By principle of mathematical induction $\cos \mathrm{\theta cos}2\mathrm{\theta cos}4\mathrm{\theta }\dots \cos \left(2^{\mathrm{n}-1}\mathrm{\theta }\right),\mathrm{\forall }\mathrm{n}\in \mathrm{N}=$

Let $\mathrm{P}\left(\mathrm{n}\right):\cos \mathrm{\theta cos}2\mathrm{\theta cos}4\mathrm{\theta }\dots \cos \left(2^{\mathrm{n}-1}\mathrm{\theta }\right)=\frac{\sin 2^{\mathrm{n}}\mathrm{\theta }}{2^{\mathrm{n}}\sin \mathrm{\theta }}\dots \text{ (i) }$

Step l :For n=1,

          $\mathrm{LHS}=\cos \mathrm{\theta }\text{ and RHS }=\frac{\sin 2\mathrm{\theta }}{2\sin \mathrm{\theta }}=\cos \mathrm{\theta }$  P(1 ) is true. 

Step ll :Let P(n) is true, then

$\mathrm{P}\left(\mathrm{k}\right):\cos \mathrm{\theta cos}2\mathrm{\theta cos}4\mathrm{\theta }\dots \cos \left(2^{\mathrm{k}-1}\mathrm{\theta }\right)=\frac{\sin 2^{\mathrm{k}}\mathrm{\theta }}{2^{\mathrm{k}}\sin \mathrm{\theta }}$

Step III: For n=k+1

$\begin{array}{l}\mathrm{P}(\mathrm{k}+1):\cos \mathrm{\theta cos}2\mathrm{\theta }\dots \cos \left(2^{\mathrm{k}}\mathrm{\theta }\right)=\frac{\sin 2^{\mathrm{k}+1}\mathrm{\theta }}{2^{\mathrm{k}+1}\sin \mathrm{\theta }}\\ \text{ LHS }=\cos \mathrm{\theta cos}2\mathrm{\theta }\dots \cos 2^{(\mathrm{k}-1)}\mathrm{\theta cos}{2}^{\mathrm{k}}\mathrm{\theta }\\ =\frac{\sin \left(2^{\mathrm{k}}\mathrm{\theta }\right)}{2^{\mathrm{k}}\sin \mathrm{\theta }}\cdot \cos \left(2^{\mathrm{k}}\mathrm{\theta }\right)\\ =\frac{2\sin \left(2^{\mathrm{k}}\mathrm{\theta }\right)\cdot \cos \left(2^{\mathrm{k}}\mathrm{\theta }\right)}{2^{\mathrm{k}+1}\sin \mathrm{\theta }}\\ =\frac{\sin 2^{\mathrm{k}+1}\mathrm{\theta }}{2^{\mathrm{k}+1}\sin \mathrm{\theta }}=\mathrm{RHS}\end{array}$

   For n =k + 1, P(n) is true. 

Hence, by principle for mathematical induction for all n $\in$ N,P(n) is true.