How should I prove that: $\sum_{i=1} ^{n}(\sin(\frac{i\pi}{n}))^2=\frac{n}{2}$

Question

$$\sum_{i=1} ^{n}\Big(\sin\big(\frac{i\pi}{n}\big)\Big)^2=\frac{n}{2}$$

An interesting conclusion and checked for validity...holds for $n\geq 2$, but yet do not know how to prove it. Are there any suggestions? I am trying too....cheers!

Answer 1

Using $\displaystyle\cos2A=1-2\sin^2A\iff2\sin^2A=\cdots$

$$\sum_{r=1}^n2\sin^2\frac{r\pi}n=\sum_{r=1}^n\left(1-\cos\frac{2r\pi}n\right)=n-\sum_{r=1}^n\cos\frac{2r\pi}n$$

Now use $\sum \cos$ when angles are in arithmetic progression to find $$\sum_{r=1}^n\cos\frac{2r\pi}n=0$$