Arccosine Logarithmic Formulation

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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 $.

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

$\blacksquare$


Also see