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प्रश्न
If \[\left( 2^n + 1 \right) x = \pi,\] then \[2^n \cos x \cos 2x \cos 2^2 x . . . \cos 2^{n - 1} x = 1\]
विकल्प
-1
1
1/2
None of these
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उत्तर
1
\[\left( 2^n + 1 \right)x = \pi \left( \text{ Given } \right)\]
\[ \Rightarrow 2^n x + x = \pi\]
\[ \Rightarrow 2^n x = \pi - x\]
\[ \Rightarrow \sin 2^n x = \sin\left( \pi - x \right)\]
\[ \Rightarrow \sin 2^n x = \sin x . . . (1) \]
\[2^n \cos x \cos 2x \cos 2^2 x . . . \cos 2^{n - 1} x = 2^n \times \frac{\sin 2^n x}{2^n \sin x}\]
\[ = \frac{\sin 2^n x}{\sin x}\]
\[ = \frac{\sin x}{\sin x} \left[ \text{ From } (1) \right]\]
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