Proving $\sum_{n=1}^m\sin^2\left(n\frac\pi{m}\right)=\frac{m}2$ for $m\geq 2$ by induction 
$$\sum_{n=1}^m \sin^2\left(n\frac{\pi}{m}\right) = \frac{m}{2}$$ for all $m\ge2$ positive integers

I'm doing a maths problem that involves proving by mathematical induction the sum of a sigma. I find it hard because the highest term is included in the sigma so it changes the actual sigma.
If anyone can help me, please respond it would be very useful!! Even just the smallest help will go a long way : )
Here is the Desmos link to show you this is true.
But I want to prove it using induction.
Thanks.
 A: HINTS
Write $$\sin^2(n \pi /m)= \frac{1 - \cos(2 n \pi/m)}{2}.$$
Then by the linearity of finite sums, we have
$$\sum_{n=1}^m \sin^2(n \pi /m) = \sum_{n=1}^m \frac{1 - \cos(2 n \pi/m)}{2}=\frac{m}{2}-\sum_{n=1}^m \frac{\cos(2n \pi/m)}{2},$$ which is almost the answer.
You can now show that $\sum_{n=1}^m \frac{\cos(2n \pi/m)}{2}=0$ for $m \ge 2$ by an induction argument or by noticing it is the real part of the finite geometric series $$\frac{1}{2}\sum_{n=1}^m e^{2 \pi i n/m}.$$
A: Well I guess I found a proof without using induction. Perhaps it will be of any use to you so I will leave it here.
$$\sin\varphi=\frac{e^{i\varphi}-e^{-i\varphi}}{2i}$$
Let $\varphi_n=\frac{n\pi}{m}$
$$ \sum_{n=1}^{m}\sin^2{\varphi_n}=\sum_{n=1}^m\left(\frac{e^{i\varphi_n}-e^{-i\varphi_n}}{2i}\right)^2=\sum_{n=1}^m\frac{e^{2i\varphi_n}+e^{-2i\varphi_n}-2}{-4}=\frac{m}{2}-\left(\sum_{n=1}^me^{2i\varphi_n} +\sum_{n=1}^me^{-2i\varphi_n}\right)=\frac{m}{2}-\left(\sum_{n=1}^me^{2i\frac{n\pi}{m}} +\sum_{n=1}^me^{-2i\frac{n\pi}{m}}\right)=\frac{m}{2}-\left( \frac{e^{2\pi i}-1}{e^\frac{2\pi i}{n}-1}+\frac{e^{-2\pi i}-1}{e^{-\frac{2\pi i}{n}}-1}\right)=\frac{m}{2}-\left( \frac{1-1}{e^\frac{2\pi i}{n}-1}+\frac{1-1}{e^{-\frac{2\pi i}{n}}-1}\right)=\frac{m}{2}$$
