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.
 A: You have found out that the unknown $y$ satisfies the equation $e^y+e^{-y}=2$. Multiply by $e^y$ and rearrange terms. You then get
$$e^{2y}-2e^y+1=0\ .$$
Now use the following trick: Put $e^y=:u$ with a new unknown $u$. This $u$ has to satisfy the quadratic equation
$$u^2-2u+1=0\ ,\quad{\rm i.e.,}\quad (u-1)^2=0\ .$$
The last equation has the unique solution $u=1$. The corresponding $y$ therefore satisfies the equation $e^y=1$, and there is only one such real $y$, namely $y=0$.
All in all we have shown that $\cosh^{-1}(1)=0$, which is corroborated by the fact that conversely  $\cosh(0)={1\over2}(e^0+e^{-0})=1$. 
A: To find the inverse for any $x$, we are looking for 
$$ y = \cosh^{-1} x, $$
i.e. $$x = \cosh y = \frac{1}{2} (e^y + e^{-y}). $$
Multiplying through by $2e^y$ gives
$$ (e^y)^2 -2x\,e^y + 1 = 0, $$
which is a quadratic in $e^y$.  You can then use the quadratic formula, or here completing the square
$$ (e^y-x)^2 - x^2 + 1 = 0, $$
$$ e^y -x = \sqrt{x^2 - 1}, $$
$$ y = \ln \left(x+\sqrt{x^2-1}\right). $$
At the start I said "for any $x$", but observe that the result is only valid for $x\ge 1$.  This is the inverse for the right-hand side of the (even) function $y=\cosh x$, with $x\ge0$ and $y\ge1$.
A: 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 $$y=\ln\left(\sqrt{x^2-1}+x\right)$$
A: It may be more helpful to consider the significant hyperbolic identities first. We have in general:    
$\small \begin{array} {rcllll} 
1)&  \exp(z) &=& \cosh(z) + \sinh(z) \\  
2)&   1 &=&  \cosh(z)^2 - \sinh(z)^2 \\ 
&&&  \implies \\
3)&\sinh(z) &=& \pm \sqrt{\cosh(z)^2-1} & \text{ using 2)}\\
4)& \exp(z)&=& \cosh(z) \pm \sqrt{\cosh(z)^2-1}  & \text{ using 1) and 3)}\\
 \end{array} $      
Now the given problem is to find another expression for $\small y=\cosh^{-1}(x)$ which means $\small x = \cosh(y) $
We use 4) and insert our current y for the general z to get    
$\small \begin{array} {rcllll} 
5)& \exp(y)&=& \cosh(y) \pm \sqrt{\cosh(y)^2-1}  & \text{ using 4)}\\
6)& \exp(y)&=& x \pm \sqrt{x^2-1}  & \text{ inserting x for } \cosh(y)\\
7)& y&=& \log(x \pm \sqrt{x^2-1} ) & \\
8)& \cosh^{-1}(x)&=& \log(x \pm \sqrt{x^2-1} ) &\text{ inserting } \cosh^{-1}(x) \text{ for } y \\
9)& \cosh^{-1}(1)&=& ??? \\
 \end{array} $      
Now 8) can be used as a new, general hyperbolic identity like that in the list from 1) to 4) and 9) is your remaining little to-do ...
A: $$
e^y+e^{-y}=2
$$
Letting $u = e^y$, this becomes
$$
u + \frac 1u = 2
$$
Multiplying both sides by $u$:
$$
u^2 + 1 = 2u
$$
That's just a quadratic equation.
