A feature of the inverse hyperbolic sine

Considering the hyperbolic sine $\sinh(x)$ and its definition, it should not be surprising that its inverse function $\mathrm{arcsin}(x)$ (also known as $\sinh^{-1}(x)$) is odd, as well. This can be analytically proved in more than one way.

Proof 1

It is presented in WikiProof. Let

$$y = \mathrm{arcsinh}(-x) \label{a} \tag{1}$$

$x \in \mathbb{R}$. Compute the hyperbolic sine of both members in $\ref{a}$ and, as regards the Right Hand Side, apply the definition of inverse hyperbolic sine: $\sinh \left[ \mathrm{arcsinh}(\alpha) \right] = \alpha$:

$$\sinh (y) = -x\\ x = - \sinh (y)$$

The hyperbolic sine is an odd function: therefore,

$$x = \sinh (-y)$$

Compute the inverse hyperbolic sine of both members (and again apply the definition of inverse hyperbolic sine as before, in the Right Hand Side):

$$\mathrm{arcsinh} (x) = -y\\ y = - \mathrm{arcsinh} (x) \label{b} \tag{2}$$

Comparing $\ref{b}$ to $\ref{a}$:

$$\mathrm{arcsinh} (-x) = - \mathrm{arcsinh} (x)$$

which proves the initial statement.

Proof 2

The inverse hyperbolic sine can be defined in terms of a logarithm:

$$\mathrm{arcsinh}(x) = \log \left( x + \sqrt{x^2 + 1} \right) \label{c} \tag{3}$$

where $\log$ is the natural logarithm. If the variable $-x$ is substituted to $x$,

$$\mathrm{arcsinh}(-x) = \log \left( -x + \sqrt{x^2 + 1} \right) \label{d} \tag{4}$$

it seems not immediate to deduce that the function is odd. However, note that:

$$\left( \sqrt{x^2 + 1} + x \right) \cdot \left( \sqrt{x^2 + 1} - x \right) = x^2 + 1 - x^2 = 1$$

Summing the members of $\ref{c}$ and $\ref{d}$:

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

Remembering the logarithmic identity for a product:

$$\mathrm{arcsinh}(x) + \mathrm{arcsinh}(-x) = \log \left[ \left( x + \sqrt{x^2 + 1} \right) \cdot \left( -x + \sqrt{x^2 + 1} \right) \right] =\\ = \log (1) = 0$$

Therefore:

$$\mathrm{arcsinh}(-x) = - \mathrm{arcsinh}(x)$$

which proves the initial statement.

–

This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License

–

The contents in WikiProof was distributed under a Creative Commons Attribution-ShareAlike 3.0 Unported License. According to this page, it is compatible with the license of the present document. For any issue, please contact thecurlingteam ( at ) gmx ( dot ) com.