How to prove that a sum of $\cosh(kx)$ is equal to a formula? I need to prove that 
$$\sum_{k=0}^{n}\cosh(kx) = \frac{\sinh((n+1/2)x) + \sinh(x/2)}{2\sinh(x/2)}$$
Can you help me out? How do I even start?
 A: $$\begin{align}
\sum_{k=0}^n \cosh(kx)&=\frac{1}{2}\sum_{k=0}^n \left(e^{kx}+e^{-kx}\right)\\\\
&=\frac12 \left(\frac{1-e^{(n+1)x}}{1-e^{x}}+\frac{1-e^{-(n+1)x}}{1-e^{-x}}\right)\\\\
&=\frac12\frac{e^{nx}\sinh\left(\frac{n+1}{2}x\right)}{\sinh\left(\frac{1}{2}x\right)}+\frac12 \frac{e^{-nx}\sinh\left(\frac{n+1}{2}x\right)}{\sinh\left(\frac{1}{2}x\right)}\\\\
&=\frac{\sinh\left(nx\right)\sinh\left(\frac{n+1}{2}x\right)}{2\sinh\left(\frac{1}{2}x\right)}\\\\
&=\frac{\sinh\left(\frac{2n+1}{2}x\right)+\sinh\left(\frac x2\right)}{2\sinh\left(\frac{1}{2}x\right)}
\end{align}$$
as was to be shown, where we used a Prosthaphaeresis Identity  to arrive at the last equality!
A: Note that $\sinh(k + 1/2)x - \sinh(k-1/2)x \equiv 2\sinh \frac{x}{2}\cosh kx $ so your sum is simply $$\sum_{k=0}^n \cosh kx = \frac{1}{2\sinh \frac{x}{2}}\sum_{k=0}^n \left(\sinh (k+1/2)x - \sinh(k -1/2)x \right)$$ 
Then, telescoping to victory we regain $$\sum_{k=0}^{n} \cosh kx = \frac{\sinh (n+1/2)x + \sinh(x/2)}{2\sinh \frac{x}{2}}$$
