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:
\[\tan 2x = \frac{\sin 2x}{\cos 2x}\]
\[=\frac{2\sin x \cos x}{\cos^2 x -\sin^2 x}\]
\[=\frac{2\frac{\sin x}{\cos x}}{1 -\frac{\sin^2 x}{\cos^2 x}}\]
\[=\frac{2\tan x}{1 -\tan^2 x}\]
\[QED\]