Statement-1: cos⁡π7cos⁡2π7cos⁡4π7=−18Statement-2: if θ=π2n−1. cos⁡θcos⁡2θcos⁡4θ…cos⁡2n−1θ=-12n

cos⁡θcos⁡2θ…cos⁡2n−1θ=12sin⁡θ2sin⁡θcos⁡θcos⁡2θ…cos⁡2n−1θ=12nsin⁡θsin⁡2nθ=12nsin⁡θsin⁡(π+θ)=12nsin⁡θ(−sin⁡θ)=−12n.

Statement-1: cos⁡π7cos⁡2π7cos⁡4π7=−18Statement-2: if θ=π2n−1. cos⁡θcos⁡2θcos⁡4θ…cos⁡2n−1θ=-12n

cos⁡θcos⁡2θ…cos⁡2n−1θ=12sin⁡θ2sin⁡θcos⁡θcos⁡2θ…cos⁡2n−1θ=12nsin⁡θsin⁡2nθ=12nsin⁡θsin⁡(π+θ)=12nsin⁡θ(−sin⁡θ)=−12n.

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Statement-1: $\cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7} = - \frac{1}{8}$
Statement-2: $if θ = \frac{π}{2^{n} - 1} .$ $\cos θ \cos 2 θ \cos 4 θ \dots \cos 2^{n - 1} θ = - \frac{1}{2^{n}}$

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STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is False

STATEMENT-1 is False, STATEMENT-2 is True

answer is A.

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Detailed Solution

$\begin{array}{l} \cos θ \cos 2 θ \dots \cos 2^{n - 1} θ \\ = \frac{1}{2 \sin θ} [2 \sin θ \cos θ \cos 2 θ \dots \cos 2^{n - 1} θ] \\ = \frac{1}{2^{n} \sin θ} (\sin 2^{n} θ) = \frac{1}{2^{n} \sin θ} \sin (π + θ) \\ = \frac{1}{2^{n} \sin θ} (- \sin θ) = - \frac{1}{2^{n}} . \end{array}$