Question

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

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

(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

Answer 1

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

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