# If n=1, 2, 3

Question:

If $n=1,2,3, \ldots$, then $\cos \alpha \cos 2 \alpha \cos 4 \alpha \ldots \cos 2^{n-1} \alpha$ is equal to

(a) $\frac{\sin 2 n \alpha}{2 n \sin \alpha}$

(b) $\frac{\sin 2^{n} \alpha}{2^{n} \sin 2^{n-1} \alpha}$

(c) $\frac{\sin 4^{n-1} \alpha}{4^{n-1} \sin \alpha}$

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

(e) None of these

Solution:

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

$\because \cos \alpha \cos 2 \alpha \cos 4 \alpha \ldots \cos 2^{n-1} \alpha=\frac{\sin 2^{n} \alpha}{2^{n} \sin \alpha}$