If

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) -1-1">

(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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