Answer
$$(\cos2x+\sin2x)^2=1+\sin4x$$
The equation is an identity. The proof is shown below.
Work Step by Step
$$(\cos2x+\sin2x)^2=1+\sin4x$$
We tackle from the left side.
$$X=(\cos2x+\sin2x)^2$$
Apply the expansion $(a+b)^2=a^2+2ab+b^2$ for $a=\cos2x$ and $b=\sin2x$.
$$X=\cos^22x+2\cos2x\sin2x+\sin^22x$$
$$X=(\cos^22x+\sin^22x)+(2\cos2x\sin2x)$$
- Pythagorean Identity: $\cos^22x+\sin^22x=1$
- Double-Angle Identity: $2\cos2x\sin2x=\sin(2\times2x)=\sin4x$ (for $\sin2\theta=2\sin\theta\cos\theta$)
$$X=1+\sin4x$$
Thus, $$(\cos2x+\sin2x)^2=1+\sin4x$$
The equation is an identity.
