2017/013) Prove that $\prod_{n=1}^{m-1} \sin \frac{n \pi}{m} = \frac{m}{2^{m-1}}$

we know $\sin \, nt = \frac{1}{2i}(e^{int} - e^{-int})=\frac{1}{2i}e^{-int}(e^{2int}-1) $
lettiing $t=\frac{\pi}{m}$ we have taking the product from n = to m-1 we get
$\prod_{n=1}^{m-1} sin \frac{n \pi}{m}= (\frac{1}{2i})^{m-1} e^{-i\sum_{n=1}^{m-1}\frac{n\pi}{m}}\prod_{n=1}^{m-1} (e^{2\frac{in\pi}{m}}-1) \cdots(1) $

Now $ e^{-i\sum_{n=1}^{m-1}\frac{n\pi}{m}} = e^\frac{-(m-1)(m)\pi\,i }{2m}= e^\frac{-(m-1)\pi\,i }{2}=  (e^\frac{-\pi\,i}{2})^{(m-1)}= (-i)^{m-1}\cdots(2)$

from (1) and (2)
$\prod_{n=1}^{m-1} sin \frac{n \pi}{m}= (\frac{1}{2i})^{m-1} (-i)^{m-1}= (\frac{-1}{2})^{m-1} \prod_{n=1}^{m-1} (e^{2\frac{in\pi}{m}}-1)$
$= (\frac{1}{2})^{m-1} \prod_{n=1}^{m-1} (1- e^{2\frac{in\pi}{m}}) = (\frac{1}{2})^{m-1} \prod_{n=1}^{m-1} (1- w^n) \dots (3)$ where w is $m^{th}$ root of 1

now because w is $n^{th}$ root of 1 we have

$x^n-1 = \prod_{n=0}^{m-1} (x- w^n) = (x-1)  \prod_{n=1}^{m-1} (x- w^n)\cdots(4)$
further $x^n-1 = (x-1) \sum_{n=0}^{m-1} x^n\cdots(5)$
so we have from (4) and (5)
$\prod_{n=1}^{m-1} (x- w^n) = \sum_{n=0}^{m-1} x^n$
putting x = 1 we get
$\prod_{n=1}^{m-1} (1- w^n) = \sum_{n=0}^{m-1} 1^n= m $

from (3) and above we get
$\prod_{n=1}^{m-1} \sin \frac{n \pi}{m} = \frac{m}{2^{m-1}}$