If $\cos x \cos 2 x \cos 2^{2} x_{\ldots} \cos 2^{n-1} x=\lambda \frac{\sin 2^{n} \lambda}{\sin x}$, then $\lambda=$ ______________________
$\cos x \cos 2 x \cos 2^{2} x \ldots \ldots \ldots \ldots \cos 2_{x}^{n-1}$
multiply and divide the above equation by $2 \sin x$ we get,
$=\frac{1}{2 \sin x}\left[2 \sin x \cos x \cos 2 x \cos 2^{2} x \ldots \ldots \ldots \cos 2^{n-1} x\right]$ (using identity, $2 \sin x \cos x=\sin 2 x$ )
$=\frac{1}{2 \sin x}\left[\sin 2 x \cos 2 x \cos 2^{2} x \ldots \ldots \ldots \ldots \cos 2^{n-1} x\right]$
multiply and divide the above equation by 2 , we get,
$=\frac{1}{2 \times 2 \sin x}\left[\begin{array}{lll}2 & \sin 2 x \cos 2 x & \left.\cos 2^{2} x \ldots \ldots \ldots \ldots \ldots \cos 2^{n-1} x\right]\end{array}\right.$
$=\frac{1}{2^{2} \sin x}\left[\sin 4 x \cos 2^{2} x \ldots \ldots \ldots \cos 2^{n-1} x\right] \quad$ (using identity, $2 \sin \theta \cos \theta=\sin 2 \theta$ )
similarly, multiply and divide by 2 , then repeat the above step.
$=\frac{1}{2^{n} \sin x}\left[2 \sin 2^{n-1} x \cos 2^{n-1} x\right]$
$=\frac{1}{2^{n} \sin x}\left[\sin 2^{n} x\right]$ (using identity : $2 \sin \theta \cos \theta=\sin 2 \theta$ )
Hence, $\lambda=\frac{1}{2^{n}}$