#### Answer

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

#### Work Step by Step

Start with the left side:
$(\cos 2x-\sin 2x)^2$
Expand:
$=\cos^2 2x-2\cos 2x\sin 2x+\sin^2 2x$
Rearrange terms:
$=\cos^2 2x+\sin^2 2x-2\cos 2x\sin 2x$
Use the identities $\cos^2 \theta+\sin^2\theta=1$ and $2\cos\theta\sin\theta=\sin 2\theta$, where $\theta=2x$:
$=1-\sin (2*2x)$
$=1-\sin 4x$