Question

The value of $\cos \frac{\pi }{11}+\cos \frac{3\pi }{11}+\cos \frac{5\pi }{11}+\cos \frac{7\pi }{11}+\cos \frac{9\pi }{11}$ is

• A. $0$
• B. $1$
• C. $\frac{1}{2}$
• D. none of these

Answer 1

Answer: (C)

$\frac{1}{2}$

Let $y=\cos \frac{\pi }{11}+\cos \frac{3\pi }{11}+\cos \frac{5\pi }{11}+\cos \frac{7\pi }{11}+\cos \frac{9\pi }{11}$
$\Rightarrow y=\frac{\left(2\cos \frac{\pi }{11}\sin \frac{\pi }{11}\right)+\left(2\cos \frac{3\pi }{11}\sin \frac{\pi }{11}\right)+\left(2\cos \frac{5\pi }{11}\sin \frac{\pi }{11}\right)+\left(2\cos \frac{7\pi }{11}\sin \frac{\pi }{11}\right)+\left(2\cos \frac{9\pi }{11}\sin \frac{\pi }{11}\right)}{2\sin \frac{\pi }{11}}$
$\Rightarrow y=\frac{\sin \frac{2\pi }{11}+\sin \frac{4\pi }{11}-\sin \frac{2\pi }{11}+\sin \frac{6\pi }{11}-\sin \frac{4\pi }{11}+\sin \frac{8\pi }{11}-\sin \frac{6\pi }{11}+\sin \frac{10\pi }{11}-\sin \frac{8\pi }{11}}{2\sin \frac{\pi }{11}}$
because $2\cos A\sin B=\sin (A+B)-\sin (A-B)$
$\Rightarrow y=\frac{\sin \frac{10\pi }{11}}{2\sin \frac{\pi }{11}}=\frac{\sin \frac{10\pi }{11}}{2\sin (\pi -\frac{\pi }{11})}=\frac{1}{2}$
Ans: C