Proposition 1

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

For every complex number $z$ and every integer $n$ one has

\sin nz = 2^{n-1} \sin z \sin \left(z + \frac{\pi}{n}\right) \sin \left(z + \frac{2\pi}{n}\right) \ldots \sin \left(z + \frac{(n-1)\pi}{n}\right).

(2)

Indeed, one can write

\sin nz = \frac{e^{niz} - e^{-niz}}{2i} = \frac{e^{-niz}(e^{2niz} - 1)}{2i} = \frac{e^{-niz}(e^{2iz} - 1)(e^{2iz} - e^{-2i\pi/n}) \ldots (e^{2iz} - e^{-2(n-1)i\pi/n})}{2i} = A \sin z \sin \left( z + \frac{\pi}{n} \right) \sin \left( z + \frac{2\pi}{n} \right) \ldots \sin \left( z + \frac{(n-1)\pi}{n} \right)

where

A = (2i)^{n-1} e^{-\frac{i\pi}{n}(1+2+\ldots+(n-1))} = (2i)^{n-1} e^{-i(n-1)\frac{\pi}{2}} = 2^{n-1}.

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

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