# Double Angle Formula (Sine, Cosine, and Tangent)

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Ray is a licensed engineer in the Philippines. He loves to write about mathematics and civil engineering.

## What Is a Double Angle Formula?

The double angle formula contains the expression 2u formed by using the sum identities, whereas identities for f(2a), where f is a trigonometric function, are easy to derive. The double angle identities are used to write a trigonometric expression in terms of a single trigonometric function. These formulas can help in evaluating trigonometric function values for angles other than 30°, 45°, and 60°, and are even multiples of these special angles. They are a good help in finding the exact values of many functions without needing to know the actual value of the special angle.

There are double angle formulas for each basic trigonometric identity - sin double angle formula, cosine double angle formula, and double angle formula for the tangent. Below is the summary of the double angle identities.

1. sin (2u) = 2 sin (u) cos (u)
2. cos (2u) = cos2 (u) - sin2 (u)
3. cos (2u) = 1 - 2 sin2 (u)
4. cos (2u) = 2 cos2 (u) - 1
5. tan (2u) = 2 tan (u) / 1 - tan2 (u)

## Double Angle Formulas Proof

We may prove each double angle formula by letting v = u in the appropriate addition formula. For instance, let us use the addition formula for sin (u + v).
sin (2u) = sin (u + u)
sin (2u) = sin (u) cos (u) + cos (u) sin (u)
sin (2u) = 2 sin (u) cos (u)

Next, let us use the formula for cos (u + v) and assume v = u.
cos (2u) = cos (u + u)
cos (2u) = cos (u) cos (u) - sin (u) sin (u)
cos (2u) = cos2 (u) - sin2 (u)

In obtaining the other two forms for the cos double angle formula, use the fundamental identity sin2 (u) + cos2 (u) = 1. Use this identity for cos (2u). Similarly, the process flows the same if we substitute sin2 (u) instead of cos2 (u).
cos (2u) = cos2 (u) - sin2 (u)
cos (2u) = (1 - sin2 (u)) - sin2 (u)
cos (2u) = 1 - 2 sin2 (u)

cos (2u) = cos2 (u) - (1 - cos2 (u))
cos (2u) = 2 cos2 (u) - 1

For the double angle formula for tangent tan (2u), we may obtain the identity by letting v = u in the formula for tan (u + v).
tan (2u) = tan (u) + tan (u) / 1 - tan (u) tan (u)
tan (2u) = 2 tan (u) / 1 - tan2 (u)

## Example 1: Using Double Angle Formulas

If sin (α) = ⅘ and α is an acute angle, find the exact values of cos (2α).

Solution

If we regard α as an acute angle of a right triangle. We obtain cos α = ⅗. We next substitute in the double angle formula for cosine.

cos (2α) = cos2 (α) - sin2 (α)

cos (2α) = cos2 (α) - sin2 (α)

cos (2α) = (3/5)2- (4/5)2

Scroll to Continue

cos (2α) = 9/25 - 16/25

cos (2α) = -7/25

Therefore, the exact value of cos (2α) is -7/25.

## Example 2: Changing the Form of Cos (3θ)

Express cos (3θ) in terms of cos (θ).

Solution

First, apply the addition formula and expand 3θ to 2θ and θ. Then, apply double angle formulas for sine and cosine. Once done, multiply each like term to shorten the equation. Notice that you can utilize the Pythagorean identity sin2θ + cos2θ = 1. Finally, simplify the equation to its lowest term.

cos (3θ) = cos (2θ) + cos (θ)

cos (3θ) = cos (2θ) cos (θ) - sin (2θ) sin (θ)

cos (3θ) = (2 cos2 (θ) - 1)cos (θ) - (2 sin (θ) cos (θ)) sin (θ)

cos (3θ) = 2 cos3 (θ) - cos (θ) - 2 cos (θ) sin2 (θ)

cos (3θ) = 2 cos3 (θ) - cos (θ) -- 2 cos (θ) (1 - cos2 (θ))

cos (3θ) = 4 cos3 (θ) - 3 cos (θ)

Therefore, the equation cos (3θ) in terms of cos (θ) is 4cos3 (θ) - 3 cos (θ).

## Example 3: Simplifying a Trigonometric Expression

Write 4 sin (6θ) cos (6θ) in terms of a single trigonometric function.

Solution

For this problem, let u = 6θ and treat them as a single term. Then, use the sin double angle formula sin (2u) = 2 sin (u) cos (u) in simplifying the trigonometric expression.

4 sin (6θ) cos (6θ) = 2 [2 sin (6θ) cos (6θ)]

4 sin (6θ) cos (6θ) = 2 [sin (2 (6θ))]

4 sin (5θ) cos (5θ) = 2 sin (12θ)

The single trigonometric expression of 4 sin (5θ) cos (5θ) is 2 sin (12θ).

## Example 4: Evaluating a Trigonometric Function Using a Double Angle Formula

If sin (α) = ⅘ and 0° < α < 90°, find the exact value of sin (2α).

