Use the Principle of Mathematical Induction to prove that $\cos \theta \cos 2 \theta \cos 2^{2} \theta \ldots \cos 2^{n-1} \theta = \frac{\sin 2^{n} \theta}{2^{n} \sin \theta}$ for all $n \in N$.

(A) $P(n): \cos \theta \cdot \cos 2 \theta \cdot \cos 2^{2} \theta \ldots \cos 2^{n-1} \theta = \frac{\sin 2^{n} \theta}{2^{n} \sin \theta}, \quad \forall n \in N$
For $n=1$,$L.H.S. = \cos \theta$
$R.H.S. = \frac{\sin 2 \theta}{2 \sin \theta} = \frac{2 \sin \theta \cos \theta}{2 \sin \theta} = \cos \theta$
$\therefore L.H.S. = R.H.S.$
$\therefore P(1)$ is true.
Assume that $P(k)$ is true for some $k \in N$,i.e.,$\cos \theta \cdot \cos 2 \theta \cdot \cos 2^{2} \theta \ldots \cos 2^{k-1} \theta = \frac{\sin 2^{k} \theta}{2^{k} \sin \theta} \quad (i)$
For $n=k+1$,we need to prove $P(k+1): \cos \theta \cdot \cos 2 \theta \ldots \cos 2^{k-1} \theta \cdot \cos 2^{k} \theta = \frac{\sin 2^{k+1} \theta}{2^{k+1} \sin \theta}$
$L.H.S. = \left( \frac{\sin 2^{k} \theta}{2^{k} \sin \theta} \right) \cdot \cos 2^{k} \theta \quad (\text{using } (i))$
$= \frac{2 \sin 2^{k} \theta \cos 2^{k} \theta}{2 \cdot 2^{k} \sin \theta} = \frac{\sin 2(2^{k} \theta)}{2^{k+1} \sin \theta} = \frac{\sin 2^{k+1} \theta}{2^{k+1} \sin \theta}$
$\therefore P(k+1)$ is true.
Hence,by the principle of mathematical induction,$P(n)$ is true for all $n \in N$.