Simplfying Inverse Hyperbolic Cosine

[phyzz]

Homework Statement

Simplify the following expression:

arccosh \left(\frac{1}{\sqrt{1 - x^2}}\right) \forall x ∈ (-1, 1)

Homework Equations

cosh(u) = \left(\frac{1}{\sqrt{1 - tanh^{2}u}}\right) u ∈ ℝ

The Attempt at a Solution

x = tanhu ∴ u = arctanhx

u ∈ (arctanh(-1), arctanh(1))

arccosh(coshu) = u = arctanhx

but that gives me the interval with infinities in it which is what's confusing me.

I have the solution but I don't understand it particularly the first line.

4. Solution

Suppose that x ∈ [0, 1) (Why are they supposing this?)

and that u = arctanhx ≥ 0

so
cosh(u) = \left(\frac{1}{\sqrt{1 - tanh^{2}u}}\right) = \frac{1}{\sqrt{1 - x^2}} so therefore

arccosh \left(\frac{1}{\sqrt{1 - x^2}}\right) = arccosh(coshu)

but arccosh(coshu) = u as long as u ≥ 0 which is the case here

From where arccosh \left(\frac{1}{\sqrt{1 - x^2}}\right) = u = arctanh(x)

Since arccosh \left(\frac{1}{\sqrt{1 - x^2}}\right) is an even function (another big ? here, don't know what an even function is) in x ∈ (-1, 1) we obtain \forall x ∈ (-1, 1), arccosh \left(\frac{1}{\sqrt{1 - x^2}}\right) = arctanh(|x|)


I appreciate the help!

 

[bossman27]

Just as a hint for the last part, an even function is one such that f(-x) = f(x), meaning that the function is symmetric across the y-axis. If a function is even, that means we can solve for it over the interval (0, L) and we've implicitly solved it over (-L, L) since it is the same function on both sides of the axis.