Double angle formulas are trigonometric identities which involve functions of twice an angle. They are got as corollaries of sum formulas like sin (A + B), cos (A + B) and tan (A + B).

These formulas are used in

1. Evaluating functions of double angles.
2. Proving trigonometric identities containing double angle functions.
3. Solving trigonometric equations involving double angles.
4. Analyzing trigonometric graphs.
5. Deriving formulas for higher multiple angles.

Let us list and prove double angle formulas and also solve few problems using them.

## Sine Double Angle Formula

The double angle formula for the sine function is,
sin 2θ = 2 sinθ cosθ which is true for all real values of θ.

Let us verify this formula for θ = 30º

We know from the table for trigonometric functions of standard angles
sin 30 = $\frac{1}{2}$ and cos 30 = $\frac{\sqrt{3}}{2}$

Using double angle formula for sin,
Sin 2(30) = 2 sin30 cos 30 = 2 x $\frac{1}{2}$ x $\frac{\sqrt{3}}{2}$ = $\frac{\sqrt{3}}{2}$

Thus sin 60 = $\frac{\sqrt{3}}{2}$ which is the value given in the table.

## Double Angle Formula for Cosine

The double angle formula for the cosine function is
cos 2θ = cos² θ - sin² θ for all real values of θ.

This formula is also given purely in one function as
cos 2θ = 2cos² θ - 1 or cos 2θ = 1 - 2sin² θ

The last two versions are derived from the first using Pythagorean identity sin² θ + cos² θ =1.

Let us verify this formula for θ = 45º.

From the table, sin 45 = cos 45 = $\frac{\sqrt{2}}{2}$

cos 2(45) = cos² 45 - sin² 45 = $\frac{\sqrt{2}}{2}$ - $\frac{\sqrt{2}}{2}$ = 0

Thus cos 90º = 0 which is also confirmed by the table.

## Tangent Double Angle Formula

The double angle formula for the tangent function is,
tan 2θ = $\frac{2tan\theta }{1-tan^{2}\theta }$ for all real values of θ.

Let us verify the formula for θ = 45º.

tan 2(45) = $\frac{2tan45}{1-tan^{2}45}$

= $\frac{2\times 1}{1-1^{2}}$ = $\frac{2}{0}$ = Division by zero is undefined.

The table also confirms tan 90º as undefined.

## Double Angle Formulas Proofs

We can prove the double angle formulas using sum identities.
sin (α + β) = sinα cosβ + cosα sinβ Sum Identity for sine function
Let α = β = θ
sin (θ + θ) = sinθ cosθ + cosθ sinθ
sin 2θ = 2 sinθ cosθ. Double angle formula for sine function.

cos (α + β) = cosα cosβ - sinα sinβ Sum identity for cosine function.
Let α = β = θ
cos (θ + θ) = cosθ cosθ - sinθ sinθ
cos 2θ = cos2 θ - sin2θ Double angle formula for cosine function.

tan (α + β) = $\frac{tan\alpha +tan\beta }{1-tan\alpha tan\beta }$ Sum identity for tangent function

Let α = β = θ
tan (θ + θ) = $\frac{tan\theta +tan\theta }{1-tan\theta tan\theta }$

tan 2θ = $\frac{2tan\theta }{1-tan^{2}\theta }$ Double angle function for tangent function.

## Double Angle Formulas Examples

1. Evaluating functions involving Double Angle Formulas:
Given cos θ = -$\frac{2}{3}$ $\frac{\pi}{2}$ < θ < π

θ is angle in the second quadrant. Hence the sin function is positive and tan function negative.
sin θ = $\sqrt{1-cos^{2}\theta }$

= $\sqrt{1-(-\frac{2}{3})^{2}}$ = $\sqrt{1-\frac{4}{9}}$ = $\frac{\sqrt{5}}{3}$

tan θ = $\frac{sin \theta}{cos \theta}$ = -$\frac{\sqrt{5}}{2}$.
Now let us evaluate the double angle functions.

sin 2θ = 2 sinθ cosθ = 2$\frac{\sqrt{5}}{3}$ x -$\frac{2}{3}$

= -$\frac{4\sqrt{5}}{9}$

cos 2θ = cos2 θ - sin2 θ

= (-$\frac{2}{3}$)$^{2}$ - ($\frac{\sqrt{5}}{3}$)$^{2}$

= $\frac{4}{9}$ - $\frac{5}{9}$

= -$\frac{1}{9}$.

tan 2θ = $\frac{2tan\theta }{1-tan^{2}\theta }$

= $\frac{2\frac{-\sqrt{5}}{2}}{1-(\frac{-\sqrt{5}}{2})^{2}}$

= $\frac{-\sqrt{5}}{1-\frac{5}{4}}$

= 4√5.

2. Verifying an Identity:
Verify that 4 cos2 θ - sin2 2θ = 4 cos2
For verifying identities we simplify the left side of the equation to get the right side.
Left side = 4 cos2 θ - sin2 2θ = 4 cos2 θ - (2 sinθ cosθ)2. Double angle formula for sin.
= 4 cos2 θ - 4 sin2 θ cos2 θ
= 4 cos2 θ (1 - sin2 θ) Factoring.
= 4 cos2 θ cos2 θ Pythagorean identity
= 4 cos4 θ Right side of the equation.
The identity is thus verified.

3. Solving a trigonometric equation:
Solve the trigonometric equation cos 2x + sin x = 0 and find the exact solution in the interval [0. 2π)
cos 2x + sin x = 0
1 - 2 sin2 x + sin x = 0 Double angle formula for cosine.
2 sin2 x - sin x - 1 = 0
(2 sin x + !)(sin x - 1) = 0 Quadratic expression factored.
Applying zero factor property
2 sin x + 1 = 0 or sin x - 1 = 0
2 sin x = -1 or sin x = 1
sin x = -$\frac{1}{2}$

x = sin-1 ($\frac{1}{2}$) or x = sin-1 1

x = $\frac{4\pi }{3}$, $\frac{5\pi }{3}$ or x = $\frac{\pi }{2}$

Thus the solutions for x in the given interval are $\frac{4\pi }{3}$, $\frac{5\pi }{3}$ and $\frac{\pi }{2}$.