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# If $\theta=\large\frac{\pi}{2^n+1}$ then $\cos \theta\cos 2\theta\cos 2^2\theta......\cos2^{n-1}\theta$ is equal to

$(a)\;\large\frac{1}{2^n}\normalsize\qquad(b)\;\cos\theta\qquad(c)\;2\qquad(d)\;2^n$

We have
$\cos\theta\cos2\theta\cos 2^2\theta......\cos 2^{n-1}\theta=\large\frac{\sin 2^n\theta}{2^n\sin\theta}$
$\theta=\large\frac{\pi}{2^n+1}$
$\Rightarrow 2^n\theta+\theta=\pi$
$\Rightarrow 2^n\theta=\pi-\theta$
$\cos\theta\cos2\theta\cos 2^2\theta......\cos 2^{n-1}\theta=\large\frac{\sin(\pi-\theta)}{2^n\sin \theta}$
$\Rightarrow \large\frac{\sin\theta}{2^n\sin\theta}$
$\Rightarrow \large\frac{1}{2^n}$
Hence (a) is the correct option.