Question

The value of $\sin \frac{\pi }{n}+\sin \frac{3\pi }{n}+\sin \frac{5\pi }{n}$..... upto $n$ terms in equal to

• A. $1$
• B. $0$
• C. $\frac{n}{2}$
• D. $\frac{n}{4}$

Answer 1

The correct option is B $0$

Using $\sin \left(x\right)+\sin (x+y)+\sin (x+2y)+...+\sin (x+(n-1)y)=\frac{\sin \left(\frac{ny}{2}\right)}{\sin \left(\frac{y}{2}\right)}\sin (x+\frac{n-1}{2}y)$
We get
$\sin \left(\frac{\pi }{n}\right)+\sin \left(\frac{3\pi }{n}\right)+\sin \left(\frac{5\pi }{n}\right)+...+\sin \left(\frac{(2n+1)\pi }{n}\right)=\frac{\sin \left(\frac{\left(2n\right)2\pi }{2n}\right)}{\sin \left(\frac{2\pi }{2n}\right)}\sin (\frac{\pi }{n}+n\frac{2\pi }{n})=0$