Finding inverse cosh I am trying to find $\cosh^{-1}1$ I end up with something that looks like $e^y+e^{-y}=2x$. I followed the formula correctly so I believe that is correct up to this point. I then plug in $1$ for $x$ and I get $e^y+e^{-y}=2$ which, according to my mathematical knowledge, is still correct. From here I have absolutely no idea what to do as anything I do gives me an incredibly complicated problem or the wrong answer.

start with
$$\cosh(y)=x$$
since
$$\cosh^2(y)-\sinh^2(y)=1$$ or $$x^2-\sinh^2(y)=1$$
then
$$\sinh(y)=\sqrt{x^2-1}$$
now add $\cosh(y)=x$ to both sides to make
$$\sinh(y)+\cosh(y) = \sqrt{x^2-1} + x $$
which the left hand side simplifies to : $\exp(y)$
so the answer is $$y=\ln\left(\sqrt{x^2-1}+x\right)$$
