – a) ] / [bc]

We have derived one half-angle formula for cosine of angle A/2. Similarly, we can derive other half angle identities of cosine using the semi perimeter. Another half angle formula of sine can be derived using the semi perimeter.

sin^{2}(A/2) = (1 − cos A)/2

= (1/2)[1−(b^{2}+c^{2}−a^{2})/2bc] (Using the law of cosines)

= (1/2)(a^{2}−(b−c)^{2})/2bc

= (1/2)(a + b − c)(a + c − b)/2bc

= (1/2){(a + b + c) − 2c}{(a + b + c) − 2b}/2bc

= (1/2)(2s − 2c)(2s − 2b)/2bc

= (s − b)(s − c)/bc

⇒ sin (A/2) = √[(s − b)(s − c)/bc]

Similarly, we can derive other half angle formulas of the sine function. Half angle formulas for tangent function can be derived using the formula tan (A/2) = sin (A/2)/cos (A/2).

## Examples Using Half Angle Formula

**Example 1:** Use an appropriate half angle formula to find the exact value of cos π/8.

**Solution:**

Using the half angle formula of cos,

cos A/2 = ±√[(1 + cos A )/ 2]

We know that π/8 = 22.5°.

Substitute A = 45° on both sides,

cos 45°/2 = ±√[(1 + cos 45°) / 2]

We know that cos 45° = √2/2.

cos 22.5° = ±√[1 + (√2/2) / 2]

cos 22.5° = ±√[(2 + √2) / (2 × 2)]

cos 22.5° = ± √(2 + √2) / 2

But 22.5° lies in quadrant I and hence cos 22.5° is positive. Thus,

cos 22.5° = √(2 + √2) / 2

**Answer: **cos π/8 = √(2 + √2) / 2.

**Example 2:** Prove that cos A / (1 + sin A) = tan [ (π/4) – (A/2) ].

**Solution:**

LHS = cos x / (1 + sin x)

Using the half angle formulas,

= [cos^{2} (A/2) – sin^{2} (A/2)] / [1 + 2 sin (A/2) cos (A/2)]

We know that 1 = cos^{2}(A/2) + sin^{2}(A/2). So

= [ (cos (A/2) + sin (A/2)) (cos (A/2) – sin (A/2)) ] / [cos^{2}(A/2) + sin^{2}(A/2) + 2 sin (A/2) cos (A/2)]

= [ (cos (A/2) + sin (A/2)) (cos (A/2) – sin (A/2)) ] / [cos (A/2) + sin (A/2)]^{2}

= [cos (A/2) – sin (A/2)] / [cos (A/2) + sin (A/2)]

= [ cos (A/2) ( 1 – sin (A/2)/cos(A/2) ) ] / [ cos (A/2) ( 1 + sin (A/2)/cos(A/2) ) ]

= (1 – tan (A/2)) / (1 + tan (A/2))

We know that 1 = tan (π/4). So

= (tan (π/4) – tan (A/2)) / (1 + tan (π/4) tan (A/2))

We have (tan A – tan B) / (1 + tan A tan B) = tan (A – B). So

= tan [ (π/4) – (A/2) ]

= RHS

Hence proved.

**Answer:** We proved the given identity.

**Example 3:** In a triangle ABC, if AB = c = 12, BC = a = 13, and CA = b = 5, then find the value of sin A/2.

**Solution:**

It is given that a = 13; b = 5; c = 12.

Then the semiperimeter is, s = (a + b + c) / 2 = (13 + 5 + 12) / 2 = 15.

Using the half angle identity of sin in terms of semi perimeter,

sin A/2 = √[(s – b) (s – c) / bc]

= √[(15 – 5) (15 – 12) /(5)(12)]

= √[(10) (3) / 60]

= √2/2

**Answer: **sin A/2 = √2/2.

## FAQs on Half Angle Formula

### What Are Half Angle Formulas in Trigonometry?

The half angle formulas give the value of half angles like A/2, x/2, etc of trigonometric ratios. The half angle formulas of sin, cos, and tan are

- sin A/2 = ±√[(1 – cos A) / 2]
- cos A/2 = ±√[(1 + cos A) / 2]
- tan A/2 = ±√[1 – cos A] / [1 + cos A]

### What Is Half Angle Formula for Sin?

The **half angle formula** of sin is sin A/2 = ±√[(1 – cos A) / 2]. We have another half angle formula of sin in terms of semiperimeter. If a, b, and c are the sides of a triangle and A, B, and C are their corresponding opposite angles, then sin A/2 = √[(s – b) (s – c)/bc].

