Question

If $\theta =\frac{\pi }{2^n+1}$, then the value of $\cos \theta \cos 2\theta \cos 2^2\theta$------$\cos 2^{n-1}\theta$ is?

• A. $1$
• B. $\frac{1}{2^n}$
• C. $2^n$
• D. None of these

Answer 1

The correct option is D None of these

$cos\theta cos2\theta cos{2}^2\theta .....cos{2}^{n-1}\theta$

$=\frac{1}{2sin\theta }\left(2sin\theta cos\theta \right)cos2\theta cos{2}^2\theta ...cos{2}^{n-1}\theta$

$=\frac{1}{2sin\theta }sin2\theta cos2\theta cos{2}^2\theta ....cos{2}^{n-1}\theta$

$=\frac{1}{2^{2}sin\theta }\left(2sin2\theta cos2\theta \right)cos{2}^2\theta ....cos{2}^{n-1}\theta$

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

$=\frac{2}{2\times 2^{n-1}sin\theta }sin{2}^{n-1}\theta cos{2}^{n-1}\theta$

$=\frac{sin{2}^n\theta }{2^{n}sin\theta }=\frac{1}{2^n}.\frac{sin\left(\frac{\pi }{2}\right)}{sin\left(\frac{\pi }{2^{n+1}}\right)}=\frac{1}{2^{n}sin\left(\frac{\pi }{2^{n+1}}\right)}$

$(\because \theta =\frac{\pi }{2^{n+1}})$