How do I write asinh in terms of $\log$?

Question

I've found in multiple places (e.g. Wikipedia) that

$$ \sinh^{-1} x = \log\left[x + \sqrt{x^2 + 1}\right] $$

Wikipedia says this can be derived via the quadratic formula. Can anyone explain how I would go about this derivation?

Answer 1

Per Gibbs, we set

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

and let $z = e^x$. We then can write

$$ y = \frac{z - \frac{1}{z}}{2} $$

$$ 2yz = z^2 - 1 $$

$$ z^2 - 2yz - 1 = 0 $$

Using the quadratic formula, we may then find $z$ in terms of $y$:

$$ z = \frac{2y \pm \sqrt{4y^2 + 4}}{2} $$

$$ z = y \pm \sqrt{y^2 + 1} $$

Replacing our definitions of $y$ and $z$, we will have

$$ e^x = \sinh x \pm \sqrt{\sinh^2 x + 1} $$

If we let $x = \sinh^{-1} u$, we will then have

$$ \sinh^{-1} u = \log\left[u + \sqrt{u^2 + 1}\right] $$

where we've dropped the possibility of subtraction with in the $\log$ because the argument to $\log$ must be positive.

Answer 2

By definition $$\sinh(x) = \frac{e^x-e^{-x}}{2}.$$ Set $$y=\frac{e^x-e^{-x}}{2}$$ and try to find $x$. Can you do it?

Answer 3

Let $y = \sinh^{-1}(x)$. Then, by definition of inverse function:

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

Let $u = e^y$. Then:

$$u - \frac{1}{u} = 2x$$ $$u^2 - 1 = 2xu$$ $$u^2 - 2xu - 1 = 0$$

This is a quadratic in terms of $u$, with $a = 1$, $b = -2x$, and $c = -1$. Plugging these numbers into the formula gives:

$$u = \frac{2x \pm \sqrt{4x^2 + 4}}{2}$$ $$u = x \pm \sqrt{x^2 + 1}$$ $$e^y = x \pm \sqrt{x^2 + 1}$$ $$y = \log\left(x \pm \sqrt{x^2 + 1}\right)$$

But note that using the $-$ root from the formula would have us taking the log of a negative number, which does not produce a real number. So we choose the $+$ root instead.

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

QED