How to prove that a sum of $\cosh(kx)$ is equal to a formula?

Question

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?

Answer 1

$$\begin{align} \sum_{k=0}^n \cosh(kx)&=\frac{1}{2}\sum_{k=0}^n \left(e^{kx}+e^{-kx}\right)\\\\ &=\frac12 \left(\frac{1-e^{(n+1)x}}{1-e^{x}}+\frac{1-e^{-(n+1)x}}{1-e^{-x}}\right)\\\\ &=\frac12\frac{e^{nx}\sinh\left(\frac{n+1}{2}x\right)}{\sinh\left(\frac{1}{2}x\right)}+\frac12 \frac{e^{-nx}\sinh\left(\frac{n+1}{2}x\right)}{\sinh\left(\frac{1}{2}x\right)}\\\\ &=\frac{\sinh\left(nx\right)\sinh\left(\frac{n+1}{2}x\right)}{2\sinh\left(\frac{1}{2}x\right)}\\\\ &=\frac{\sinh\left(\frac{2n+1}{2}x\right)+\sinh\left(\frac x2\right)}{2\sinh\left(\frac{1}{2}x\right)} \end{align}$$

as was to be shown, where we used a Prosthaphaeresis Identity to arrive at the last equality!

Answer 2

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 $$\sum_{k=0}^{n} \cosh kx = \frac{\sinh (n+1/2)x + \sinh(x/2)}{2\sinh \frac{x}{2}}$$