Question

If $S_n={\sin }^2\frac{\pi }{n}+{\sin }^2\frac{2\pi }{n}+\dots +{\sin }^2\frac{(n-1)\pi }{n}$, then the value of $S_{100}$ is

• A. $49$
• B. $50$
• C. $\frac{99}{2}$
• D. $\frac{101}{2}$

Answer 1

The correct option is B $50$

$S_n={\sin }^2\frac{\pi }{n}+{\sin }^2\frac{2\pi }{n}+\dots +{\sin }^2\frac{(n-1)\pi }{n}=\frac{1}{2}[1-\cos \frac{2\pi }{n}+1-\cos \frac{4\pi }{n}+\cdots +1-\cos \frac{2(n-1)\pi }{n}]=\frac{1}{2}\left[\right(n-1)-(\cos \frac{2\pi }{n}+\cos \frac{4\pi }{n}+\cdots +\cos \frac{2(n-1)\pi }{n})]$

We know that,
$(\cos \frac{2\pi }{n}+\cos \frac{4\pi }{n}+\cdots +\cos \frac{2(n-1)\pi }{n})=\cos 0+\cos \frac{2\pi }{n}+\cos \left(\frac{4\pi }{n}\right)+\cdots +\cos \left(\frac{2(n-1)\pi }{n}\right)-\cos 0=\frac{\sin \pi }{\sin \frac{\pi }{n}}\left[\cos \left(\frac{2\times 0+(n-1)\frac{2\pi }{n}}{2}\right)\right]-1=-1$

Therefore,
$\Rightarrow S_n=\frac{1}{2}[n-1-(-1\left)\right]=\frac{n}{2}$
$\therefore S_{100}=50$