Inverse of cosh

**circuscircus** · September 27th 2007, 10:55 AM

$cosh^{-1}x=ln(x+\sqrt{x^2-1}), x>=1$

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

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

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

**topsquark** · September 27th 2007, 11:04 AM

Originally Posted by circuscircus

$cosh^{-1}x=ln(x+\sqrt{x^2-1}), x>=1$

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

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

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

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

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

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

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

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

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

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

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

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

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

-Dan

**Krizalid** · September 27th 2007, 11:28 AM

Another way to see this

Recall

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

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

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

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

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

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

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

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