# Inverse of cosh

• Sep 27th 2007, 09:55 AM
circuscircus
Inverse of cosh
$\displaystyle cosh^{-1}x=ln(x+\sqrt{x^2-1}), x>=1$

I need to derive the formula above by using the definition of cosh (which I think is below) and then solving for the inverse

$\displaystyle cosh x = \frac{e^{-x} + e^x}{2}$

How do I get the e^-x and e^x to join to create like 1 variable so I can solve for the inverse?
• Sep 27th 2007, 10:04 AM
topsquark
Quote:

Originally Posted by circuscircus
$\displaystyle cosh^{-1}x=ln(x+\sqrt{x^2-1}), x>=1$

I need to derive the formula above by using the definition of cosh (which I think is below) and then solving for the inverse

$\displaystyle cosh x = \frac{e^{-x} + e^x}{2}$

How do I get the e^-x and e^x to join to create like 1 variable so I can solve for the inverse?

Let
$\displaystyle y = cosh(x) = \frac{e^x + e^{-x}}{2}$

Now switch the roles of x and y:
$\displaystyle x = \frac{e^{y} + e^{-y}}{2}$

$\displaystyle e^y + e^{-y} = 2x$

Multiply both sides by $\displaystyle e^y$:
$\displaystyle e^{2y} + 1 = 2xe^{y}$

$\displaystyle e^{2y} - 2xe^y + 1 = 0$

This is a quadratic in $\displaystyle e^x$, so use the quadratic formula:
$\displaystyle e^y = \frac{2x \pm \sqrt{4x^2 - 4}}{2}$

$\displaystyle e^y = \frac{2x \pm 2\sqrt{x^2 - 1}}{2}$

$\displaystyle e^y = x \pm \sqrt{x^2 - 1}$

Note that the "-" solution makes $\displaystyle e^y$ negative, which is impossible. Thus
$\displaystyle e^y = x + \sqrt{x^2 - 1}$

$\displaystyle y = ln (x + \sqrt{x^2 - 1} )$

Then this y is the inverse function.
$\displaystyle cosh^{-1}(x) = ln (x + \sqrt{x^2 - 1} )$

-Dan
• Sep 27th 2007, 10:28 AM
Krizalid
Another way to see this

Recall

$\displaystyle \int\frac1{\sqrt{x^2-1}}\,dx=\mathrm{arccosh}\, x+\tau$, where $\displaystyle \tau$ is a constant

Since $\displaystyle \frac1{\sqrt{x^2-1}}$ can be integrated by makin' a substitution defined by $\displaystyle u=x+\sqrt{x^2-1}$, it yields

$\displaystyle \int\frac1{\sqrt{x^2-1}}\,dx=\ln\left|x+\sqrt{x^2-1}\right|+\tau$

Since two antiderivatives of a function can differ at most by a constant, there must exist a constant $\displaystyle k$ such that

$\displaystyle \mathrm{arccosh}\, x=\ln\left|x+\sqrt{x^2-1}\right|+k,\,\forall x\ge1$

Evaluating this equality for $\displaystyle x=1\implies k=0$

Therefore $\displaystyle \mathrm{arccosh}\, x=\ln\left(x+\sqrt{x^2-1}\right)\,\blacksquare$