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:

\[\tan 2x = \frac{\sin 2x}{\cos 2x}\]

\[=\frac{2\sin x \cos x}{\cos^2 x -\sin^2 x}\]

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

\[=\frac{2\tan x}{1 -\tan^2 x}\]

\[QED\]

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