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

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

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This is a complex analysis problem. Let $z=cos(2x) + isin(2x)$, we have $$|z^n – 1 |=|2sin(nx)|$$

On the other hand,
$$z^n – 1 = (z-1)(z-a)(z-a^2)…(z-a^{n-1}),$$
where $a=cos(2\pi/n) + i\sin(2\pi/n)$ is the $n$-th unit root. If we take norms on both sides, we get the other half of the formula.

There is a sign problem, which is easy to hand. ( for example, just look at value for $x\in (0, \pi/n)$; or instead of taking norms, write the terms as the product of a real number times a unit complex number)