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 $-\dfrac \pi 2 \le y \le \dfrac \pi 2$.

\(\displaystyle y\) \(=\) \(\displaystyle \arcsin x\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle \sin y\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle \frac 1 {2 i} \paren {e^{i y} - e^{-i y} }\) Sine Exponential Formulation
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle 2 i x\) \(=\) \(\displaystyle e^{i y} - e^{-i y}\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle 2 i x e^{i y}\) \(=\) \(\displaystyle e^{2 i y} - 1\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle e^{2 i y} -2 i x e^{i y}\) \(=\) \(\displaystyle -1\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle e^{2 i y} - 2 i x e^{i y} - x^2\) \(=\) \(\displaystyle 1 - x^2\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \paren {e^{i y} - i x}^2\) \(=\) \(\displaystyle 1 - x^2\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle e^{i y} - i x\) \(=\) \(\displaystyle \sqrt {1 - x^2}\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle i y\) \(=\) \(\displaystyle \map \ln {i x + \sqrt {1 - x^2} }\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle y\) \(=\) \(\displaystyle \dfrac 1 i \map \ln {i x + \sqrt {1 - x^2} }\)

$\blacksquare$


Also see