Prove that $\sin(nx) \cos((n+1)x)-\sin((n-1)x)\cos(nx) = \sin(x) \cos(2nx)$

Question

Question:
Prove that $\sin(nx) \cos((n+1)x)-\sin((n-1)x)\cos(nx) = \sin(x) \cos(2nx)$ for $n \in \mathbb{R}$.

My attempts:

I initially began messing around with the product to sum identities, but I couldn't find any way to actually use them.
I also tried compound angles to expand the expression, but it became too difficult to work with.

Any help or guidance would be greatly appreciated

Answer 1

The left-hand side is$$\begin{align}&\sin nx(\cos nx\cos x-\sin nx\sin x)-(\sin nx\cos x-\cos nx\sin x)\cos nx\\&=(\cos^2nx-\sin^2nx)\sin x\\&=\cos 2nx\sin x.\end{align}$$