# Arcsine Logarithmic Formulation

## 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 \ \$ $\displaystyle y$ $=$ $\displaystyle \dfrac 1 i \map \ln {i x + \sqrt {1 - x^2} }$

$\blacksquare$