Question

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

• A. 1
• B. 0
• C. $\frac{n}{2}$
• D. none of these

Answer 1

Answer: (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
$sin\left(\frac{\pi }{n}\right)+sin\left(\frac{3\pi }{n}\right)+sin\left(\frac{5\pi }{n}\right)+...+sin\left(\frac{(2n+1)\pi }{n}\right)=\frac{sin\left(\frac{\left(2n\right)2\pi }{2n}\right)}{sin\left(\frac{2\pi }{2n}\right)}sin(\frac{\pi }{n}+n\frac{2\pi }{n})=0$