Sum Cosine Mod? interval $-\pi:\pi$ split into M equal intervals.
midpoints are $y_K$
but i dont understand how to show
$$
\frac{1}{M}\sum_{j=1}^{M}\cos(mx_{j})=\begin{cases} 1, & \ m \equiv 0\pmod{M}\\ 0, & \text{else} \end{cases}$$
thank you very much
 A: It is convenient to push the interval forward by $\pi$.  Since we are working over a full period of the cosine function, the sum does not change.
Now use the fact that 
$$\cos x=\frac{e^{ix}+e^{-ix}}{2}.$$
Then our sum turns out to be the sum of two finite geometric series. The case $m\equiv 0\pmod{M}$ corresponds to the trivial geometric series of all $1$'s. 
Some detail: After the shift, the midpoints $x'_k$ at which we evaluate $\cos(mx)$ are 
$x'_k=\frac{2\pi(2k+1)}{2M}$, where $k$ ranges from $0$ to $M-1$. So one of the geometric series that we calculate the sum of is
$$\sum_{k=0}^{M-1}\exp\left(\frac{2\pi i(2k+1)m}{2M}\right).\tag{$1$}$$
The other is the conjugate, obtained by replacing $i$ by $-i$.
For the sum $(1)$, by taking out the obvious common factor, we can see that all we need is $\sum_{k=0}^{M-1} \exp(2\pi i m k/M)$. This reduces to finding $\sum_{k=0}^{M-1} t^k$, where $t=2\pi i m/M$. If $1-t\ne 0$, then by the usual formula for the sum of a finite geometric series, the sum is 
$$\frac{1-t^M}{1-t}.$$
This sum (if $1-t\ne 0$) is $0$, since $t^M=1$.  The only situation in which $1-t=0$ is when $m/M$ is an integer, meaning that $m \equiv 0\pmod M$.  
Remark: If we do not push forward by $\pi$, again we obtain a geometric series, with somewhat nicer symmetries. We went to the interval $[0,2\pi]$ mainly because it is more familiar.   
The relevant facts can be also proved without appealing to complex numbers. What we need then is a formula for the sum of the cosines of numbers in arithmetic progression. Please see Wikipedia for the 
relevant formula. The formula can be proved by a telescoping series argument. 