### What Is Half Angle Formula for Cosine?

The half angle formula of cos is cos A/2 = ±√[(1 + cos A)/2]. We have another half angle formula of cos in terms of semiperimeter. If a, b, and c are the sides of a triangle and A, B, and C are their corresponding opposite angles, then cos (A/2) = √[ s (s – a)/bc].

### What Is Half Angle Formula for Tangent?

The half angle formula of tangent is tan (A/2) = ±√[1 – cos A] / [1 + cos A] = (1 – cos A) / sin A = sin A / (1 + cos A). We have another half angle formula of tan in terms of semiperimeter. If a, b, and c are the sides of a triangle and A, B, and C are their corresponding opposite angles, then sin A/2 = √[(s – b) (s – c) ] / [s(s – a)].

### Why To Use Half Angle Formulas?

We use half angle formulas in finding the trigonometric ratios of the half of the standard angles, for example, we can find the trigonometric ratios of angles like 15°, 22.5°, etc using the half angle identities. They can be used in proving various trigonometric identities. They are also used in solving integrals.

### How To Derive Half Angle Formula of Cos?

Using the double angle formula of cos,

cos 2x = 2cos^{2}x – 1

By replacing x with (A/2), we get

cos A = 2 cos^{2}(A/2) – 1

We will solve this for cos (A/2).

2 cos^{2}(A/2) = 1 + cos A

cos^{2} (A/2) = (1 + cos A) / 2

cos A/2 = ±√(1 + cos A) / 2

### What Is tan 15° Using Half Angle Identities?

Using the half angle identity of tan,

tan (A/2) = (1 – cos A) / sin A

Substitute A = 30°,

tan (30°/2) = (1 – cos 30°) / sin 30°

= [1 – (√3/2)] / (1/2)

= [ (2 – √3) / 2] / (1/2)

= 2 – √3

Therefore, tan 15° = 2 – √3.

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

Learning Objectives

- 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

Bicycle ramps made for competition (see Figure 9.3.1) 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* such that 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*. Deriving the double-angle formula for sine begins with the sum formula,

**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.

## 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.

Learning Objectives

- 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

Bicycle ramps made for competition (see Figure 9.3.1) 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* such that tan *θ*=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.

: Bicycle ramps for advanced riders have a steeper incline than those designed for novices.

## 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*

. Deriving the double-angle formula for sine begins with the sum formula,sin(*α*+*β*)=sin*α*cos*β*+cos*α*sin*β*(9.3.1)

If we let *α*=*β*=*θ*

, then we havesin(*θ*+*θ*)sin(2*θ*)=sin*θ*cos*θ*+cos*θ*sin*θ*=2sin*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*,and letting *α*=*β*=*θ*

, we havecos(*θ*+*θ*)cos(2*θ*)=cos*θ*cos*θ*−sin*θ*sin*θ*=cos2*θ*sin2*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*)=cos2*θ*−sin2*θ*=(1−sin2*θ*)−sin2*θ*

The second variation is:cos(2*θ*)=cos2*θ*−sin2*θ*=cos2*θ*−(1−cos2*θ*)=2cos2*θ*−1

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

in the sum formula givestan(*α*+*β*)tan(*θ*+*θ*)tan(2*θ*)=tan*α*+tan*β*1−tan*α*tan*β*=tan*θ*+tan*θ*1−tan*θ*tan*θ*=2tan*θ*1−tan2*θ*

DOUBLE-ANGLE FORMULAS

The *double-angle formulas* are summarized as follows:sin(2*θ*)cos(2*θ*)tan(2*θ*)=2sin*θ*cos*θ*=cos2*θ*−sin2*θ*=1−2sin2*θ*=2cos2*θ*−1=2tan*θ*1−tan2*θ*(9.3.2)(9.3.3)(9.3.4)

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

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

Example 9.3.1

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

Given that tan *θ*=−34 and *θ*

is in quadrant II, find the following:

- sin(2
*θ*)

cos(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*=−34,such that *θ* is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because *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)216+925*c*=*c*2=*c*2=*c*2=5

Now we can draw a triangle similar to the one shown in Figure 9.3.2

.

