My goal is to find the inverse of $y=\cosh(x)$
Therefore: $$x=\cosh(y)=\frac{e^y+e^{-y}}{2}=\frac{e^{2y}+1}{2e^y}$$ If we define $k=e^y$ then: $$k^2-2xk+1=0$$ $$k=e^{y}=x\pm\sqrt{x^2-1}$$ $$y=\ln(x\pm\sqrt{x^2-1})=\cosh^{-1}(x)$$
However, apparently: $\cosh^{-1}(x)=\ln(x+\sqrt{x^2-1})$ is right, but NOT $\cosh^{-1}(x)=\ln(x-\sqrt{x^2-1})$
What step did I miss?
$\endgroup$ 41 Answer
$\begingroup$The function $\cosh$ is even, so formally speaking it does not have an inverse, for basically the same reason that the function $g(t)=t^2$ does not have an inverse.
But if we restrict the domain of $\cosh$ suitably, then there is an inverse. The usual definition of $\cosh^{-1}x$ is that it is the non-negative number whose $\cosh$ is $x$.
Note that for $x\gt 1$, we have$$x-\sqrt{x^2-1}=\frac{1}{x+\sqrt{x^2-1}}\lt 1,$$and therefore $\ln(x-\sqrt{x^2-1})\lt 0$ whereas we were looking for the non-negative $y$ which would satisfy the inverse equation. Thus, $y=\ln(x+\sqrt{x^2-1})$ is not the non-negative number whose $\cosh$ is $x$.
Remark: If one defined $\cosh^{-1}(x)$ as the non-positive number whose $\cosh$ is $x$, then the answer $\ln(x-\sqrt{x^2-1})$ would be the right one.
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