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?

More generic : http://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro — lab bhattacharjee, Apr 09 '14 at 14:35

score 1 · Accepted Answer · answered Apr 09 '14 at 09:30

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.)

answered Apr 09 '14 at 09:30

Matt L.

10,636

The questioner has the imaginary unit $i$ in the argument of the trig functions .... – Andrea Mori Apr 09 '14 at 09:32
@AndreaMori No, it's the sum index. – Matt L. Apr 09 '14 at 09:33
oh my! :) bad notation! – Andrea Mori Apr 09 '14 at 09:35
That's why I've changed it in my answer ... :) – Matt L. Apr 09 '14 at 09:35
yeah i thought i migth be confusing, but didnt thought we would need complex numbers – user141899 Apr 09 '14 at 09:36
It's no real problem (no pun intended). As soon as we need $i$ as the imaginary unit (which is the case in my solution), we simply change the summation index. – Matt L. Apr 09 '14 at 09:37
could you elaborate $Re\sum_{k=0}^{n-1}e^{i2\pi k/n}= Re\left{\frac{1-e^{i2\pi n/n}}{1-e^{i2\pi/n}}\right}$, i'm afraid it's a bit too fast for me. – user141899 Apr 09 '14 at 09:51
I've used the formula for a geometric series: $\sum_{k=0}^{n-1}r^k=(1-r^n)/(1-r)$. – Matt L. Apr 09 '14 at 10:23
thank you, it's so simple :-). – user141899 Apr 09 '14 at 11:00

Andrea Mori · Answer 2 · 2014-11-01T23:48:40.313

1

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.

edited Nov 01 '14 at 23:48

answered Apr 09 '14 at 09:40

Andrea Mori

26,969

it should be $\zeta^k=e^{2k\pi i/n}$, right? – user141899 Apr 09 '14 at 09:44

Proof for value of sum of sine and cosine

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