- #1
phyzz
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Homework Statement
Simplify the following expression:
[tex] arccosh \left(\frac{1}{\sqrt{1 - x^2}}\right) \forall x ∈ (-1, 1)[/tex]
Homework Equations
[tex] cosh(u) = \left(\frac{1}{\sqrt{1 - tanh^{2}u}}\right) u ∈ ℝ [/tex]
The Attempt at a Solution
[tex] x = tanhu ∴ u = arctanhx [/tex]
[tex] u ∈ (arctanh(-1), arctanh(1)) [/tex]
[tex] arccosh(coshu) = u = arctanhx [/tex]
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 [tex] x ∈ [0, 1) [/tex] (Why are they supposing this?)
and that [tex] u = arctanhx ≥ 0 [/tex]
so
[tex] cosh(u) = \left(\frac{1}{\sqrt{1 - tanh^{2}u}}\right) = \frac{1}{\sqrt{1 - x^2}} [/tex] so therefore
[tex] arccosh \left(\frac{1}{\sqrt{1 - x^2}}\right) = arccosh(coshu) [/tex]
but [tex] arccosh(coshu) = u [/tex] as long as [tex] u ≥ 0 [/tex] which is the case here
From where [tex] arccosh \left(\frac{1}{\sqrt{1 - x^2}}\right) = u = arctanh(x) [/tex]
Since [tex] arccosh \left(\frac{1}{\sqrt{1 - x^2}}\right) [/tex] is an even function (another big ? here, don't know what an even function is) in [tex] x ∈ (-1, 1) [/tex] we obtain [tex] \forall x ∈ (-1, 1), arccosh \left(\frac{1}{\sqrt{1 - x^2}}\right) = arctanh(|x|) [/tex]
I appreciate the help!