Calculate $sum_{r=0}^n cosh(alpha+2rbeta)$

I try to calculate $sum_{r=0}^ncos(alpha+2rbeta)$ first to get some insights.

First we prove that $sum_{r=0}^inftycos(alpha+2rbeta)=0$. (The statement and the proof is flawed, I will edit it later.)

$Proof$: For $sum_{r=0}^inftycos(alpha+2rbeta)$ is the real part of $sum_{r=0}^inftyexp i(alpha+2rbeta)$, and the latter (denoted by $X$), as addition of complex numbers, can be viewed as addition of vectors (of module 1), we have:

when $2beta=2pi/n$, $X$ equals to sum of vectors that corresponds to edges of a n-polygon, and so $X=0$;

when $2betaneq2pi/n$, $2pi/n = kcdot2beta+rho$ where $0<rho<2beta$, and so by summing $k$ vectors we get a vector $z_{1,j_1}$, i.e. $sum_{r=j_1}^{j_1+k-1}exp i(alpha+2rbeta)$, that is of smaller module and smaller ‘phase shift’, say $beta_1$.

Then we have $2pi/n = k_1cdot2beta_1+rho_1$, and so by summing $k_1$ vectors we get a new vector $z_{2, j_2} = sum_{r=j_2}^{j_2+k-1}z_{1,j_1}$, that is of smaller module and smaller ‘phase shift’, say $beta_2$.

Repeating the process we will finally get smaller and smaller module and ‘phase shift’, the sequence of modules will tend to $0$ (the proof is left out), which completes the proof. $blacksquare$

But I have no idea yet how to do similar things to $sum_{r=0}^ncosh(alpha+2rbeta)$. The latter can be viewed as a component of $sum_{r=0}^n e^{alpha+2rbeta}$. But one cannot easily separate the former from the latter as we do above (for here is no clear distinction between components like one between real and imaginary parts). And the sum itself can’t be calculated conveniently as sum of vectors.

So is this method workable for $sum_{r=0}^n cosh(alpha+2rbeta)$? If not, can anyone just give me a hint about how to calculate it?

(Thanks for pointing out that the sum doesn’t converge as $nrightarrowinfty$.)

A graph representation:
https://www.desmos.com/calculator/nih1sg4fwm