Question

If \( n = 1, 2, 3, \dots \), then \( \cos \alpha \cos 2\alpha \cos 4\alpha \dots \cos^{2n - 1} \alpha \) 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) $\frac{\sin 2^n\alpha }{2^n\sin \alpha }$

Answer 1

The Correct Option is D

Explanation:
Konsider the expression $\Rightarrow \cos \alpha \cos 2\alpha \cos 4\alpha \dots \dots \cos \left(2^{n-1}\right)\alpha$Multiply and divide the expression by $2^n\sin \alpha$ $\Rightarrow \frac{2^{n-1}}{2^n\sin \alpha }[2\sin \alpha \cos \alpha \cos 2\alpha \cos 4\alpha \dots \dots \cos \left(2^{n-1}\right)\alpha ]$$\Rightarrow \frac{2^{n-2}}{2^n\sin \alpha }[2\sin 2\alpha \cos 2\alpha \cos 4\alpha \dots \dots \cos \left(2^{n-1}\right)\alpha ]$$\Rightarrow \frac{2^{n-3}}{2^n\sin \alpha }[2\sin 4\alpha \cos 4\alpha \dots \dots \cos \left(2^{n-1}\right)\alpha ]$$\Rightarrow \frac{1}{2^n\sin \alpha }[2\sin 2^{n-1}\alpha \cos 2^{n-1}\alpha ]$$\Rightarrow \frac{1}{2^n\sin \alpha }\sin \left({2.2}^{n-1}\alpha \right)$$\Rightarrow \frac{\sin 2^n\alpha }{2^n\sin \alpha }$