Question

Prove that $\cos 4x=1-8{\sin }^{2}x{\cos }^{2}x$

Answer 1

Identities:
$\cos 2\theta =1-2{\sin }^2\theta$ (1)
$\sin 2\theta =2\sin \theta \cos \theta$ (2)
using (1) :-

$\cos 4x$ = $1-2(\sin 2x{)}^2$

using (2) :-
$=1-2(2\sin x\cos x{)}^2$
$=1-8{\sin }^{2}x{\cos }^{2}x$
Hence, proved.