Arccosine Logarithmic Formulation
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Theorem
For any real number $x: -1 \le x \le 1$:
- $\arccos x = \dfrac 1 i \map \ln {x + \sqrt {x^2 - 1} }$
where $\arccos x$ is the arccosine and $i^2 = -1$.
Proof
Assume $y \in \R$ such that $0 \le y \le \pi$.
| \(\ds y\) | \(=\) | \(\ds \arccos x\) | ||||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(=\) | \(\ds \cos y\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(=\) | \(\ds \frac 1 2 \paren {e^{-i y} + e^{i y} }\) | Euler's Cosine Identity | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds 2 x\) | \(=\) | \(\ds e^{-i y} + e^{i y}\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds 2 x e^{i y}\) | \(=\) | \(\ds 1 + e^{2 i y}\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds e^{2 i y} - 2 x e^{i y}\) | \(=\) | \(\ds -1\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds e^{2 i y} - 2 x e^{i y} + x^2\) | \(=\) | \(\ds x^2 - 1\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \paren {e^{i y} - x}^2\) | \(=\) | \(\ds x^2 - 1\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds e^{i y} - x\) | \(=\) | \(\ds \sqrt {x^2 - 1}\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds i y\) | \(=\) | \(\ds \map \ln {x + \sqrt {x^2 - 1} }\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds y\) | \(=\) | \(\ds \dfrac 1 i \map \ln {x + \sqrt {x^2 - 1} }\) |
$\blacksquare$