# Evaluation of a product of sines [duplicate]

Possible Duplicate:
Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$

I am looking for a closed form for this product of sines:

$$\sin \left(\frac{\pi}{n}\right)\,\sin \left(\frac{2\pi}{n}\right)\dots\sin \left(\frac{(n-1)\pi}{n}\right),$$

where $n$ is a fixed integer. I would like to see here a strategy that hopefully can be generalized to similar cases, not just the result (which probably can be easily found).

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Maybe using polynomial which involve $n$-th roots of unity. – Davide Giraudo Jun 16 '12 at 21:12
...which probably can be easily found ... $n/2^n$ – GEdgar Jun 16 '12 at 21:37
@GEdgar I had a proof of that somewhere. I think anon proved it already, too. – Peter Tamaroff Jun 16 '12 at 21:38
This is duplicate I have seen before. But am unable to find it. – user17762 Jun 16 '12 at 21:44
$n/2^{n-1}$ is correct – GEdgar Jun 16 '12 at 21:56

## marked as duplicate by Hans Lundmark, Martin Sleziak, t.b., Nate Eldredge, Zev ChonolesJun 18 '12 at 2:17

Use the formula $\sin(x) = \frac{1}{2i}(e^{ix}-e^{-ix})$ to get $$\prod_{k=1}^{n-1} \sin(k\pi/n) = \left(\frac{1}{2i}\right)^{n-1}\prod_{k=1}^{n-1} \left(e^{k\pi i/n} - e^{-k\pi i/n}\right) = \left(\frac{1}{2i}\right)^{n-1}\left(\prod_{k=1}^{n-1} e^{k\pi i/n} \right) \prod_{k=1}^{n-1} \left(1-e^{-2k\pi i/n} \right).$$ The first product simplifies to $$e^{\sum_{k=1}^n k\pi i/n} = e^{(n-1)\pi i/2} = i^{n-1}$$ which cancels out with the $i^{n-1}$ in the denumerator. The second product can be recognized as the polynomial $f(X) = \prod_{k=1}^{n-1} (X-e^{-2k\pi i/n})$ evaluated at $X = 1$. The roots of this polynomial are the non-trivial $n$-th roots of unity, so $f(X) = \frac{X^n-1}{X-1} = 1+X+X^2+\ldots+X^{n-1}$. Plugging in $1$ for $X$ yields $$\prod_{k=1}^{n-1} \left(1-e^{-2k\pi i/n} \right) = f(1) = n.$$ Altogether, we have $$\prod_{k=1}^{n-1} \sin(k\pi/n) = \frac{n}{2^{n-1}}.$$