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Mathematics Question on Trigonometric Equations

cos⁡ A cos⁡ 2A cos⁡ 4A...cos⁡ 2n−1A\cos \, A \, \cos \, 2A \, \cos \, 4A ... \cos \, 2^{n -1} A equals

A

sin⁡ 2nA2n sin⁡ A\frac{\sin \, 2^n A}{2^n \, \sin \, A}

B

2n sin⁡ 2nAsin⁡ A\frac{2^n \, \sin \, 2^n A}{ \sin \, A}

C

2n sin⁡ Asin⁡ 2n A\frac{2^n \, \sin \, A}{ \sin \, 2^n \, A}

D

sin⁡ A2n sin⁡ 2n A\frac{\sin \, A}{2^n \, \sin \, 2^n \, A}

Answer

sin⁡ 2nA2n sin⁡ A\frac{\sin \, 2^n A}{2^n \, \sin \, A}

Explanation

Solution

It is a standard result.
cos⁡ A cos⁡ 2A cos⁡ 22 A.....cos⁡ 2n−1A\cos \, A \, \cos \, 2A \, \cos \, 2^2 \, A..... \cos \, 2^{n - 1} A
=sin⁡ 2n A2n sin⁡ A= \frac{\sin \, 2^n \, A}{2^n \, \sin \, A}