# 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

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

-
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 $Q(x)=1$.

-
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... –  Guess who it is. 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. $x^2-1=(x+1)(x-1)$. –  Guess who it is. 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, $P=n\cdot2^{1−n}$

-

Here is a more "1st principles" pf. I use a hint in Marsden's book.
1st, $\cos(A-B)-\cos(A+B)=2\sin A \sin B$ (1), which follows by angle summation formulas.
Next, we use Marsden's hint to consider roots of $(1-z)^n-1$. These satisfy $(1-z)^n=1 \leftrightarrow (1-z) \in \{\cos \frac{2 \pi k}{n}+i \sin \frac{2\pi k}{n}:k=0,...,n-1 \}$ (the set of nth roots of 1) $\leftrightarrow z \in \{z_k= 1-\cos \frac{2 \pi k}{n}-i \sin \frac{2\pi k}{n}:k=0,...,n-1\}$ (2).
Since $z_0,....,z_{n-1}$ are the roots of $(1-z)^n-1$, we have by factorization that $(1-z)^n-1=\prod_{k=0}^{n-1}(z_k-z)=-z \prod_{k=1}^{n-1}(z_k-z)$ (3) (since, by (2), $z_0=0$)
In (3), the LHS and RHS are polynomials in z. Equating the coeffs in front of z, we get $-n=-\prod_{k=1}^{n-1}z_k \leftrightarrow n=\prod_{k=1}^{n-1}z_k$. Note $\prod_{k=1}^{n-1} \bar{z}_k=\overline{\prod_{k=1}^{n-1}z_k}=n$ (since $n\in \mathbb{R}$), so $\prod_{k=1}^{n-1}|z_k|^2=\prod_{k=1}^{n-1} z_k \bar{z}_k=\prod_{k=1}^{n-1} z_k \prod_{k=1}^{n-1} \bar{z}_k=n^2$ (4).
Next, $|z_k|^2=(1-\cos \frac{2 \pi k}{n})^2+ \sin^2 \frac{2\pi k}{n}=2(1-\cos \frac{2 \pi k}{n})$; using this in (4) gives $2^{n-1} \prod_{k=1}^{n-1}(1-\cos \frac{2 \pi k}{n})=n^2$ (5).
Next, $(\prod_{k=1}^{n-1} \sin \frac{k \pi}{n})^2=\prod_{k=1}^{n-1} \sin \frac{k \pi}{n} \prod_{k=1}^{n-1} \sin \frac{(n-k) \pi}{n}=\prod_{k=1}^{n-1} \sin \frac{k \pi}{n} \sin \frac{(n-k) \pi}{n}$ (where in the last 2 steps, we exploit that the order of taking a product doesn't matter) =$\frac{1}{2^{n-1}} \prod_{k=1}^{n-1} (\cos \frac{(n-2k) \pi}{n}-\cos \pi)$ (by (1))=$\frac{1}{2^{n-1}} \prod_{k=1}^{n-1} (1-\cos \frac{2k \pi}{n})$ (using $\cos (\pi -x)=-\cos x$) =$n^2 /2^{2(n-1)}$. Applying a sqrt to everything gives the desired result.

-