Corollary 1

Tag 08ID, permanent. Cite it as [Bourbaki, Tag 08ID].

For every integer n one has

\sin \frac{\pi}{n} \sin \frac{2\pi}{n} \ldots \sin \frac{(n-1)\pi}{n} = \frac{n}{2^{n-1}}.

(3)

It suffices to divide both sides of (2) by $\sin z$ and let $z$ tend to 0.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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