If cos x cos 2x cos2

$\cos x \cos 2 x \cos 2^{2} x \ldots \ldots \ldots \ldots \cos 2_{x}^{n-1}$

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, $\lambda=\frac{1}{2^{n}}$