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

We see that we to need to find sin *θ* and cos *θ*. Based on Figure 9.3.2, we see that the hypotenuse equals 5, so sin*θ*=35, sin *θ*=35, and cos*θ*=−45

. Substitute these values into the equation, and simplify.

Thus,sin(2*θ*)=2(35)(−45)=−2425 Write the double-angle formula for cosine.

cos(2*θ*)=cos2*θ*−sin2*θ*

Again, substitute the values of the sine and cosine into the equation, and simplify.cos(2*θ*)=(−45)2−(35)2=1625−925=725 Write the double-angle formula for tangent.

tan(2*θ*)=2 tan *θ*1−tan2*θ*

In this formula, we need the tangent, which we were given as tan *θ*=−34

. Substitute this value into the equation, and simplify.tan(2*θ*)=2(−34)1−(−34)2=−321−916=−32(167)=−247

Exercise 9.3.1

Given sin *α*=58,with *θ* in quadrant I, find cos(2*α*)

. Answer

Example 9.3.2

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

Use the double-angle formula for cosine to write cos(6*x*) in terms of *c**o**s*(3*x*)

.

**Solution**cos(6*x*)=cos(3*x*+3*x*)=cos3*x*cos3*x*−sin3*x*sin3*x*=cos23*x*−sin23*x*

**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 9.3.3

: Using the Double-Angle Formulas to Verify an Identity

Verify the following identity using double-angle formulas:1+sin(2*θ*)=(sin*θ*+cos*θ*)2

**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=sin2*θ*+2sin*θ*cos*θ*+cos2*θ*=(sin2*θ*+cos2*θ*)+2sin*θ*cos*θ*=1+2sin*θ*cos*θ*=1+sin(2*θ*)

**Analysis**

This process is not complicated, as long as we recall the perfect square formula from algebra:(*a*±*b*)2=*a*2±2*a**b*+*b*2

where *a*=sin *θ* and *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.

Exercise 9.3.2

Verify the identity: cos4*θ*−sin4*θ*=cos(2*θ*)

. Answer

cos4*θ*−sin4*θ*=(cos2*θ*+sin2*θ*)(cos2*θ*−sin2*θ*)=cos(2*θ*)

Example 9.3.4

: Verifying a Double-Angle Identity for Tangent

Verify the identity: tan(2*θ*)=2cot*θ*−tan*θ*

**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*)=2tan*θ*1−tan2*θ*Double-angle formula=2tan*θ*(1tan*θ*)(1−tan2*θ*)(1tan*θ*)Multiply by a term that results in desired numerator=21tan*θ*−tan2*θ*tan*θ*=2cot*θ*−tan*θ*Use reciprocal identity for 1tan*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 show2tan*θ*1−tan2*θ*Lets work on the right side2cot*θ*−tan*θ*=2cot*θ*−tan*θ*=21tan*θ*−tan*θ*(tan*θ*tan*θ*)=2tan*θ*1tan*θ*(tan*θ*)−tan*θ*(tan*θ*)=2tan*θ*1−tan2*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.

Exercise 9.3.3

Verify the identity: cos(2*θ*)cos*θ*=cos3*θ*−cos*θ*sin2*θ*

. Answer

cos(2*θ*)cos *θ*=(cos2*θ*−sin2*θ*)cos *θ*=cos3*θ*−cos *θ*sin2*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*. Solve for sin^{2}*θ*:

**REDUCTION FORMULAS**

The reduction formulas are summarized as follows:

**Example 9.3.5**

: Writing an Equivalent Expression Not Containing Powers Greater Than 1

Write an equivalent expression for cos4*x* that does not involve any powers of sine or cosine greater than 1

.**Solution**

We will apply the reduction formula for cosine twice.

Howto: 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.

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

## DOUBLE-ANGLE, POWER-REDUCING, AND HALF-ANGLE FORMULAS

✅ Math Formulas ⭐️⭐️⭐️⭐️⭐

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