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.