Question

If n = 1, 2, 3, …., then cos α cos 2 α cos 4 α … cos 2^{n – 1} α is equal to

A. \(\frac{\text{sin 2n α}}{\text{2n sin α}}\)

B. \(\frac{\text{sin 2}^n\text{α}}{\text{2}^n\text{sin 2}^{n-1}\text{α}}\)

C. \(\frac{\text{sin 4}^{n-1}\text{α}}{\text{4}^{n-1}\text{sin α}}\)

D. \(\frac{\text{sin 2}^n\text{α}}{\text{2}^n\text{sin α}}\)

Answer 1

Correct answer is D.

Given expression

cos α cos 2 α cos 4 α … cos 2^{n – 1} α

multiplying and dividing the expression by 2 sin α , we get,

Now multiplying and dividing the expression with 2.

Continuing this process for n - 1 times we will get