trigonometry - Proof of a summation

Here is the problem I am stuck with: Is it true that for every positive integer $n > 1$,
$$\sum\limits_{k=1}^n \cos \left(\frac {2 \pi k}{n} \right) =0= \sum \limits_{k=1}^n \sin \left(\frac {2 \pi k}{n} \right)$$
I'm imagining the unit circle and adding up the value of both trig functions separately but I cannot picture see how their sum add up to $0$.

EDIT I updated the question from earlier and realized that it was my fault because of a major typo. At least it was better to point it out late than never. So how would I go about solving this question as of now?

Originally the question was: $$\sum\limits_{k=1}^n \frac {\cos(2πk)}{n} =0= \sum\limits_{k=1}^n \frac {\sin(2πk)}{n}$$

Answer

Old answer to old question:

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Multiply both sides by $n$ to get

$$\sum_{k=1}^n\cos(2\pi k)\stackrel?=0\stackrel?=\sum_{k=1}^n\sin(2\pi k)$$

It's easy enough to see that for positive integers $k$, $\cos(2\pi k)=1$ and $\sin(2\pi k)=0$, thus,

$$\sum_{k=1}^n\sin(2\pi k)=0$$

$$\sum_{k=1}^n\cos(2\pi k)=n\ne0$$