Inverse hyperbolic functions (logarithmic form)

1. Jun 18, 2015

To express the $\cosh^{-1}$ function as a logarithm, we start by defining the variables $x$ and $y$ as follows:
$$y = \cosh^{-1}{x}$$
$$x = \cosh{y}$$
Where $y ∈ [0, \infty)$ and $x ∈ [1, \infty)$.
Using the definition of the hyperbolic cosine function, rearranging, and multiplying through by $e^y$, we get:
$$y = \cosh^{-1}{x} = \ln{(x \pm \sqrt{x^2 -1})}$$
How exactly do we get rid of the minus sign?

2. Jun 18, 2015

SteamKing

Staff Emeritus
You haven't shown all the steps in your derivation of the inverse cosh function, especially how you came to have a "±" in your logarithmic expression.

3. Jun 18, 2015

$$x = \cosh{y} = \frac{e^y + e^{-y}}{2}$$
$$e^y - 2x + e^{-y} = 0$$
$$e^{2y} - 2x e^y + 1 = 0$$
$$e^y = \frac{2x \pm \sqrt{4x^2 - 4}}{2}$$
$$e^y = x \pm \sqrt{x^2 - 1}$$
$$y = \cosh^{-1}{x} = \ln{(x \pm \sqrt{x^2 - 1})}$$

4. Jun 18, 2015

SteamKing

Staff Emeritus
5. Jun 18, 2015

If $x > \sqrt{x^2 - 1}$ for all $x \geq 1$, why should we ignore the second solution ($e^y = x - \sqrt{x^2 - 1}$)?

6. Jun 18, 2015

SteamKing

Staff Emeritus
7. Jun 19, 2015