trigonometry - Express the trigonometric identity $cos nx$ as an infinite series of $sin^2 x/2$.

The trigonometric identity $\cos nx$ is expressed as an infinite series only in terms of $\sin^2 \frac{x}{2}$ as follows.

$$\cos nx = 1 + \sum_{l=1}^n (-1)^l \frac{2^{2l}}{(2l)!} \prod_{k=0}^{l-1} (n^2 - k^2) \sin^{2l} \frac{x}{2}$$
This is given in the literature but the authors have not provided any proof to this equation. I have tried the proof using the Euler formula and Binomial theorem, but could not succeed. Can anyone please provide the proof for this equation, or at least a guide on how to prove this ?

Thank You.

Answer

The Chebyshev polynomials verify $T_n(\cos x) = \cos nx$. This means that $\cos nx$ can be written as a finite sum of powers of $\cos x$. Now simply use the identity $\cos x = 1-2\sin^2(x/2)$ to obtain that $\cos nx$ is a polynomial of $\sin^2(x/2)$.