Question

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

• A. $\frac{1}{2^{n-1}}$
• B. $\frac{1}{2^{n+1}}$
• C. $\frac{1}{2^n}$
• D. $\frac{1}{2^{2n}}$

Answer 1

The correct option is C $\frac{1}{2^n}$

$\text{Given:}\cos \theta .\cos 2\theta .\cos 2^2\theta \cdots \cos 2^{n-1}\theta ,\theta =\frac{\pi }{2^n+1}$

Using the result:
$\therefore \cos \theta .\cos 2\theta .\cos 2^2\theta \cdots \cos 2^{n-1}\theta =\frac{\sin 2^n\theta }{2^n\sin \theta }$

$\Rightarrow \cos \theta .\cos 2\theta .\cos 2^2\theta \cdots \cos 2^{n-1}\theta =\frac{\sin (\frac{\pi }{2^n+1})2^n}{2^n\sin (\frac{\pi }{2^n+1})}$

$\Rightarrow \cos \theta .\cos 2\theta .\cos 2^2\theta \cdots \cos 2^{n-1}\theta =\frac{\sin (\frac{2^n}{2^n+1})\pi }{2^n\sin (\frac{\pi }{2^n+1})}$

$\Rightarrow \cos \theta .\cos 2\theta .\cos 2^2\theta \cdots \cos 2^{n-1}\theta =\frac{\sin (\frac{2^n+1-1}{2^n+1})\pi }{2^n\sin (\frac{\pi }{2^n+1})}$

$\Rightarrow \cos \theta .\cos 2\theta .\cos 2^2\theta \cdots \cos 2^{n-1}\theta =\frac{\sin (\pi -\frac{\pi }{2^n+1})}{2^n\sin (\frac{\pi }{2^n+1})}$

$\Rightarrow \cos \theta .\cos 2\theta .\cos 2^2\theta \cdots \cos 2^{n-1}\theta =\frac{\sin (\frac{\pi }{2^n+1})}{2^n\sin (\frac{\pi }{2^n+1})}$

$\Rightarrow \cos \theta .\cos 2\theta .\cos 2^2\theta \cdots \cos 2^{n-1}\theta =\frac{1}{2^n}$