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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:

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

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).


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

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

Use the formula $\sin(x) = \frac{1}{2i}(e^{ix}-e^{-ix})$ to get \begin{align*} \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). \end{align*} The first product simplifies to $$e^{\sum_{k=1}^{n-1} k\pi i/n} = e^{(n-1)\pi i/2} = i^{n-1}$$ which cancels out with the $i^{n-1}$ in the numerator. 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 $$\prod_{k=1}^{n-1} \sin(k\pi/n) = \frac{n}{2^{n-1}}.$$


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