Question

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

Answer 1

$\cos 4x=1-8si{n}^{2}x{\cos }^{2}x(\because \cos 2A=2{\cos }^{2}A-1)$
$L.H.S=\cos 4x=\cos 2\left(2x\right)=2{\cos }^{2}2x-1$
$=2(2{\cos }^{2}x-1{)}^2-1$
$=2(4{\cos }^{4}x+1-4{\cos }^{2}x)-1$
$=8{\cos }^{4}x+2-8{\cos }^{2}x-1$
$=8{\cos }^{4}x-8{\cos }^{2}x+1$
$=-8{\cos }^{2}x(1-{\cos }^{2}x)+1=1-8{\cos }^{2}x{\sin }^{2}x$

$\Rightarrow L.H.S=R.H.S.$
Hence proved.