Question

If $\theta =\frac{\pi }{2^n+1}$, then $\cos \theta \cos 2\theta 2\cos 2^2\theta ...\cos 2^{n-1}\theta =\frac{1}{a^n}$. Find $a$

Answer 1

$\cos \theta \cos 2\theta \cos 2^2\theta ...\cos 2^{n-1}\theta =\frac{1}{a^n}$
$\frac{2^{n-1}\left(2\sin \theta \cos \theta \right)\cos 2\theta \cos 2^2\theta ...\cos 2^{n-1}\theta }{2^n\sin \theta }=\frac{2^{n-2}\left(2\sin 2\theta \cos 2\theta \right)\cos 2^2\theta ...\cos 2^{n-1}\theta }{2^n\sin \theta }=\frac{\sin 2^n\theta }{2^n\sin \theta }=\frac{1}{a^n}$
Substitute $\theta =\frac{\pi }{2^n+1}$
$\frac{\sin 2^n\left(\frac{\pi }{2^n+1}\right)}{2^n\sin \left(\frac{\pi }{2^n+1}\right)}=\frac{\sin 2^n(\pi -\frac{\pi }{2^n+1})}{2^n\sin \left(\frac{\pi }{2^n+1}\right)}=\frac{1}{2^n}=\frac{1}{a^n}$
Ans: 2