Answer
$\cos 4x=1-8\sin^2 x+8\sin^4 x$
Work Step by Step
Start with the left side:
$\cos 4x$
Rewrite $4x$ as $2*2x$:
$=\cos (2*2x)$
Use the Double-Angle Formula $\cos (2u)=1-2\sin^2 u$ where $u=2x$:
$=1-2\sin^2(2x)$
Use the Double-Angle Formula $\sin(2x)=2\sin x\cos x$:
$=1-2(2\sin x\cos x)^2$
Simplify:
$=1-2(4\sin^2 x\cos^2 x)$
$=1-8\sin^2 x\cos ^2 x$
Rewrite $\cos^2 x$ as $1-\sin^2 x$:
$=1-8\sin^2 x(1-\sin^2 x)$
$=1-8\sin^2 x+8\sin^4 x$
Since this equals the right side, the identity has been proven.
