I've not done it yet, but what I am going to try is writing the series as:
\[
\frac 1 2 (e^x+e^{2x}+ \cdots + e^{nx} + e^{-x}+e^{-2x}+ \cdots + e^{-nx})
\]
and then summing the two series. I will also try fiddling the required expression to see if that gives me any clues.

Start by summing the two series:
\[
\frac 1 2 (e^x+e^{2x}+ \cdots + e^{nx} + e^{-x}+e^{-2x}+ \cdots + e^{-nx})
\\
=\frac 1 2 (e^x \times \frac {e^{nx}-1}{e^x-1}+e^{-x} \times \frac {1-e^{-nx}}{1-e^{-x}})
\]

Then you can combine the fractions to get:
\[
\frac 1 2 \times \frac 1 {(e^x-1)(1-e^{-x})}(e^x(e^{nx}-1)(1-e^{-x})+e^{-x}(1-e^{-nx})(e^{x}-1))
\]

The denominator can be written as $(e^{x/2}-e^{-x/2})^2$ and the numerator can be written as $(e^{nx}-1)(e^{x}-1)+(1-e^{-nx})(1-e^{-x})$ which can then be written as $e^{x/2}(e^{nx}-1)(e^{x/2}-e^{-x/2})+e^{-x/2}(1-e^{-nx})(e^{x/2}-e^{-x/2})$.

Some cancelling and factorising and you are there! (I expanded out $\cosh \frac{(n+1)x}2 \times \sinh \frac{nx} 2$ to help me with this...)