Inverse Hyperbolic Functions

Since $\cosh x$ is defined in terms of the exponential function, you should not be surprised that its inverse function can be expressed in terms of the logarithmic function:

Let's set $\displaystyle y=\cosh(x)$ , keep in mind that we restrict to $x\geq 0$ , and try to solve for x:

$\begin{eqnarray*}&&y=\cosh(x)\\ &\Longleftrightarrow&2y=e^x+e^{-x}\\ &\Longlef... ...rrow&2ye^x=e^{2x}+1\\ &\Longleftrightarrow&(e^x)^2-2y (e^x)+1=0 \end{eqnarray*}$

This is a quadratic equation with e^x instead of x as the variable. y will be considered a constant.

So using the quadratic formula, we obtain

$\begin{displaymath}e^x=y\pm\sqrt{y^2-1}\end{displaymath}$

Since $x\geq 0$ we have that $e^x\geq 1$ for all x, and since $y-\sqrt{y^2-1}$ fails to exceed 1 for some y, we have to discard the solution with the minus sign, so

$\begin{displaymath}e^x=y+\sqrt{y^2-1},\end{displaymath}$

and consequently

$\begin{displaymath}x=\ln\left(y+\sqrt{y^2-1}\right).\end{displaymath}$

Read that last sentence again slowly!

We have found out that

$\displaystyle \cosh^{-1} x=\ln\left(x+\sqrt{x^2-1}\right).$