Let $y=\cosh^{-1}{x}$. Then, $x=\cosh{y}=\frac{1}{2}\left(e^y+e^{-y}\right)$. Multiplying by $2e^y$, we get $2xe^y=e^{2y}+1$. Solving $e^{2y}-2xe^y+1=0$ by the quadratic formula, we have $e^y=\frac{2x\pm\sqrt{4x^2-4}}{2}=x\pm\sqrt{x^2-1}$. We find that both roots are possible. Thus, $y=\ln\left(x\pm\sqrt{x^2-1}\right)$. Why did we fail to eliminate the minus?
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Because $\operatorname{arccosh}$ is the inverse of the restriction of $\cosh$ to $[0,+\infty)$, so you want $y\geq0$... and since $x-\sqrt{x^2-1}\leq1$, we need to take $x+\sqrt{x^2-1}$. – gniourf_gniourf Dec 31 '16 at 13:31
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The reason why we can dispose of the minus sign is that, firstly, $$(x+\sqrt{x^2-1})^{-1}=(x-\sqrt{x^2-1})$$
Therefore $$\ln(x\pm \sqrt{x^2-1})=\ln(x+\sqrt{x^2-1})^{\pm1}=\pm\ln(x+\sqrt{x^2-1})$$
But $\operatorname{arcosh}(x)$ is defined as non-negative, so the minus sign is not required
David Quinn
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The following is a graph of $y=\cosh(x):$
But the next one is the graph of $y=\operatorname{arccosh}(x)$, considering both real branches$:$
Applying domain restrictions or branch restrictions, one can remove the $\pm$, but I think the graphs make it clear why one would have the $\pm$ in the first place.
Simply Beautiful Art
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The graphs remind me of the square root function. It seems conventional to choose the positive output. – W. Zhu Dec 31 '16 at 14:23
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since the definition of the $$\sqrt{x^2-1}$$ we get $$|x|>1$$ which is for $$x-\sqrt{x^2-1}$$ impossible
Dr. Sonnhard Graubner
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Nop. It is always true that $;x>\sqrt{x^2-1};$ . Perhaps you should read other answers/commetns before you post: this already was addressed. – DonAntonio Dec 31 '16 at 13:33
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1@DonAntonio No, I think he was pointing out that it must be the case that $x\ge1$, but that still doesn't explain why we would have the $\pm$. – Simply Beautiful Art Dec 31 '16 at 13:34
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@SimpleArt Then you were able to understand way more than I was from this cryptic answer... – DonAntonio Dec 31 '16 at 13:35
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As $y\ge 0$, $\;\mathrm e^y\ge 1$. Now $\mathrm e^y$ is a root of the quadratic polynomial $\;P(t)=t^2-2xt+1$. Just observe that, as $P(1)=2(1-x) < 0$ for $x>1$, $1$ separates the roots. Hence $\mathrm e^y$ is the greatest root — with a + sign.
Bernard
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If $y\ge0$, since I see nowhere claiming that $y$ must be greater than $0$. – Simply Beautiful Art Dec 31 '16 at 13:39
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It's in the definition of $\operatorname{argcosh}$. As $\cosh$ is not bijective, we have to consider its restriction to an interval on which it is bijective. One has the choice between $(-\infty,0]$ andr $[0,+\infty)$. The normal choice has been the second for centuries. This is the same problem as for the inverse trigonoùetric functions or the square root function. – Bernard Dec 31 '16 at 13:46