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