How do I prove that $\sum_{i=1}^m \cos^2\left(\frac{2\pi i}{m}\right) = \sum_{i=1}^m \sin^2\left(\frac{2\pi i}{m}\right) = \frac{m}{2}$?

Question

I haven't had any good ideas nor found any helpful identities so far, so I'd appreciate some help.

Also, here $m > 2$.

Update: Thanks to the hints and to this previous post I managed to get to the conclusion I wanted. Thanks guys.

Hint: show the two sums are equal by using the identity $\cos^2\theta-\sin^2\theta=\cos2\theta$. — Barry Cipra, May 18 '18 at 12:02

score 2 · Accepted Answer · answered May 18 '18 at 12:06

Hint:

Use $\cos2\theta=2\cos^2\theta-1$ to get

$$\sum_{i=1}^{m}\cos^2\bigg(\frac{2\pi i}{m}\bigg)=\frac{1}{2}\sum_{i=1}^{m}1+\cos\bigg(\frac{4\pi i}{m}\bigg)\ .$$

Now you can see where the $m/2$ comes from. For the $\cos(4\pi i/m)$ term you could do a little cancelling by symmetry.

score 2 · Answer 2 · answered May 18 '18 at 12:12

2

Once you prove the first equality, you can get the second by adding together the cosine and sine terms and using the identity $\sin^2\theta + \cos^2\theta = 1$.

answered May 18 '18 at 12:12

Jalex Stark

659

How do I prove that $\sum_{i=1}^m \cos^2\left(\frac{2\pi i}{m}\right) = \sum_{i=1}^m \sin^2\left(\frac{2\pi i}{m}\right) = \frac{m}{2}$?

2 Answers2