# How do you prove sin4x = 4sinxcos^3x - 4sin^3xcosx?

May 28, 2015

I'll start from the double angle identities:

$\cos 2 \theta = {\cos}^{2} \theta - {\sin}^{2} \theta$

$\sin 2 \theta = 2 \sin \theta \cos \theta$

Then:

$\sin 4 x = 2 \sin 2 x \cos 2 x$

$= 2 \left(2 \sin x \cos x\right) \left({\cos}^{2} x - {\sin}^{2} x\right)$

$= 2 \left(2 \sin x \cos x {\cos}^{2} x - 2 \sin x \cos x {\sin}^{2} x\right)$

$= 4 \sin x {\cos}^{3} x - 4 {\sin}^{3} x \cos x$