Imaginary $\cos^{-1}$ value significance? When I was bored in AP Psych last year, I jokingly asked myself if there was a cosine inverse of $2$. Curious about it, I tried calculating it as follows:
$$
\begin{align*}
\cos (x) &= 2 \\
\sin (x) &= \sqrt{1 - \cos^2(x)} = \sqrt{1 - 4} = \pm i \sqrt{3} 
\end{align*}
$$
Then, by Euler's formula, you have
$$
\begin{align*}
e^{ix} &= \cos (x) + i \sin (x) \\
e^{ix} &= 2 \pm\sqrt{3} \\
ix  &= \ln (2 \pm  \sqrt{3}) \\
x &= \boxed{-i \ln (2 \pm  \sqrt{3})}
\end{align*}
$$
So, there was a way to calculate the inverse cosine of numbers whose magnitude is greater than $1$ (this was verified on Wolfram Alpha). To what extent is this kind of calculation valid? Does it have any interesting applications/implications in math, or any other subjects? Thanks. :)
Edit I just realized this is very easily explained by $2\cos (x) = e^{ix} + e^{-ix}$, but I'm still curious if this has any significance/intuition.
 A: The trigonometric and hyperbolic functions are interchangeable by switching from real to imaginary.
$$\cos(ix)=\frac{e^{i^2x}+e^{-i^2x}}2=\frac{e^{-x}+e^x}2=\cosh(x),\\
\sin(ix)=\frac{e^{i^2x}-e^{-i^2x}}{2i}=-i\frac{e^{-x}-e^x}2=i\sinh(x),$$
and conversely
$$\cosh(ix)=\cos(x),\\\sinh(ix)=i\sin(x).$$
These are just two facets of the complex exponential.
Also consider the unit circle constraint $c^2+s^2=1$. If you pass it witn $|c|>1$,
$$s=\pm\sqrt{1-c^2}=\pm i\sqrt{c^2-1}$$
is an hyperbola in the imaginary plane $(c,is)$ that you can see as perpendicular to the plane $(c,s)$.
A: I should point out there is an error in your work.  Your equation:
$$e^{ix}=2\pm i\sqrt{3}$$ should instead be
$$e^{ix}=2+ i(\color{red}{\pm i\sqrt{3}}).$$
In fact, the function $f(z)=\arccos(z)$ is purely imaginary for $\Re(z)>1.$  To see why, first let's consider the fact that $\arccos(x)$ has range $[0,\pi].$  So this leads to the unique value $\arcsin 2=\color{red}+i\sqrt{3}.$  That makes
$$\arccos2 = -i\ln(2+i^2\sqrt{3})=\fbox{$-i\ln(2-\sqrt{3})$}.$$
Now for $x\in\mathbb{R},$ we have
$$\arccos{|x|}=-i\ln\left(\cos|x|+i\sqrt{1-\cos^2|x|}\right),$$
where $|\cos x|>1$ returns a pure imaginary value and $|\cos x|\le 1$ returns a real value.
For the case where $|\cos x|\le1$ consider the complex logarithm defined as
$$\ln(a+bi)=\ln\sqrt{a^2+b^2}+i\arctan(b/a),$$
which can be derived via Euler's Formula (note $\arctan(b/a)$ may need to be adjusted for what quadrant the angle lies in).  Letting $a=\cos x$ and $b=\sin x$ under this restriction will eliminate the first term on the RHS.
