Question

If $S=\sin \frac{\pi }{n}+\sin \frac{3\pi }{n}+\sin \frac{5\pi }{n}+...n$ terms, then $nS=$

• A. 0.0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is A 0.0

Given series $S=\sin \frac{\pi }{n}+\sin \frac{3\pi }{n}+\sin \frac{5\pi }{n}+...\sin (\alpha +(n-1\left)\beta \right)$

Comparing the given series with standard sine series:

$\sin \alpha +\sin (\alpha +\beta )+\sin (\alpha +2\beta )+\sin (\alpha +3\beta )+...$

$\alpha =\frac{\pi }{n}$

$\beta =\frac{2\pi }{n}$

Sum of the sine series $=\frac{\sin \frac{n\beta }{2}}{\sin \frac{\beta }{2}}\sin (\alpha +\frac{(n-1)\beta }{2})$

$=\frac{\sin \left(\frac{n\times 2\pi }{2\times n}\right)}{\sin \left(\frac{2\pi }{2\times n}\right)}\sin (\frac{\pi }{n}+\frac{(n-1)2\pi }{2.n})$

$=\frac{\sin \pi }{\sin \left(\frac{\pi }{n}\right)}\times \sin (\frac{\pi }{n}+\frac{(n-1)\pi }{n})$

$=\frac{0}{\sin \left(\frac{\pi }{n}\right)}\sin (\frac{\pi }{n}+\frac{(n-1)\pi }{n})$

$=0$

$\Rightarrow nS=n\times 0=0$