Question

Value of $sin\frac{\pi }{2n+1}sin\frac{2\pi }{2n+1}sin\frac{3\pi }{2n+1}.....sin\frac{n\pi }{2n+1}$.

• A. $\frac{\sqrt{2n+1}}{2^n}$
• B. 1
• C. $\frac{n(n+1)}{2}$
• D. None of these

Answer 1

The correct option is A $\frac{\sqrt{2n+1}}{2^n}$

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