Proof of $\sin nx=2^{n-1}\prod_{k=0}^{n-1} \sin\left( x + \frac{k\pi}{n} \right)$

Solution 1:

There is a nice argument based on Weierstrass products.

The sine function has its (simple) zeroes at $\pi\mathbb{Z}$ and

$$\frac{\sin x}{x}=\prod_{m\geq 1}\left(1-\frac{x^2}{n^2\pi^2}\right)\tag{1}$$ holds. The $\sin(n x)$ function has its zeroes at $\frac{\pi}{n}\mathbb{Z}=\left(\pi Z\right)\cup\left(\pi \mathbb{Z}+\frac{1}{n}\right)\cup\ldots\cup\left(\pi\mathbb{Z}+\frac{n-1}{n}\right)$, so by separating the zeroes according to their residue class $\pmod{\pi}$ and using $(1)$ and $$ \prod_{k=1}^{n-1}\sin\frac{k\pi}{n}=\frac{2n}{2^n}\tag{2}$$ your identity easily follows.

With a similar argument, you may check that your RHS and LHS have the same value at $x=0$ and by applying $\frac{d}{dz}\log(\cdot)$ to both sides you get two meromorphic functions with the same Eisenstein series, by Herglotz' trick.