Question

The value of $\sum _{k=1}^{n-1}{\sin }^2\frac{k\pi }{n}$ is

• A. $\frac{n}{2}$
• B. $\frac{n-1}{2}$
• C. $0$
• D. $\frac{n+1}{2}$

Answer 1

Answer: (A)

$\frac{n}{2}$

$\sum _{k=1}^{n-1}{\sin }^2\frac{k\pi }{n}$

$k^{th}$ term $=t_k=\frac{1-\cos \frac{2k\pi }{n}}{2}$

$\sum _{k=1}^{n-1}{t}_k=\frac{1}{2}(n-1)-\frac{1}{2}\{\cos \frac{2\pi }{n}+\cos \frac{4\pi }{n}+...+\cos \frac{(n-1)2\pi }{n}\}$

$=\frac{1}{2}(n-1)-\frac{1}{2}\frac{\cos \{\frac{2\pi }{n}+(n-2)\frac{\pi }{n}\}\sin (n-1)\frac{\pi }{n}}{\sin \frac{\pi }{n}}$

$=\frac{1}{2}(n-1)-\frac{1}{2}(-1)1=\frac{n}{2}$
Hence, option 'A' is correct.