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 ex 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


Helmut Knaust

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