Question:
If $\left(2^{n}+1\right) x=\pi$, then $2^{n} \cos x \cos 2 x \cos 2^{2} x \ldots \cos 2^{n-1} x=1$
(a) -1
(b) 1
(c) 1/2
(d) None of these
Solution:
(b) 1
$\left(2^{n}+1\right) x=\pi \quad$ (Given)
$\Rightarrow 2^{n} x+x=\pi$
$\Rightarrow 2^{n} x=\pi-x$
$\Rightarrow \sin 2^{n} x=\sin (\pi-x)$
$\Rightarrow \sin 2^{n} x=\sin x$
$2^{n} \cos x \cos 2 x \cos 2^{2} x \ldots \cos 2^{n-1} x=2^{\mathrm{n}} \times \frac{\sin 2^{n} x}{2^{n} \sin x}$
$=\frac{\sin 2^{n} x}{\sin x}$
$=\frac{\sin x}{\sin x}$ [ Form 1 ]
$=1$
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