Question

If$n=1,2,3....,$then $\cos \left(\alpha \right)\cos \left(2\alpha \right)\cos \left(3\alpha \right)·····\cos \left(2^{n-1}\alpha \right)$ is equal to.

• A. $\frac{\sin \left(2n\alpha \right)}{2n\sin \left(\alpha \right)}$
• B. $\frac{\sin \left(2^n\alpha \right)}{2^n\sin \left(2^{n-1}\alpha \right)}$
• C. $\frac{\sin \left(4^{n-1}\alpha \right)}{4^{n-1}\sin \left(4^{n-1}\alpha \right)}$
• D. $\frac{\sin \left(2^n\alpha \right)}{2^n\sin \left(\alpha \right)}$

Answer 1

The correct option is D

$\frac{\sin \left(2^n\alpha \right)}{2^n\sin \left(\alpha \right)}$

Explanation for correct option

Given trigonometric expression is $\cos \left(\alpha \right)\cos \left(2\alpha \right)\cos \left(3\alpha \right)·····\cos \left(2^{n-1}\alpha \right)$

$$\begin{gathered} \therefore \cos \left(\alpha \right)\cos \left(2\alpha \right)\cos \left(4\alpha \right)·····\cos \left(2^{n-1}\alpha \right) \\ =\frac{1}{2\sin \left(\alpha \right)}\left[\left[2\sin \left(\alpha \right)\cos \left(\alpha \right)\right]\cos \left(2\alpha \right)\cos \left(4\alpha \right)·····\cos \left(2^{n-1}\alpha \right)\right] \\ =\frac{1}{2\sin \left(\alpha \right)}\left[\sin \left(2\alpha \right)\cos \left(2\alpha \right)\cos \left(4\alpha \right)·····\cos \left(2^{n-1}\alpha \right)\right]\left[\because 2\sin \left(\theta \right)\cos \left(\theta \right)=\sin \left(2\theta \right)\right] \\ =\frac{1}{2^2\sin \left(\alpha \right)}\left[\left[2\sin \left(2\alpha \right)\cos \left(2\alpha \right)\right]\cos \left(4\alpha \right)·····\cos \left(2^{n-1}\alpha \right)\right] \\ =\frac{1}{2^2\sin \left(\alpha \right)}\left[\sin \left(4\alpha \right)\cos \left(4\alpha \right)·····\cos \left(2^{n-1}\alpha \right)\right]\left[\because 2\sin \left(\theta \right)\cos \left(\theta \right)=\sin \left(2\theta \right)\right] \end{gathered}$$

Proceeding in this way we get $\frac{\sin \left(2^n\alpha \right)}{2^n\sin \left(\alpha \right)}$

Hence, option (D) i.e. $\frac{\sin \left(2^n\alpha \right)}{2^n\sin \left(\alpha \right)}$ is correct