Proof for value of sum of sine and cosine

Question

I have come across those sums of sine and cosine, while trying to show that windmills dont move without external force. Although it is clear that should be the case, i'm stuck in proofing it. I wonder if there is some way to proof $$\sum_{i=0}^{n-1} \cos\left(\frac{2\pi i}{n}\right) = 0 \tag{1} \label{1} $$

In case of sine the proof is easy because of simple symmetry $$\sum_{i=0}^{n-1} \sin\left(\frac{2\pi i}{n}\right) = 0 \tag{2}$$

Also if n is even $\sum_{i=0}^{n-1} \cos\left(\frac{2\pi i}{n}\right) = 0$ is easy to show. So my question is: How do you prove or at least argue that $\eqref{1}$ holds for all n?

Answer 1

Write the sum as

$$\sum_{k=0}^{n-1}\cos(\frac{2\pi k}{n})=Re\sum_{k=0}^{n-1}e^{i2\pi k/n}= Re\left\{\frac{1-e^{i2\pi n/n}}{1-e^{i2\pi/n}}\right\}=0,\quad n>1$$

(Note that I used $k$ as the summation index, and $i$ as the imaginary unit.)

Answer 2

Let $$ \zeta=e^{2\pi i/n}=\cos\left(\frac{2\pi}n\right)+i\sin\left(\frac{2\pi}n\right). $$ then for all $k=0,...,n-1$, $$ \zeta^k=e^{2\pi ik/n}=\cos\left(\frac{2k\pi}n\right)+i\sin\left(\frac{2k\pi}n\right). $$ Since $\zeta^n=1$ but $\zeta\neq1$ we must have $$ 1+\zeta+\zeta^2+\cdots+\zeta^{n-1}=0. $$ Take real and imaginary parts.