Question

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

• A. $\frac{\sin \, 2^n A}{2^n \, \sin \, A}$
• B. $\frac{2^n \, \sin \, 2^n A}{ \sin \, A}$
• C. $\frac{2^n \, \sin \, A}{ \sin \, 2^n \, A}$
• D. $\frac{\sin \, A}{2^n \, \sin \, 2^n \, A}$

Answer 1

Answer: (A)

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