Solution

For this problem, use the double angle formula for sine which is sin (2u) - 2 sin (u) cos (u). Find the exact value of cos (α) by substituting for sin (α) in the equation sin2 (α) + cos2 (α) = 1. Take note that the question mentions a parameter on the value of α which is 0° < α < 90°. It means that cos (α) > 0 if α is in Quadrant I. Substitute the value of α = ⅘ to the equation.

sin2 (α) + cos2 (α) = 1

cos2 (α) = 1 - sin2 (α)

cos (α) = √1 - sin2 (α)

cos (α) = √1 - (⅘)2

cos (α) = ⅗

Substitute the values of sin (α) and cos (α) in the double angle formula for sin 2α.
sin (2α) = 2 sin (α) cos (α)
sin (2α) = 2 (⅘) (⅗)
sin (2α) = 24/25

The exact value of sin (2α) is 24/25.

## Example 5: Verifying an Identity Using the Double Angle Formula for Cosecant

Verify the identity csc (2α) = ½ (tan α + cot α).

Solution

In verifying the given identity, let us work on the right-hand side of the equation. Recall than tan (α) = sin (α) / cos (α) and cot (α) = cos (α) / sin (α).

½ (tan α +cot α) = ½ (sin α / cos α + cos α / sin α)

½ (tan α +cot α) = ½ (sin2 α + cos2 α / cos α sin α)

½ (tan α +cot α) = 1 / (2 cos α sin α)

½ (tan α +cot α) = 1 / sin (2α)

½ (tan α +cot α) = csc (2α)

Therefore, by verifying the identity we can conclude that csc (2α) is equal to ½ (tan α +cot α).

## Example 6: Using Double Angle Identities in Finding the Exact Values of Trigonometric Functions

If cos (α) = 2/5, find the values of sin (2α) given sin (α) < 0.

Solution

Given the value of cos (α), solve for sin (α) using the Pythagorean identity and substitute it to the sin double angle formula. Given the parameter, sin (α) < 0, make sure to use the negative result.

sin2 (α) + cos2 (α) = 1

sin2 (α) + (⅖)2 = 1

sin2 (α) = 1 - (⅖)2

sin (α) = √1 - (⅖)2

sin (α) = -√21 / 5

Use the double angle formula for the sine function and substitute the obtained value of sin (α) and the given value of cos (α) to the equation. Simplify the equation.

sin (2α) = 2 sin α cos α

sin (2α) = 2 (-√21 / 5) (⅖)

sin (2α) = -4√21 / 25

The exact value of sin (2α) given that cos (α) = ⅖ is -4√21 / 25.

## Example 7: Finding Exact Values Using Double Angle Identities

If sin (β) = -⅘ and cos (β) < 0, find sin (2β), cos (2β), and tan (2β).

Solution

Given sin (β) = -⅘, solve for the value of cos (β) using the Pythagorean identity. This step will serve as the preliminary requirement in solving for the required values. Substitute the value -⅘ and simplify the resulting equation. Take note that given the parameter cos (β) < 0, the value of cos (β) is negative.

sin2 (β) + cos2 (β) = 1

(-⅘)2 + cos2 (β) = 1

cos2 (β) = 1 - (-⅘)2

cos2 (β) = 1 - 16/25

cos (β) = √1 - 16/25

cos (β) = -⅗

Solve for the value of sin (2β) using the double angle identity for the sine function. Substitute sin (β) = -⅘ and cos (β) = -⅗.

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

sin (2β) = 2 sin (β) cos (β)

sin (2β) = 2 (-⅘) (-⅗)

sin (2β) = 24/25

Solve for the value of cos (2β) using the double angle formula for the cosine function. Again, substitute sin (β) = -⅘ and cos (β) = -⅗ and simplify.

cos (2x) = cos2 (x) - sin2 (x)

cos (2β) = cos2 (β) - sin2 (β)

cos (2β) = (-⅗)2- (-⅘)2

cos (2β) = -7/25

Finally, solve for the exact value of tan (2β) using the tangent trigonometric identity tan (x) = sin (x) / cos (x). For this part, let θ = 2β and then substitute sin (2β) = 24/25 and cos (2β) = -7/25. Aside from this method, you can also use the double angle formula for a tangent.

tan (θ) = sin (θ) / cos (θ)

tan (2β) = sin (2β) / cos (2β)

tan (2β) = 24/25 / -7/25

You can also try to solve for the value of angle β and solve it using a double angle formula calculator.

The exact values of sin (2β), cos (2β), and tan (2β) are 24/25, -7/25, and -24/7, respectively.

## Example 8: Verifying Trigonometric Identities

Verify the identity (sin (x) - cos (x))2 = 1 - sin (2x).

Solution

In verifying the given identity, start simplifying the left-hand side of the equation. Expand the equation by using the formula in solving squares of binomials. Once expanded, group the sin2 (x) and cos2 (x) terms.

