How to use Euler formula to prove the following conclusion？

Question

I thought about it for a long time $$\prod_{k=1}^{n}\sin(\frac{k\pi}{2n})=\frac{\sqrt{n}}{2^{n-1}}$$

Answer 1

Let $ n\geq 2 \cdot $

Consider the following polynomial : $ P_{n}\left(z\right)=z^{2n}-1 \cdot $

$ P_{n}\left(z\right)=0 \iff z^{2n}=1 \iff z=\mathrm{e}^{\mathrm{i}\frac{k \pi}{n}} ,\ \left(k\in\left[\!\left[0,2n-1\right]\!\right]\right) \cdot $

Thus $ P_{n} $ has $ 2n $ complex zeros : $ z_{k}=\mathrm{e^{\mathrm{i}\frac{k\pi}{n}}},\ \left(k\in\left[\!\left[0,2n-1\right]\!\right]\right) \cdot $ Hence it can be written as follows : $$ P_{n}\left(z\right)=\displaystyle\prod_{k=0}^{2n-1}{\left(z-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}\right)} $$

We have : \begin{align*} P_{n}\left(z\right)&=\displaystyle\prod_{k=0}^{2n-1}{\left(z-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}\right)}\\ &=\left(z-1\right)\left(\displaystyle\prod_{k=1}^{n-1}{\left(z-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}\right)}\right)\left(z+1\right)\left(\displaystyle\prod_{k=n+1}^{2n-1}{\left(z-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}\right)}\right)\\ &=\left(z^{2}-1\right)\displaystyle\prod_{k=1}^{n-1}{\left(z-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}\right)}\displaystyle\prod_{k=n+1}^{2n-1}{\left(z-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}\right)}\\ &=\left(z^{2}-1\right)\displaystyle\prod_{k=1}^{n-1}{\left(z-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}\right)}\displaystyle\prod_{k=1}^{n-1}{\left(z-\mathrm{e}^{\mathrm{i}\frac{\left(n+k\right)\pi}{n}}\right)} \\ &=\left(z^{2}-1\right)\displaystyle\prod_{k=1}^{n-1}{\left(z-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}\right)}\displaystyle\prod_{k=1}^{n-1}{\left(z-\mathrm{e}^{\mathrm{i}\frac{\left(2n-k\right)\pi}{n}}\right)}\\ &=\left(z^{2}-1\right)\displaystyle\prod_{k=1}^{n-1}{\left(z-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}\right)\left(z-\mathrm{e}^{-\mathrm{i}\frac{k\pi}{n}}\right)}\\ P_{n}\left(z\right)&=\left(z^{2}-1\right)\displaystyle\prod_{k=1}^{n-1}{\left(z^{2}-2z\cos{\left(\displaystyle\frac{k\pi}{n}\right)}+1\right)}\end{align*}

Conclusion : $$\fbox{$\begin{array}{rcl}\left(\forall n\geq 2\right)\left(\forall z\in\mathbb{C}\right),\ z^{2n}-1=\left(z^{2}-1\right)\displaystyle\prod_{k=1}^{n-1}{\left(z^{2}-2z\cos{\left(\displaystyle\frac{k\pi}{n}\right)}+1\right)}\end{array} $}$$

Thus : $$ \left(\forall n\geq 2\right)\left(\forall x\in\mathbb{R}\setminus\left\lbrace 1\right\rbrace\right),\ \displaystyle\frac{x^{2n}-1}{x^{2}-1}=\displaystyle\prod_{k=1}^{n-1}{\left(x^{2}-2x\cos{\left(\displaystyle\frac{k\pi}{n}\right)}+1\right)} $$

When $ x $ is, in particular, approaching $ 1 $, we get that : $$ \displaystyle\lim_{x\to 1}{\displaystyle\frac{x^{2n}-1}{x^{2}-1}}=\displaystyle\lim_{x\to 1}{\displaystyle\sum_{k=0}^{n-1}{x^{2k}}}=n $$

and that : \begin{align*}\displaystyle\lim_{x\to 1}{\displaystyle\prod_{k=1}^{n-1}{\left(x^{2}-2x\cos{\left(\displaystyle\frac{k\pi}{n}\right)}+1\right)}}&=\displaystyle\prod_{k=1}^{n-1}{\left(2-2\cos{\left(\displaystyle\frac{k\pi}{n}\right)}\right)}\\&=\displaystyle\prod_{k=1}^{n}{2^{2}\sin^{2}{\left(\displaystyle\frac{k\pi}{2n}\right)}}\\\displaystyle\lim_{x\to 1}{\displaystyle\prod_{k=1}^{n-1}{\left(x^{2}-2x\cos{\left(\displaystyle\frac{k\pi}{n}\right)}+1\right)}}&=2^{2n-2}\displaystyle\prod_{k=1}^{n-1}{\sin^{2}{\left(\displaystyle\frac{k\pi}{2n}\right)}}\end{align*}

Conclusion : $$\left(\forall n\geq 2\right),\ \displaystyle\prod_{k=1}^{n-1}{\sin^{2}{\left(\displaystyle\frac{k\pi}{2n}\right)}}=\displaystyle\frac{n}{2^{2n-2}} $$

Meaning : $$ \fbox{$ \begin{array}{rcl}\left(\forall n\geq 2\right),\ \displaystyle\prod_{k=1}^{n-1}{\sin{\left(\displaystyle\frac{k\pi}{2n}\right)}}=\displaystyle\frac{\sqrt{n}}{2^{n-1}}\end{array}$} $$