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

Using $\text{n}^{\text{th}}$ root of unity

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

Prove that

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

\begin{align*} 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 \frac{n(n-1)}{2}\frac{\pi}{n}} \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) \\ \end{align*}
where $$\xi=e^{2i\pi/n}$$.

Now note that $$x^n-1=(x-1)\sum_{k=0}^{n-1}x^k$$ and $$x^n-1=\prod_{k=0}^{n-1} (x-\xi^k)$$.

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 $$\prod_{k=1}^{n-1} (1-\xi^k)=n$$. $$\therefore \boxed{P=n2^{1-n}}$$

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 we find $$Q(x)$$ has degree $$0$$. Comparing highest coefficients we conclude $$Q(x)=1$$.

Edit:

We may instead use the identity $$\left\lvert 1 – e^{2ik\pi/n} \right\rvert = 2\sin(k\pi/n), k = 1, …, n – 1,$$ to establish immediately that $$P \equiv \prod_{k=1}^{n-1}\sin(k\pi/n)= 2^{1-n}\prod_{k=1}^{n-1}\left\lvert 1 – e^{2ik\pi/n} \right\rvert = 2^{1 – n}\left\lvert \prod_{k=1}^{n-1}(1 – e^{2ik\pi/n}) \right\rvert$$, and continue by applying the foregoing logic to the product to obtain $$P=n2^{1-n}$$.