Given cos 2 m θ cos 2 m + 1 θ .............. cos 2 n θ = 2^ n +1 2^ n - m +1 2^ m , where…
Mathematics · JEE Advanced · NTA Exams — Trigonometry
Given cos 2mθ cos 2m + 1 θ .............. cos 2nθ
\(=\frac{\sin 2^{\mathrm{n}+1} \theta}{2^{\mathrm{n}-\mathrm{m}+1} \sin 2^{\mathrm{m}} \theta},\) where 2m θ ≠ kπ, n, m, k ∈ I
sin \(\frac{9 \pi}{14}\). sin \(\frac{11 \pi}{14}\) sin \(\frac{13 \pi}{14}=\)
- \(\frac{1}{64}\)
- \(-\frac{1}{64}\)
- \(\frac{1}{8}\)
- \(-\frac{1}{8}\)
Answer
(C) 1 8
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