Question:medium

\( \cos A \cos 2A \cos 4A \cdots \cos 2^{\,n-1}A \) equals

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Standard result: $\prod_{k=0}^{n-1} \cos(2^k A) = \frac{\sin(2^n A)}{2^n \sin A}$.
Updated On: Apr 10, 2026
  • $\frac{\sin 2^{n}A}{2^{n}\sin A}$
  • $\frac{2^{n}\sin 2^{n}A}{\sin A}$
  • $\frac{2^{n}\sin A}{\sin 2^{n}A}$
  • $\frac{\sin A}{2^{n}\sin 2^{n}A}$
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The Correct Option is A

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