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
cos 23 \(\frac{\pi}{10}\) cos 24 \(\frac{\pi}{10}\) cos25 \(\frac{\pi}{10}\) ........ cos 210 \(\frac{\pi}{10}\)=
- \(\frac{1}{128}\)
- \(\frac{1}{256}\)
- \(\frac{1}{512} \sin \frac{\pi}{10}\)
- \(\frac{\sqrt{5}-1}{512} \sin \frac{3 \pi}{10}\)
Answer
(B) 1 256
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