Question

$sin\frac{\pi }{2n}sin\frac{2\pi }{2n}sin\frac{3\pi }{2n}.........sin\frac{(n-1)\pi }{2n}=\frac{\sqrt{n}}{2^{n-1}}$=

Answer 1

The roots of the equation $x^{2n}-1=0$ are
$1,-1,cos\frac{\pi }{n}+isin\frac{\pi }{n},cos\frac{2\pi }{n}+isin\frac{2\pi }{n},cos\frac{3\pi }{n}+isin\frac{3\pi }{n},.....,cos\frac{(2n-1)\pi }{n}+isin\frac{(2n-1)\pi }{n}$
Therefore
$x^{2n}-1=(x-1)(x+1)(x-cos\frac{\pi }{n}-isin\frac{\pi }{n})(x-cos\frac{2\pi }{n}-isin\frac{2\pi }{n}).......(x-cos\frac{(2n-1)\pi }{n}-isin\frac{(2n-1)\pi }{n})$
Further since
$cos\frac{(2n-r)\pi }{n}=cos\frac{r\pi }{n}$ and $sin\frac{(2n-r)\pi }{n}=-sin\frac{r\pi }{n}$
it follows that
$(x-cos\frac{\pi }{n}-isin\frac{\pi }{n})(x-cos\frac{(2n-1)\pi }{n}-isin\frac{(2n-1)\pi }{n})$
$=x^2-2xcos\frac{\pi }{n}+1$
$(x-cos\frac{2\pi }{n}-isin\frac{2\pi }{n})(x-cos\frac{(2n-2)\pi }{n}-isin\frac{(2n-2)\pi }{n})$
$=x^2-2xcos\frac{2\pi }{n}+1$
$(x-cos\frac{(n-1)\pi }{n}-isin\frac{(n-1)\pi }{n})(x-cos\frac{(n+1)\pi }{n}-isin\frac{(n+1)\pi }{n})$
$=x^2-2xcos\frac{(n-1)\pi }{n}+1$
then the polynomial $x^{2n}-1$ can be rewritten thus.
$x^{2n}-1=(x^2-1)(x^2-2xcos\frac{\pi }{n}+1)(x^2-2xcos\frac{2\pi }{n}+1).........(x^2-2xcos\frac{(n-1)\pi }{n}+1)$
or $\frac{x^{2n}-1}{x^{x^2-1}}=(x^2-2xcos\frac{\pi }{n}+1)(x^2-2xcos\frac{2\pi }{n}+1).........(x^2-2xcos\frac{(n-1)\pi }{n}+1)$
Taking $\underset{x\to 1}{lim}$ on both sides, we have
$n=4^{n-1}si{n}^2\frac{\pi }{2n}si{n}^2\frac{2\pi }{2n}......si{n}^2\frac{(2n-1)\pi }{2n}$
Hence, $sin\frac{\pi }{2n}sin\frac{2\pi }{2n}.....sin\frac{(n-1)\pi }{2n}=\frac{\sqrt{n}}{2^{2n-1}}$
Ans: 1