If $(2^{n} + 1) x = π,$ then $2^{n} \cos x \cos 2 x \cos 2^{2} x . . . \cos 2^{n - 1} x = 1$
(a) $- 1$
(b) 1
(c) 1/2
(d) None of these

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Solution

(b) 1

$(2^{n} + 1) x = π (Given) \Rightarrow 2^{n} x + x = π \Rightarrow 2^{n} x = π - x \Rightarrow \sin 2^{n} x = \sin (π - x) \Rightarrow \sin 2^{n} x = \sin x . . . (1)$

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

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(2^{n} + 1) θ = π,

2^{n} \cos θ \cos 2 θ \cos 2^{2} θ . . . \cos 2^{n - 1} θ = 1

- 1

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