Proving that the sum of $\cos k \theta_\mu \cos\theta_\nu$ from k=0(this term is to be halved) to n-1 equals to $\frac{1}{2}n\delta_{\nu\mu}$.

Question

Been trying a while to prove that
\begin{equation*}\sideset{}{'} \sum_{k=0}^{n-1} \cos k\theta_\nu\hspace{0.05cm} \cos k\theta_\mu=\frac{1}{2}n\delta_{\nu\mu}, \hspace{0.3cm}\nu,\mu=1,2,\dots,n, \end{equation*} where the prime means that the first term (for $k=0$) is to be halved. The $\theta_\nu$-s are defined by \begin{equation}\theta_\nu=\frac{2\nu-1}{2n}\pi,\hspace{0.3cm}\nu=1,2,\dots,n.\end{equation} I tried to use the Euler's formula for the cosine (eg $\cos k\theta_\nu=\frac{1}{2}(e^{i\theta_\nu}+e^{-i\theta_\nu}$)) and then summing over $k$ but it didn't work. Any ideas?