What is the simplest accurate approximation of $\sin^{-1}x$ using a (polynomial) function of $x$ and $\sqrt{1-x^2}$? 
What is the simplest accurate approximation of $\sin^{-1}x$ using a (polynomial) function of $x$ and $\sqrt{1-x^2}$? 

The most well-known definition in this subject is Euler's formula:
$$\sin^{-1}x=-i\ln\left(\sqrt{1-x^2}+ix\right) \tag{1}$$
I'm looking for a bivariate polynomial approximation 
$$\sin^{-1}x\approx p_N(x,\sqrt{1-x^2}) \tag{2}$$
where $N$ is the degree of the polynomial. For example, for $N=3$, I discovered a formula
$$\sin^{-1}x\approx x+(1-\sqrt{1-x^2})^2(1-0.43x) \tag{3}$$
which gives a uniformly good approximation over $x\in[0,1]$. More discussions of the problem can be found here.
Are there general ways to achieve such a polynomial approximation?
 A: To express $\theta\in[0,\frac{\pi}{2}]$ as a polynomial of $\sin\theta$ and $\cos\theta$, consider
$$\theta\approx a_0+\sum_{n=1}^N(a_n\cos n\theta+b_n\sin n\theta),\quad\theta\in[0,\frac{\pi}{2}].$$
The fitting problem becomes linear this way. However, there is no requirement on the value of the right-hand side for $\theta\in(\frac{\pi}{2},2\pi)$. So the coefficients $a_n,b_n$ are not determined straightforwardly by a Fourier integral, but needs to be found numerically using perhaps a least-square fitting technique. After the coefficients are found, use trignometric identities to convert $\cos n\theta$ and $\sin n\theta$ into polynomials of $\sin\theta$ and $\cos\theta$. The degree of the polynomial will be truncated by $N$. At the end of the calculation use $\sin\theta=x$ and $\cos\theta=\sqrt{1-x^2}$. A good approximation
$$\sin^{-1}x\approx\frac{\pi}{4}+0.9776\;(x-\sqrt{1-x^2})+0.1931\;(1-2x^2),\quad x\in[0,1]$$
can be found using this method by setting $N=2$. The root-mean-square error (RMSE) is under $5\times 10^{-4}$. If $N=3$, the RMSE can be made lower than $5\times 10^{-5}$. If $\theta=\sin^{-1}x\in[-\frac{\pi}{2},\frac{\pi}{2}]$ is required on the full domain, symmetry would cancel all $a_n$ terms. Then for $N=3$ one obtains
$$\sin^{-1}x\approx 1.6434\;x-0.8719\;x\sqrt{1-x^2}+0.0804\;(3x-4x^3),\quad x\in[-1,1].$$
This formula has the right symmetry of being an odd function, but is much less accurate (RMSE = $3\times 10^{-3}$). Since $(\sqrt{1-x^2})^2=1-x^2$ is a polynomial, the general approximant takes the form $\sin^{-1}x\approx P(x)+Q(x)\sqrt{1-x^2}$, with $P(x)$ and $Q(x)$ being polynomials of desired degree.
A: From the classic
"Approximations for Digital Computers"
by Hastings,
published in 1955:
$\arcsin(x)
=\dfrac{\pi}{2}-\sqrt{1-x}f_n(x)
$
where
$f_n(x)
=\sum_{k=0}^n a_{n, k}x^k
$
are polynomials of degree $n$
for $n=3$ to $7$.
The simplest is
$f_{3}(x)$
for which
$a_{3, 0..3}
=1.5707288,
-.2121144,
.0742610,
-0.187293
$
with a maximum error of
about $0.00007$.
Using
$f_7(x)$,
the maximum error is about
$0.00000002$.
Note:
I find it fascinating
to read the first section,
which describes how
the approximations were gotten.
In 1955,
the computer power
we are so accustomed to
was not available.
