Editing Inverse trigonometric functions (section)

===Infinite series===
Similar to the sine and cosine functions, the inverse trigonometric functions can also be calculated using [[power series]], as follows. For arcsine, the series can be derived by expanding its derivative, <math display="inline">\tfrac{1}{\sqrt{1-z^2}}</math>, as a [[binomial series]], and integrating term by term (using the integral definition as above). The series for arctangent can similarly be derived by expanding its derivative <math display="inline">\frac{1}{1+z^2}</math> in a [[geometric series]], and applying the integral definition above (see [[Leibniz series]]).

: <math>
\begin{align}
\arcsin(z) & = z + \left( \frac{1}{2} \right) \frac{z^3}{3} + \left( \frac{1 \cdot 3}{2 \cdot 4} \right) \frac{z^5}{5} + \left( \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \right) \frac{z^7}{7} + \cdots \\[5pt]
& = \sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!!}\frac{z^{2n+1}}{2n+1} \\[5pt]
& = \sum_{n=0}^\infty \frac{(2n)!}{(2^n n!)^2} \frac{z^{2n+1}}{2n+1} \, ; \qquad |z| \le 1
\end{align}
</math>

:<math>\arctan(z)
= z - \frac{z^3}{3} +\frac{z^5}{5} - \frac{z^7}{7} + \cdots
= \sum_{n=0}^\infty \frac{(-1)^n z^{2n+1}}{2n+1} \, ; \qquad |z| \le 1 \qquad z \neq i,-i</math>

Series for the other inverse trigonometric functions can be given in terms of these according to the relationships given above.  For example, <math>\arccos(x) = \pi/2 - \arcsin(x)</math>, <math>\arccsc(x) = \arcsin(1/x)</math>, and so on.  Another series is given by:<ref name="Borwein_2004"/>

:<math>2\left(\arcsin\left(\frac{x}{2}\right) \right)^2 = \sum_{n=1}^\infty \frac{x^{2n}}{n^2\binom {2n} n}.</math>

[[Leonhard Euler]] found a series for the arctangent that converges more quickly than its [[Taylor series]]:

: <math>\arctan(z) = \frac z {1 + z^2} \sum_{n=0}^\infty \prod_{k=1}^n \frac{2k z^2}{(2k + 1)(1 + z^2)}.</math><ref>{{citation|title= An elementary derivation of Euler's series for the arctangent function|journal = The Mathematical Gazette | author = Hwang Chien-Lih | doi = 10.1017/S0025557200178404 | year = 2005 | volume = 89 | issue = 516|pages = 469–470 |s2cid = 123395287 }}</ref>
(The term in the sum for ''n'' = 0 is the [[empty product]], so is 1.)

Alternatively, this can be expressed as

:<math>\arctan(z) = \sum_{n=0}^\infty \frac{2^{2n} (n!)^2}{(2n + 1)!} \frac{z^{2n + 1}}{(1 + z^2)^{n + 1}}.</math>

Another series for the arctangent function is given by

:<math>\arctan(z) = i\sum_{n=1}^\infty\frac{1}{2n - 1}\left(\frac{1}{(1 + 2i/z)^{2n-1}} - \frac{1}{(1 - 2i/z)^{2n - 1}}\right),</math>

where <math>i=\sqrt{-1}</math> is the [[imaginary unit]].<ref>{{citation | title = A formula for pi involving nested radicals | journal = The Ramanujan Journal | author = S. M. Abrarov and B. M. Quine | doi = 10.1007/s11139-018-9996-8 | year = 2018 | volume = 46 | issue = 3 | pages = 657–665 | arxiv = 1610.07713 | s2cid = 119150623 }}</ref>

====Continued fractions for arctangent====
Two alternatives to the power series for arctangent are these [[generalized continued fraction]]s:

: <math>\arctan(z) =
\frac z {1 + \cfrac{(1z)^2}{3 - 1z^2 + \cfrac{(3z)^2}{5 - 3z^2 + \cfrac{(5z)^2}{7 - 5z^2 + \cfrac{(7z)^2}{9-7z^2 + \ddots}}}}} =
\frac{z}{1 + \cfrac{(1z)^2}{3 + \cfrac{(2z)^2}{5 + \cfrac{(3z)^2}{7 + \cfrac{(4z)^2}{9 + \ddots}}}}}
</math>

The second of these is valid in the cut complex plane. There are two cuts, from &minus;'''i''' to the point at infinity, going down the imaginary axis, and from '''i''' to the point at infinity, going up the same axis. It works best for real numbers running from −1 to 1. The partial denominators are the odd natural numbers, and the partial numerators (after the first) are just (''nz'')<sup>2</sup>, with each perfect square appearing once. The first was developed by [[Leonhard Euler]]; the second by [[Carl Friedrich Gauss]] utilizing the [[Gaussian hypergeometric series]].