Question

Prove that $cos4x=1-8si{n}^{2}x.co{s}^{2}x$

Answer 1

L.H.S. = $\cos 4x=\cos 2\left(2x\right)$

as, $\cos 2A=1-2{\sin }^{2}A$

$\therefore$ $\cos 2\left(2x\right)=1-2{\sin }^{2}2x$ = $1-2(\sin 2x{)}^2$

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

$\Rightarrow$ $1-2\left(4{\sin }^{2}x{\cos }^{2}x\right)$
$\Rightarrow$ $1-8({\sin }^{2}x.{\cos }^{2}x)$ =R.H.S