Answer
$$(\cos2x-\sin2x)^2=1-\sin4x$$
Since 2 sides are equal as proved above, the equation is an identity.
Work Step by Step
$$(\cos2x-\sin2x)^2=1-\sin4x$$
The left side would be tackled first here.
$$X=(\cos2x-\sin2x)^2$$
Apply the expansion $(a-b)^2=a^2-2ab+b^2$ for $a=\cos2x$ and $b=\sin2x$.
$$X=\cos^22x-2\cos2x\sin2x+\sin^22x$$
$$X=(\cos^22x+\sin^22x)-(2\cos2x\sin2x)$$
- From Pythagorean Identity: $\cos^22x+\sin^22x=1$
- And from Double-Angle Identity: $2\cos2x\sin2x=\sin(2\times2x)=\sin4x$ (for $\sin2\theta=2\sin\theta\cos\theta$)
$$X=1-\sin4x$$
Therefore, $$(\cos2x-\sin2x)^2=1-\sin4x$$
Since 2 sides are equal, the equation is an identity.