1. ## Hyperbolic Function proof

I'm required to prove that:

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

The furthest I can get is this:

$\displaystyle \sinh ^{-1}\left( x \right)\; =\; \frac{1}{\frac{1}{2}\left( e^{x}-e^{-x} \right)}$

I have no idea what to do Any kind of nudge in the right direction would be greatly appreciated.

2. Originally Posted by DJ Hobo
I'm required to prove that:

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

The furthest I can get is this:

$\displaystyle \sinh ^{-1}\left( x \right)\; =\; \frac{1}{\frac{1}{2}\left( e^{x}-e^{-x} \right)}$

I have no idea what to do Any kind of nudge in the right direction would be greatly appreciated.
$\displaystyle \sinh(x)=\frac{e^x-e^{-x}}{2}$. So, use the normal method. $\displaystyle x=\frac{e^{y}-e^{-y}}{2}\implies xe^y=\frac{e^{2y}-1}{2}$. Call $\displaystyle z=e^{y}$ and see if anything pops out at you.

3. Originally Posted by DJ Hobo
I'm required to prove that:

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

The furthest I can get is this:

$\displaystyle \sinh ^{-1}\left( x \right)\; =\; \frac{1}{\frac{1}{2}\left( e^{x}-e^{-x} \right)}$

I have no idea what to do Any kind of nudge in the right direction would be greatly appreciated.
Incorrect.

The notation $\displaystyle \sinh^{-1}{x}$ does NOT mean $\displaystyle \frac{1}{\sinh{x}}$. It stands for the inverse hyperbolic sine function.

It's the same sort of notation used in trigonometric functions to show the inverse trigonometric functions. E.g. $\displaystyle \sin^{-1}x = y$ implies $\displaystyle x = \sin{y}$.

$\displaystyle y = \sinh{x}$

$\displaystyle = \frac{1}{2}(e^x - e^{-x})$.

Its inverse function is

$\displaystyle x = \sinh{y}$ (in other words, $\displaystyle y = \sinh^{-1}{x}$)

$\displaystyle x = \frac{1}{2}(e^y - e^{-y})$

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

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

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

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

$\displaystyle 0 = e^{2y} - 2x\,e^y - 1$

$\displaystyle 0 = e^{2y} - 2x\,e^y + (-x)^2 - (-x)^2 - 1$

$\displaystyle 0 = \left(e^y - x\right)^2 - x^2 - 1$

$\displaystyle x^2 + 1 = \left(e^y - x\right)^2$

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

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

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

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

4. ah yes, I made a very stupid mistake to start with. Thanks for the help, guys