Q. 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) $\frac{\sin{2n\alpha}}{2n\sin\alpha}$
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) $\frac{\sin 2^{n}\alpha}{2^{n}\sin\alpha}$
Explanation
(d)
$\cos{\alpha\cos{2\alpha\cos{4\alpha\ldots\cos{2^{n - 1}\alpha}}}}$
$= \frac{1}{2\sin\alpha}\left\lbrack 2\sin{\alpha\cos{\alpha \times \cos{2\alpha\cos{4\alpha\ldots\cos{2^{n - 1}\alpha}}}}} \right\rbrack$
$= \frac{1}{2\sin\alpha}\left\lbrack \frac{2}{2}\sin{2\alpha\cos{2\alpha\cos{4\alpha\ldots\cos{2^{n - 1}\alpha}}}} \right\rbrack$
$= \frac{1}{2^{2}\sin\alpha}\left\lbrack \frac{2}{2}\sin{4\alpha\cos{4\alpha\ldots\cos{2^{n - 1}\alpha}}} \right\rbrack$
$= \frac{1}{2^{3}\sin\alpha}\lbrack\sin{8\alpha\cos{8\alpha\ldots\cos{2^{n - 1}\alpha}}}\rbrack$
Similarly, we can write
$= \frac{\sin{2^{n}\alpha}}{2^{n}\sin\alpha}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.