- #1

PFuser1232

- 479

- 20

$$y = \cosh^{-1}{x}$$

$$x = \cosh{y}$$

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?