Arccosine Logarithmic Formulation

Theorem

For any real number $x: -1 \le x \le 1$:

$\displaystyle \arccos x = -i \ln \left({i \sqrt{1-x^2} + x}\right)$

where $\arccos x$ is the arccosine and $i^2 = -1$.

Proof

Assume $y \in \R$, $0 \le y \le \pi$.

 $\displaystyle y$ $=$ $\displaystyle \arccos x$ $\displaystyle \iff \ \$ $\displaystyle x$ $=$ $\displaystyle \cos y$ $\displaystyle \iff \ \$ $\displaystyle x$ $=$ $\displaystyle \frac 1 2 \left({ e^{-iy} + e^{iy} }\right)$ Cosine Exponential Formulation $\displaystyle \iff \ \$ $\displaystyle 2x$ $=$ $\displaystyle e^{-iy} + e^{iy}$ $\displaystyle \iff \ \$ $\displaystyle 2x e^{iy}$ $=$ $\displaystyle 1 + e^{2iy}$ $\displaystyle \iff \ \$ $\displaystyle 2x e^{iy} - e^{2iy}$ $=$ $\displaystyle 1$ $\displaystyle \iff \ \$ $\displaystyle 2x e^{iy} - e^{2iy} - x^2$ $=$ $\displaystyle 1 - x^2$ $\displaystyle \iff \ \$ $\displaystyle \left({ e^{iy} - x }\right)^2$ $=$ $\displaystyle -\left({ 1 - x^2 }\right)$ $\displaystyle \iff \ \$ $\displaystyle e^{iy} - x$ $=$ $\displaystyle i \sqrt{ 1 - x^2 }$ $\displaystyle \iff \ \$ $\displaystyle y$ $=$ $\displaystyle -i \ln \left({ i \sqrt{ 1 - x^2 } + x }\right)$

$\blacksquare$