Dear Uncle Colin,

Can you simplify $\cosh(\arsinh(x))$?

- Clumsy Oaf Shunning Hyperbolics

Hi, COSH, and thanks for your message!

I can see two ways of doing this that I think are ok.

In both cases, let’s let $y = \cosh(\arsinh(x))$.

Let’s do the clumsy way first:

Let $\sinh(A) = x$, so $e^{A} - e^{-A} = 2x$, by definition.

Rearranging gives $e^{2A} - 2xe^{A} - 1$, which is quadratic in $e^A$; it results in $e^{A} = x \pm \sqrt{1+x^2}$. Since $e^A$ is positive, we need the positive root of this.

Now, $y = \cosh(A)$, so $2y = e^{A} + e^{-A}$.

That works out to be $x + \sqrt{1+x^2} + \frac{1}{x+\sqrt{1+x^2}}$, and I hope you pulled as disgusted a face as I just did.

Let’s introduce a letter to save us some writing: $S = \sqrt{1+x^2}$

$2y = x + S + \frac{1}{x+S} = \frac{(x+S)^2 + 1}{x+S}$

$= \frac{x^2 + 2xS + (x^2 + 1) + 1}{x+S}$

$= \frac{2\left(x^2 + xS + 1\right)}{x+S}$

Now rationalise the denominator, noting that $(x+S)(x-S) = -1$:

$y = \left(x^2 + xS + 1\right)\left(S-x\right)$

$= \left( x^2S + xS^2 + S - x^3 - x^2S - x\right)$

$= \left( x(x^2+1) + S - x^3 - x\right)$

$= S = \sqrt{x^2+1}$

Phew.

### More elegantly

$y^2 = \cosh^2(\arsinh(x))$

$= 1 + \sinh^2(\arsinh(x))$

$= 1 + x^2$

So (since $\cosh(t)>0$ for all real $t$), $y = \sqrt{1+x^2}$.

Hope that helps!

- Uncle Colin

• Edited 2021-08-25 to fix LaTeX