<s>[INST] How to prove those "curious identities"? How to prove
$$ \prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}}$$
and
$$ \prod_{k=1}^{n-1} \cos\left(\frac{k\pi}{n}\right) = \frac{\sin(\pi n/2)}{2^{n-1}}$$
 [/INST] Denote $w = e^{i \pi/n}$. We have
$$\prod_{k = 1}^{n-1} \sin \left(\frac{k\pi}{n}\right)= \prod_{k = 1}^{n-1} \frac{w^k - w^{-k}}{2i} = \frac{1}{2^{n-1}} \prod_{k = 1}^{n-1} \frac{w^k}{i} (1-w^{-2k})$$
Since we have
$$\sum_{k = 0}^{n-1} x^k = \prod_{k = 1}^{n-1} (x-w^{2k})$$
Setting $x=1$ yields
$$\prod_{k = 1}^{n-1} (1-w^{2k}) = n$$
So we get
$$\prod_{k = 1}^{n-1} \sin \left(\frac{k\pi}{n}\right)= \frac{n}{2^{n-1}} \frac{w^{n(n-1)/2}}{i^{n-1}} = \frac{i^{n-1}}{i^{n-1}} \frac{n}{2^{n-1}} = \frac{n}{2^{n-1}}$$
I guess (but did not check) that the same kind of reasoning gives the one with $\cos$.
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