(sin (x) - cos (x))2 = sin2 (x) - 2 sin (x) cos (x) + cos2 (x)

(sin (x) - cos (x))2 = sin2 (x) + cos2 (x) - 2 sin (x) cos (x)

Recognize the Pythagorean identity from the obtained equation. Since a part of the resulting equation resembles the Pythagorean identity sin2 (x) + cos2 (x) = 1. Then, simplify the equation by using the double angle formula for sine and simplify.

(sin (x) - cos (x))2 = 1 - 2 sin (x) cos (x)

(sin (x) - cos (x))2 = 1 - sin (2x)

Therefore, the identity (sin (x) - cos (x))2 = 1 - sin (2x) is valid.

## Example 9: Verifying Multiple Angle Identities Using Double Angle Identities

Verify the identity cos (3x) = (1 - 4sin2 (x)) cos(x).

Solution

For this identity, simplify the left-hand side of the equation using multiple angle identities and double angle identities. Write the cosine of a sum identity and let A = 2x and B = x.

cos (A + B) = cos (A) cos (B) - sin (A) sin (B)

cos (2x + x) = cos (2x) cos (x) - sin (2x) sin (x)

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

Simplify the equation by using a double angle formula.

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

cos (3x) = (1 - 2sin2 (x)) cos (x) - 2 sin (x) cos (x) sin (x)

cos (3x) = cos (x) - 2 sin2 (x) cos (x) - 2 sin2 (x) cos (x)

cos (3x) = cos (x) - 4sin2 (x) cos (x)

Therefore, the identity cos (3x) = (1 - 4sin2 (x)) cos(x) is valid.

## Example 10: Solving a Trigonometric Equation Using Double Angle Formulas

Solve the given equation using a double angle formula.
sin (θ) cos (θ) = -1/2 where 0 ≤ θ＜2π

Solution

It is a double angle formula example, showing that the left-hand side of the equation is in the form 2 sin (θ) cos (θ) = sin (2θ), except for a factor of two. To eliminate this, multiple each side by two.

sin (θ) cos (θ) = -½

2 sin (θ) cos (θ) = -1

sin (2θ) = -1

The argument from the obtained equation is 2θ. Therefore, you must write all the solutions of this equation and pick those inside the interval [0, 2π).

sin (3π/2 + 2πk) = -1

2θ = 3π/2 + 2πk

θ = 3π/4 + πk

Substitute k values of k = -1, k = 0, k = 1, and k = 2 to the resulting theta equation.

θ = 3π/4 + πk

θ = 3π/4 + π(-1)

θ = -π/4

θ = 3π/4 + πk

θ = 3π/4 + π(0)

θ = 3π/4

θ = 3π/4 + πk

θ = 3π/4 + π(1)

θ = 7π/4

θ = 3π/4 + πk

θ = 3π/4 + π(2)

θ = 11π/4

The solutions in the interval [0, 2π) are θ = 3π/4 and θ = 7π/4.

The solution set of the equation sin(θ) cos(θ) = -1/2 given the parameter 0 ≤ θ＜2π is {3π/4, 7π/4}.

## Example 11: Projectile Motion Using Double Angle Identities

An object is propelled upward at an angle θ to the horizontal with an initial velocity of v0 feet per second. If you ignore the air resistance, then the range R, the horizontal distance that the object travels, is given by the function shown below.

R(θ) = 1/16 v02 sin (θ) cos (θ)

a. Show that R(θ) = 1/32 v02 sin (2θ)

b. Find the angle θ for which R is a maximum.

Solution

In showing that R(θ) = 1/32 v02 sin (2θ), start by rewriting the given equation for the range using the sin double angle formula sin (2u) = 2 sin (u) cos (u).

R(θ) = 1/16 v02 sin (θ) cos (θ)

R(θ) = 1/16 v02 (2 sin (θ) cos (θ) / 2)

R(θ) = 1/32 v02 (2 sin (θ) cos (θ))

R(θ) = 1/32 v02 sin (2θ)

For letter b, the largest value for the range R is solvable using an analogy on the double angle formulas. For a fixed initial speed v0, the angle θ of inclination to the horizontal determines the value of the range. Since a sine function has the large value at 1, occurring when the argument 2θ is 90°, it follows that for the maximum R, we must have 2θ = 90° or equivalent to θ = 45°.

## Example 12: Derivation of Triple Angle Formula

Rewrite sin 3x in terms of sin x.

Solution

Rewrite the given equation as a sum using the sum formula in trigonometry. Apply and simplify the equations using the different trig identities such as double angle formulas and Pythagorean identity.

sin 3x = sin (2x + x)

sin 3x = sin 2x cos x + cos 2x sin x

sin 3x = 2sin x cos x cos x + (1 - 2sin2 x) sin x

sin 3x = 2sin x cos2 x + sin x - 2 sin3 x

sin 3x = 2sin x (1 - sin2 x) + sin x - 2 sin3 x

sin 3x = 2sin x - 2 sin3 x + sin x - 2sin3 x

sin 3x = 3 sin x - 4 sin3 x