## Double Angle Formulae Proof using Euler's Equation

The double angle formulae are:

$\sin 2x = 2\sin x \cos x$

$\cos 2x = \cos^2 x - \sin^2 x$

$\tan 2x = \frac{2\tan x}{1-\tan^2 x}$

## Proof for sin and cos

Using Euler's Equation (a proof of which can be found here), $$e^{ix}=\cos x +i\sin x$$, substitute $$2x$$ into the equation:

$①:e^{2ix}=\cos 2x +i\sin 2x$

But notice that if we square Euler's Equation, we get:

$(e^{ix})^2=(\cos x +i\sin x)^2$

$e^{2ix}=\cos^2 x +2i\cos x \sin x + (i\sin x)^2$

$②:e^{2ix}=\cos^2 x +2i\cos x \sin x - \sin^2 x$

We can now equate the right-hand sides of $$①$$ and $$②$$ since both equal $$e^{2ix}$$:

$\cos 2x +i\sin 2x=\cos^2 x +2i\cos x \sin x - \sin^2 x$

Comparing real and imaginary parts gives:

$③:\cos 2x=\cos^2 x - \sin^2 x$

$④:\sin 2x=2\cos x \sin x$

$QED$

## Proof for tan

Using the results that we just proved:

$\tan 2x = \frac{\sin 2x}{\cos 2x}$

$=\frac{2\sin x \cos x}{\cos^2 x -\sin^2 x}$

$=\frac{2\frac{\sin x}{\cos x}}{1 -\frac{\sin^2 x}{\cos^2 x}}$

$=\frac{2\tan x}{1 -\tan^2 x}$

$QED$