Inverse hyperbolic functions (logarithmic form)

472
18

Main Question or Discussion Point

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?
 

Answers and Replies

SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,794
1,665
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?
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.
 
472
18
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.
$$x = \cosh{y} = \frac{e^y + e^{-y}}{2}$$
$$e^y - 2x + e^{-y} = 0$$
$$e^{2y} - 2x e^y + 1 = 0$$
$$e^y = \frac{2x \pm \sqrt{4x^2 - 4}}{2}$$
$$e^y = x \pm \sqrt{x^2 - 1}$$
$$y = \cosh^{-1}{x} = \ln{(x \pm \sqrt{x^2 - 1})}$$
 
472
18
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}##)?


image.jpg
 

Related Threads for: Inverse hyperbolic functions (logarithmic form)

Replies
5
Views
2K
  • Last Post
Replies
6
Views
20K
  • Last Post
Replies
7
Views
2K
Replies
3
Views
546
  • Last Post
Replies
7
Views
4K
Top