Question

If $\theta =\frac{\pi }{2^{100}+1},$ then $\cos \theta \cos 2\theta \cos 2^2\theta \cdots \cos 2^{99}\theta$ is

• A. $\frac{1}{2^{100}}$
• B. $\frac{1}{2^{99}}$
• C. $\frac{1}{2^{101}}$
• D. $\frac{1}{2^{200}}$

Answer 1

The correct option is A $\frac{1}{2^{100}}$

We know that
$\cos \theta \cos 2\theta \cos 2^2\theta .....\cos 2^{n-1}\theta =\frac{\sin 2^n\theta }{2^n\sin \theta }$
Now putting $\theta =\frac{\pi }{2^{100}+1}$
We have
$\cos \theta \cos 2\theta \cos 2^2\theta .....\cos 2^{99}\theta =\frac{\sin 2^{100}\theta }{2^{100}\sin \theta }=\frac{\sin 2^{100}\left(\frac{\pi }{2^{100}+1}\right)}{2^{100}\sin \frac{\pi }{2^{100}+1}}=\frac{\sin \left(\frac{2^{100}\pi }{2^{100}+1}\right)}{2^{100}\sin \frac{\pi }{2^{100}+1}}$

We know that,
$\frac{2^{100}\pi }{2^{100}+1},\frac{\pi }{2^{100}+1}$ are supplementary,

$=\frac{\sin (\pi -\frac{\pi }{2^{100}+1})}{2^{100}\sin \frac{\pi }{2^{100}+1}}=\frac{\sin \left(\frac{\pi }{2^{100}+1}\right)}{2^{100}\sin \frac{\pi }{2^{100}+1}}[\because \sin (\pi -\theta )=\sin \theta ]=\frac{1}{2^{100}}$