Given cos 2 m θ cos 2 m + 1 θ .............. cos 2 n θ = 2^ n +1 2^ n - m +1 2^ m , where 2 m θ ≠ kπ, n, m, k ∈ I cos 11 cos 2 11 cos 3 11 .... cos 11 11 =

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 2^mθ cos 2^{m + 1}θ .............. cos 2ⁿθ
\(=\frac{\sin 2^{\mathrm{n}+1} \theta}{2^{\mathrm{n}-\mathrm{m}+1} \sin 2^{\mathrm{m}} \theta},\) where 2^mθ ≠ kπ, n, m, k ∈ I
cos \(\frac{\pi}{11}\) cos \(\frac{2 \pi}{11}\) cos \(\frac{3 \pi}{11}\) .... cos \(\frac{11 \pi}{11}=\)

\(-\frac{1}{32}\)
\(\frac{1}{512}\)
\(\frac{1}{1024}\)
\(-\frac{1}{2048}\)

Answer

(C) 1 1024

Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.

Given cos 2 m θ cos 2 m + 1 θ .............. cos 2 n θ = 2^ n +1 2^ n - m +1 2^ m , where…

Answer

Related practice questions