Question

The value of $\sum _{n=1}^{10}(\sin \frac{2n\pi }{11}-\cos \frac{2n\pi }{11})$ is equal to

• A. $2$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

The correct option is B $1$

Given series
$=(\sin \frac{2\pi }{11}-\cos \frac{2\pi }{11})+(\sin \frac{4\pi }{11}-\cos \frac{4\pi }{11})+(\sin \frac{6\pi }{11}-\cos \frac{6\pi }{11})+....+(\sin \frac{20\pi }{11}-\cos \frac{20\pi }{11})$
$=(\sin \frac{2\pi }{11}+\sin \frac{4\pi }{11}+....+\sin \frac{20\pi }{11})$$-(\cos \frac{2\pi }{11}+\cos \frac{4\pi }{11}+....+\cos \frac{20\pi }{11})$
$=\frac{\sin \pi .\sin \frac{10.\pi }{11}}{\sin \frac{\pi }{11}}-\frac{\cos \pi .\sin \frac{10\pi }{11}}{\sin \frac{\pi }{11}}$
$=0+\frac{\sin (\pi -\frac{\pi }{11})}{\sin \frac{\pi }{11}}=1$