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

detailed solution

Correct option is A

Let P(n):cos⁡θcos⁡2θcos⁡4θ…cos⁡2n−1θ=sin⁡2nθ2nsin⁡θ… (i) Step l :For n=1,          LHS=cos⁡θ and RHS =sin⁡2θ2sin⁡θ=cos⁡θ  P(1 ) is true. Step ll :Let P(n) is true, thenP(k):cos⁡θcos⁡2θcos⁡4θ…cos⁡2k−1θ=sin⁡2kθ2ksin⁡θStep III: For n=k+1P(k+1):cos⁡θcos⁡2θ…cos⁡2kθ=sin⁡2k+1θ2k+1sin⁡θ LHS =cos⁡θcos⁡2θ…cos⁡2(k−1)θcos⁡2kθ=sin⁡2kθ2ksin⁡θ⋅cos⁡2kθ=2sin⁡2kθ⋅cos⁡2kθ2k+1sin⁡θ=sin⁡2k+1θ2k+1sin⁡θ=RHS   For n =k + 1, P(n) is true. Hence, by principle for mathematical induction for all n ∈ N,P(n) is true.