# Derivation of the Double Angle Formulas

The Double Angle Formulas can be derived from Sum of Two Angles listed below:

$\sin (A + B) = \sin A \, \cos B + \cos A \, \sin B$ → Equation (1)

$\cos (A + B) = \cos A \, \cos B - \sin A \, \sin B$ → Equation (2)

$\tan (A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \, \tan B}$ → Equation (3)

Let θ = A = B; Equation (1) will become

$\sin (\theta + \theta) = \sin \theta \, \cos \theta + \cos \theta \, \sin \theta$

Let θ = A = B; Equation (2) will become

$\cos (\theta + \theta) = \cos \theta \, \cos \theta - \sin \theta \, \sin \theta$

$\cos 2\theta = \cos^2 \theta - \sin^2 \theta$ → Equation (4)

The Pythagorean Identity sin^{2}θ + cos^{2}θ = 1 can be taken as sin^{2}θ = 1 - cos^{2}θ and Equation (4) will become...

$\cos 2\theta = \cos^2 \theta - (1 - \cos^2 \theta)$

$\cos 2\theta = 2\cos^2 \theta - 1$

sin^{2}θ + cos^{2}θ = 1 can also be taken as cos^{2}θ = 1 - sin^{2}θ and Equation (4) will become...

$\cos 2\theta = (1 - \sin^2) - \sin^2 \theta$

$\cos 2\theta = 1 - 2\sin^2 \theta$

For easy reference, below is the summary for cos 2θ

$\cos 2\theta = 2\cos^2 \theta - 1$

$\cos 2\theta = 1 - 2\sin^2 \theta$

Let θ = A = B; Equation (3) will become

$\tan (\theta + \theta) = \dfrac{\tan \theta + \tan \theta}{1 - \tan \theta \, \tan \theta}$