How to show $\displaystyle\sum_{j=1}^n \cos\left((k+l)\frac{\pi (j-1/2)}{n}\right) = 0$

Question

This result seems trivial, how would I show $$\sum_{j=1}^n \cos\left((k+l)\frac{\pi (j-1/2)}{n}\right) = 0$$ Where $0 \le k,l \le n-1$ And $k \neq l \neq 0$.

I tried to use the exponential definition and geo sum formula but I didn't get any trivial simplifications. Any ideas?

Answer 1

$\begin{equation} \begin{split} &\displaystyle \sum_{j=1}^N \cos \left( \dfrac{k \pi}{N} \left({j-\frac{1}{2} }\right) \right) \\ &= \displaystyle \cos \left(\dfrac{k \pi}{2N} \right)\sum_{j=1}^N \cos\left( \dfrac{ jk \pi}{N} \right) + \sin \left(\dfrac{k \pi}{2N}\right) \sum_{j=1}^N \sin\left( \dfrac{ j k\pi}{N}\right) \end{split}\end{equation}$

Now we have,

$\begin{equation} \begin{split} \displaystyle & \sum_{j=1}^N \cos \left(\dfrac{j k \pi}{N} \right)\\ &= \displaystyle Re \left(\sum_{j=1}^N \exp \left( \dfrac{j ik \pi}{N} \right) \right) \\ &= \exp \left( \dfrac{i k \pi}{N} \right) \cdot \dfrac{ \exp\left( i k \pi \right) - 1 }{ \exp \left(\dfrac{i k \pi}{N} \right) - 1 } = 0 \\ \end{split} \end{equation} $

and similarly for the $\sin$ term.

Answer 2

The identity holds only for even $k+l$. Suppose $k+l$ is even. Using $$\sum_{k=0}^{n-1}=\frac{1-x^n}{1-x}$$ one has \begin{eqnarray} &&\sum_{j=1}^n \cos\left((k+l)\frac{\pi (j-1/2)}{n}\right) \\ &=&\Re\sum_{j=1}^n e^{(k+l)\frac{\pi (j-1/2)}{n}i} \\ &=&\Re \bigg[e^{-(k+l)\frac{\pi}{2n}i}\sum_{j=1}^n e^{(k+l)\frac{\pi j}{n}i} \bigg]\\ &=&\Re \bigg[e^{-(k+l)\frac{\pi}{2n}i}\sum_{j=1}^n \bigg(e^{(k+l)\frac{\pi}{n}i} \bigg)^j\bigg]\\ &=&\Re\bigg[e^{-(k+l)\frac{\pi}{2n}i}e^{(k+l)\frac{\pi}{n}i}\frac{1-(e^{(k+l)\frac{\pi}{n}i})^n}{1-e^{(k+l)\frac{\pi}{n}i}}\bigg]\\ &=&0. \end{eqnarray}