By principle of mathematical induction cos⁡θcos⁡2θcos⁡4θ…cos⁡2n−1θ,∀n∈N=

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sin⁡2nθ2nsin⁡θ

cos⁡2nθ2nsin⁡θ

sin⁡2nθ2n−1sin⁡θ

None of these

answer is A.

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Detailed Solution

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.

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