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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).

  • 1
    $\begingroup$ Maybe using polynomial which involve $n$-th roots of unity. $\endgroup$ Jun 16, 2012 at 21:12
  • $\begingroup$ ...which probably can be easily found ... $n/2^n$ $\endgroup$
    – GEdgar
    Jun 16, 2012 at 21:37
  • $\begingroup$ @GEdgar I had a proof of that somewhere. I think anon proved it already, too. $\endgroup$
    – Pedro
    Jun 16, 2012 at 21:38
  • 1
    $\begingroup$ This is duplicate I have seen before. But am unable to find it. $\endgroup$
    – user17762
    Jun 16, 2012 at 21:44
  • $\begingroup$ $n/2^{n-1}$ is correct $\endgroup$
    – GEdgar
    Jun 16, 2012 at 21:56

1 Answer 1


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 denominator. 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 $$\prod_{k=1}^{n-1} \sin(k\pi/n) = \frac{n}{2^{n-1}}.$$

  • $\begingroup$ Excellent answer, thank you. $\endgroup$ Feb 20, 2021 at 21:17

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