# Computation of the Arcsin/Arccos

The MacLaurin (Taylor series at x=0) series for the arcsin

${\displaystyle{ \arcsin(x) = \sum_{k=0}^{\infty} \frac{(2k-1)!{\kern-1.25pt}!}{(2k)!{\kern-1.25pt}!} \frac{x^{2k+1}}{2k+1} }}$

is a very slowly converging one, especially close to the branchpoints at -1 and 1.
The MacLaurin series for arccos is the same as for arcsin because of

${\displaystyle{ \arccos x = \frac{\pi}{2} - \arcsin x }}$

so it has the very same problem.

The series for the arctan on the other side is simpler to compute.
Another good reason: I have it already implemented 😉

The arcsin is related to the atan by

${\displaystyle{ \arcsin x = 2\arctan\biggl(\frac{x}{1 + \sqrt{1 - x^2} } \biggr) }}$

which seems to be of no help at all because the fraction does not get very small and the series for atan is also quite slow near -1 and 1. We can use another relation for this range.

${\displaystyle{ \arcsin x = \mathop{\mathrm{sgn}}(x)\arctan\Biggl(\sqrt{\frac{x^2}{1 - x^2 }} \Biggr) }}$

Seems not of much use but we can expand the root

${\displaystyle{ \arcsin x = \mathop{\mathrm{sgn}}(x)\arctan\biggl(\frac{x}{\sqrt{1 - x^2 }} \biggr) }}$

The closer x comes to the branchpoint the larger the fraction, but with

${\displaystyle{ \arctan x = \frac{\pi}{2} - \arctan \frac{1}{x} }}$

the actual value gets close to zero when x goes close to one and $\arctan 0 = 0$ so
$\arcsin 1 = \frac{\pi}{2}$ which is correct and as intended. The point where $\frac{x}{\sqrt{1 - x^2 }} = 1$ is at $x = \frac{1}{\sqrt{2}}$ and at this point $\arctan 1 = \arcsin \frac{1}{\sqrt{2}} = \frac{\pi}{4}$.

My cutoff points will be $\arcsin = 2\arctan \bigl(\frac{x}{1 + \sqrt{1 - x^2} } \bigr)$ for $|x| \le 0.8$ and $\arctan\bigl(\frac{x}{\sqrt{1 - x^2 }} \bigr)$ for $|x| > 0.8$.

No code yet, just twiddling with math, sorry 😉

For $x \in\mathbb{R} \land |x| > 1$ the real part of the result is always $\frac{\pi}{2}$ and the imaginary part $\mathop{\mathrm{sgn}}(z) \log\bigl(x + \sqrt{x^2-1}\bigr)$. This comes from one of the definitions

${\displaystyle{ \arcsin x = -i\log\biggl(ix + \sqrt{1 - x^2 }\biggr) }}$

I will use this definition for complex arguments.