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

Using n_th root of unity

$$\Large\left(e^{\frac{2ki\pi}{n}}\right)^{n} = 1$$

Prove that

$$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$$

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Incidentally, the proof given in fiktor's answer below can be modified to show that $\sin nx=2^{n-1}\prod_{k=0}^{n-1} \sin\left( x + \frac{k\pi}{n} \right)$, a very pretty multiple-angle identity which is not as widely know as it deserves to be. Dividing by $\sin x$ and letting $x\to 0$ reduces that identity to the one in the question. –  Hans Lundmark Oct 31 '10 at 14:38
And here's a kill-a-mosquito-with-a-cannon proof of the identity in my previous comment: combine Gauss's multiplication formula for the gamma function, $\Gamma(nx) = \frac{n^{nx-1/2}}{(2\pi)^{(n-1)/2}} \prod_{k=0}^{n-1} \Gamma(x+\frac{k}{n})$, with Euler's reflection formula $\Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin(\pi x)}$. –  Hans Lundmark Oct 31 '10 at 16:43
And another comment... I just ran into this on Wikipedia: en.wikipedia.org/wiki/Morrie%27s_law –  Hans Lundmark Nov 4 '10 at 8:55

$$P=\prod_{k=1}^{n-1}\sin(k\pi/n)=(2i)^{1-n}\prod_{k=1}^{n-1}(e^{ik\pi/n}-e^{-ik\pi/n})=(2i)^{1-n}e^{-i\pi n(n-1)/(2n)}\prod_{k=1}^{n-1}(e^{2ik\pi/n}-1)=$$ $$(-2)^{1-n}\prod_{k=1}^{n-1}(\xi^k-1)=2^{1-n}\prod_{k=1}^{n-1}(1-\xi^k),$$ where $\xi=e^{2i\pi/n}$. Now note, that $x^n-1=(x-1)\sum_{k=0}^{n-1}x^k$ and $x^n-1=\prod_{k=0}^{n-1} (x-\xi^k)$, thus cancelling $x-1$ we have $\prod_{k=1}^{n-1} (x-\xi^k) =\sum_{k=0}^{n-1}x^k$. Substituting $x=1$ we have $\prod_{k=1}^{n-1} (1-\xi^k)=n$. Therefore $P=n2^{1-n}$.

Edit:
In order to note that $x^n-1=\prod_{k=0}^{n-1} (x-\xi^k)$, note that $1,\xi,\dots,\xi^{n-1}$ are roots of $x^n-1$. Therefore by polynomial reminder theorem we have $x^n-1=Q(x) \prod_{k=0}^{n-1} (x-\xi^k)$. Comparing degrees of lhs and rhs we can find, that $Q(x)$ has degree 0. Comparing highest coefficients we can conclude $Q(x)=1$.

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Can you explain a little bit more why $x^n-1=\prod_{k=0}^n (x-\xi^k)$? –  Robert Smith Oct 30 '10 at 21:09
@Robert: well, the $\xi^k$ are the nth roots of unity... –  Ｊ. Ｍ. Oct 30 '10 at 22:50
@J.M. Yes, I know. But I wanted to know about the validity of the equality. –  Robert Smith Oct 30 '10 at 23:36
It's just an expansion of $x^n-1$ in terms of its zeroes, e.g. $x^2-1=(x+1)(x-1)$. –  Ｊ. Ｍ. Oct 30 '10 at 23:43
@J.M: I think it should be $\prod_{k=0}^n (x-\xi^k)=(x-1)(x^{n}-1)$ instead of $\prod_{k=0}^n (x-\xi^k)=x^n-1$ –  Robert Smith Oct 31 '10 at 0:55

\begin{align*}=\prod_{k=1}^{n−1}\sin\frac{kπ}n&=(2i)^{1−n}\cdot\prod_{k=1}^{n−1}\left(e^{^\tfrac{ikπ}n}−e^{^{−\tfrac{ikπ}n}}\right)\\&=(2i)^{1−n}\cdot e^{^{−\tfrac{iπn(n−1)}{2n}}}\cdot\prod_{k=1}^{n−1}\left(e^{^\tfrac{2ikπ}n}−1\right)\\&=(−2)^{1−n}\cdot\prod_{k=1}^{n−1}\left(ξ^k−1\right)=2^{1−n}\cdot\prod_{k=1}^{n−1}\left(1−ξ^k\right),\qquad\text{where}\qquad ξ=e^{^\tfrac{2iπ}n}\end{align*}

Now, note that $\quad x^n−1=(x−1)\displaystyle\sum^{n−1}_{k=0}x^k,\quad$ and $\quad x^n−1=\displaystyle\prod^{n−1}_{k=\color{blue}0}\left(x−ξ^k\right).\quad$

Thus, cancelling $x−1$, we have $\quad\displaystyle\prod^{n−1}_{k=\color{blue}1}\left(x−ξ^k\right)=\sum^{n−1}_{k=0}x^k.\quad$

By substituting $x=1$ we have $\quad\displaystyle\prod^{n−1}_{k=1}\left(1−ξ^k\right)=n$. Therefore, $P=n\cdot2^{1−n}$

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This answer is very difficult to read. Could you format it with MathJax so that it’s helpful to other readers? –  alexwlchan Jul 20 '13 at 11:07
@alexwlchan $\LaTeX$ed. –  Ahaan S. Rungta Nov 19 at 18:20