Question

State true or false:

$cos\frac{\pi }{2n+1}cos\frac{2\pi }{2n+1}cos\frac{3\pi }{2n+1}.....cos\frac{n\pi }{2n+1}=\frac{1}{2^n}$.

• A. True
• B. False

Answer 1

The correct option is A True

$\frac{2^{2n+1}-1}{x-1}=(x^2-2xcos\frac{2\pi }{2n+1}+1)(x^2-2xcos\frac{4\pi }{2n+1}+1).......(x^2-2xcos\frac{2n\pi }{2n+1}+1)$
Taking $\underset{x\to -1}{lim}$ on both sides, we have
$1=4^{n}co{s}^2\frac{\pi }{2n+1}co{s}^2\frac{2\pi }{2n+1}.....co{s}^2\frac{n\pi }{2n+1}$
$\therefore cos\frac{\pi }{2n+1}cos\frac{2\pi }{2n+1}.....cos\frac{n\pi }{2n+1}=\frac{1}{2^n}$