Arcsine Logarithmic Formulation
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Theorem
For any real number $x: -1 \le x \le 1$:
- $\arcsin x = \dfrac 1 i \map \ln {i x + \sqrt {1 - x^2} }$
where $\arcsin x$ is the arcsine and $i^2 = -1$.
Proof
Assume $y \in \R$ where $-\dfrac \pi 2 \le y \le \dfrac \pi 2$.
| \(\ds y\) | \(=\) | \(\ds \arcsin x\) | ||||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(=\) | \(\ds \sin y\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(=\) | \(\ds \frac 1 {2 i} \paren {e^{i y} - e^{-i y} }\) | Euler's Sine Identity | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds 2 i x\) | \(=\) | \(\ds e^{i y} - e^{-i y}\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds 2 i x e^{i y}\) | \(=\) | \(\ds e^{2 i y} - 1\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds e^{2 i y} -2 i x e^{i y}\) | \(=\) | \(\ds 1\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds e^{2 i y} - 2 i x e^{i y} - x^2\) | \(=\) | \(\ds 1 - x^2\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \paren {e^{i y} - i x}^2\) | \(=\) | \(\ds 1 - x^2\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds e^{i y} - i x\) | \(=\) | \(\ds \sqrt {1 - x^2}\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds i y\) | \(=\) | \(\ds \map \ln {i x + \sqrt {1 - x^2} }\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds y\) | \(=\) | \(\ds \dfrac 1 i \map \ln {i x + \sqrt {1 - x^2} }\) |
$\blacksquare$