I have found that the Fourier cosine series from $({-\pi},{\pi})$ of the function $f(x)=\cosh(x)$ is
$$ \frac{2\sinh({\pi})}{\pi}\left[\frac{1}{2}+ \sum_{n\: =\: 1}^{\infty}\:\ \frac{(-1)^n}{n^2+1}\cos(nx)\right]$$
How do I use this to show:
$$ \sum_{n\: =\: 1}^{\infty}\:\ \frac{1}{n^2+1}= \frac{{\pi}\coth({\pi})-1}{2}$$
I have no idea really, the $\coth{\pi}$ has thrown me off.
Based on your calculations we advance as
$$ \cosh(x) = \frac{2\sinh({\pi})}{\pi}\left[1+ \sum_{n\: =\: 1}^{\infty}\:\ \frac{(-1)^n}{n^2+1}cos(nx)\right ]. $$
Substituting $x=\pi$ in the above identity gives
$$ \cosh(\pi) = \frac{2\sinh({\pi})}{\pi}\left[1+ \sum_{n\: =\: 1}^{\infty}\:\ \frac{(-1)^n}{n^2+1}(-1)^n\right ] \\ \iff \cosh(\pi) = \frac{2\sinh({\pi})}{\pi}\left[1+ \sum_{n\: =\: 1}^{\infty}\:\ \frac{1}{n^2+1}\right ] $$
and then simplifying to get the result.
Using Parseval's identity you get that $\frac{2\sinh \pi}{\pi}\left(1 + \sum\limits_{n=1}^n\frac{1}{n^2+1}\right) = \frac{1}{2\pi}\int\limits_{-\pi}^\pi \cosh^2 x dx$.
You can calculate that $\int\limits_{-\pi}^\pi \cosh^2 x dx = \pi + \sinh x \cosh x$, substituting this into Parseval's identity above gives you the result.
For any $x\in(-\pi,\pi)$ we have: $$ f(x)=\cosh(x) = \frac{2\sinh \pi}{\pi}\left(\frac{1}{2}+\sum_{n\geq 1}\frac{(-1)^n}{n^2+1}\cos(nx)\right)\tag{1}$$ and: $$\frac{2\sinh \pi}{\pi}\left(\frac{1}{2}+\sum_{n\geq 1}\frac{(-1)^n}{n^2+1}\cos(\pi n)\right)=\frac{f(\pi)+f(-\pi)}{2}=\cosh\pi\tag{2}$$ Since $\cos(\pi n)=(-1)^n$ it is sufficient to rearrange $(2)$ to get the identity: $$ \sum_{n\geq 1}\frac{1}{n^2+1}=\frac{\pi\coth\pi-1}{2}.\tag{3}$$
