AP EAMCET · Maths · Trigonometric Ratios & Identities
\(\cos A \cos 2 A \cos 4 A \ldots \cos 2^{n-1} A\) 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 & Solution
Correct Answer
(A) \(\frac{\sin 2^n A}{2^n \sin A}\)
Step-by-step Solution
Detailed explanation
It is a standard result. \(\begin{gathered} \cos A \cos 2 A \cos 2^2 A \ldots \cos 2^{n-1} A \\ =\frac{\sin 2^n A}{2^n \sin A} \end{gathered}\)
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