$\cos(y\,\operatorname{acosh}(\exp(x)))$ is real for all real $x,y$ even though $\operatorname{acosh}$ is complex for $x<1$. I found it empirically but still can't prove it yet. Can someone please offer or point me to a proof? My search yielded nothing though it is hard to believe that this result isn't published.
Obviously I already tried Euler's formula and the logarithmic representation of inverse hyperbolic functions. They didn't help yet.
To see a normalized plot of this function on Wolfram Alpha click here
For $x \in\mathbb{R}$ we have $e^x>0 \in \mathbb{R}$ and, $0<e^x<1$ for $x<0$ and $e^x\ge 1$ for $x \ge 0$. So, for $x \ge 0$ $\mbox{acosh}( e^x) \in \mathbb{R}$ and, if $y \in \mathbb{R}$, also $\cos (y \,\mbox{acosh}( e^x)) \in \mathbb{R}$.
If $x<0$ and $0<e^x<1$ than $\mbox{acosh}( e^x)$ is a pure imaginary number and, since $\cos iz=\cosh z \in \mathbb{R}$ for $z\in \mathbb{R}$, we have $\cos (y \,\mbox{acosh}( e^x)) \in \mathbb{R}$.