If \((2^n+1)x=\pi\), Find \(2^n\cos x\cos2x\cos2^2x\cdots\cos2^{n-1}x\)
Question
If
\[ (2^n+1)x=\pi, \]
then find
\[ 2^n\cos x\cos2x\cos2^2x\cdots\cos2^{n-1}x \]
(a) \(-1\)
(b) \(1\)
(c) \(\frac12\)
(d) none of these
Solution
Use the standard identity:
\[ \sin(2^n x) = 2^n \sin x \cos x \cos2x \cos2^2x \cdots \cos2^{n-1}x \]
Therefore,
\[ 2^n\cos x\cos2x\cdots\cos2^{n-1}x = \frac{\sin(2^n x)}{\sin x} \]
Given
\[ (2^n+1)x=\pi \]
\[ 2^n x=\pi-x \]
Hence,
\[ \sin(2^n x) = \sin(\pi-x) = \sin x \]
Substituting,
\[ 2^n\cos x\cos2x\cdots\cos2^{n-1}x = \frac{\sin x}{\sin x} = 1 \]
Final Answer
\[ \boxed{1} \]
Hence, the correct option is (b) 1.