Question

If  \[\left( 2^n + 1 \right) x = \pi,\] then \[2^n \cos x \cos 2x \cos 2^2 x . . . \cos 2^{n - 1} x = 1\]

 

Options

• -1

• 1

• 1/2

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Answer 1

1

\[\left( 2^n + 1 \right)x = \pi \left( \text{ Given } \right)\]

\[ \Rightarrow 2^n x + x = \pi\]

\[ \Rightarrow 2^n x = \pi - x\]

\[ \Rightarrow \sin 2^n x = \sin\left( \pi - x \right)\]

\[ \Rightarrow \sin 2^n x = \sin x . . . (1) \]

\[2^n \cos x \cos 2x \cos 2^2 x . . . \cos 2^{n - 1} x = 2^n \times \frac{\sin 2^n x}{2^n \sin x}\]
\[ = \frac{\sin 2^n x}{\sin x}\]
\[ = \frac{\sin x}{\sin x} \left[ \text{ From }  (1) \right]\]