Question

Simplify the expressions of the products.

$cos\frac{\pi }{2n}cos\frac{2\pi }{2n}cos\frac{3\pi }{2n}......cos\frac{n\pi }{2n}$

Answer 1

Given, $f\left(n\right)=\cos \left(\frac{\pi }{2n}\right)\cos \left(\frac{2\pi }{2n}\right)\cos \left(\frac{3\pi }{2n}\right)...\cos \left(\frac{n\pi }{2n}\right)$.
In general,
$f\left(n\right)$
${\prod }_{m=1}^{m=n}\cos \left(\frac{m\pi }{2n}\right)$
But at $m=n$ for some $mϵN$
$\cos \left(\frac{n\pi }{2n}\right)$
$=\cos \left(\frac{\pi }{2}\right)$
$=0$.
Hence ${\prod }_{n=1}^{n=n}cos\left(\frac{n\pi }{2n}\right)$
$=0$
Hence the above expression yields 0.