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