Question

$\cos \frac{\pi }{11}+\cos \frac{3\pi }{11}+\cos \frac{5\pi }{11}+\cos \frac{7\pi }{11}+\cos \frac{9\pi }{11}=?$

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $1$
• D. $-1$

Answer 1

The correct option is B

$\frac{1}{2}$

Explanation for the correct option:

Given, $\cos \frac{\pi }{11}+\cos \frac{3\pi }{11}+\cos \frac{5\pi }{11}+\cos \frac{7\pi }{11}+\cos \frac{9\pi }{11}$

Multiply and divide it by $2\sin \frac{\mathrm{\pi }}{11}$,

$=\frac{1}{2\sin \frac{\mathrm{\pi }}{11}}\left[2\sin \frac{\pi }{11}\cos \frac{\pi }{11}+2\sin \frac{\pi }{11}\cos \frac{3\pi }{11}+2\sin \frac{\pi }{11}\cos \frac{5\pi }{11}+2\sin \frac{\pi }{11}\cos \frac{7\pi }{11}+2\sin \frac{\mathrm{\pi }}{11}\cos \frac{9\pi }{11}\right]$ use the identity $\mathbf{2}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{A}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{B}\mathbf{}\mathbf{=}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\left(A+B\right)\mathbf{+}\mathit{s}\mathit{i}\mathit{n}\left(A-B\right)$

$=\frac{1}{2\sin {\displaystyle \frac{\mathrm{\pi }}{11}}}\left[\sin \frac{2\pi }{11}+\sin \frac{4\pi }{11}–\sin \frac{2\pi }{11}+\sin \frac{6\pi }{11}–\sin \frac{4\pi }{11}+\sin \frac{8\pi }{11}–\sin \frac{6\pi }{11}+\sin \frac{10\pi }{11}-\sin \frac{8\mathrm{\pi }}{11}\right]$

$=\frac{1}{2\sin {\displaystyle \frac{\mathrm{\pi }}{11}}}\times \sin \frac{10\pi }{11}$

$=\frac{1}{2\sin {\displaystyle \frac{\mathrm{\pi }}{11}}}\times \sin \left(\pi –\frac{\pi }{11}\right)$

$=\frac{1}{2\sin {\displaystyle \frac{\mathrm{\pi }}{11}}}\times \sin \frac{\mathrm{\pi }}{11}$ ; $\left[\mathbf{\because }\mathbf{sin}\mathbf{(}\mathbf{\pi }\mathbf{-}\mathbf{\theta }\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{\theta }\right]$

$=\frac{\mathbf{1}}{\mathbf{2}}$

Hence, Option ‘B’ is Correct.