Well, I'll show you where I was able to get, but I wasn't able to push through it all to the end. Maybe it will get you close enough that you could go from there.
The paper by Cohen, Villegas, and Zagier (linked above) works on series
of the form
$$
S = \sum_{k=0}^\infty (-1)^k a_k.
$$
Though they state that "$a_k$ is a reasonably well-behave function, which
goes slowly to $0$ as $k\rightarrow\infty$." Whether $\operatorname{acos}()$ behaves
like this, well,
that's another question that I'm blatantly disregarding at the moment.
Van Wijngaarden has a transform to convert a positive series
summation into an alternating one
$$
\sum_{r=1}^\infty v_r = \sum_{r=1}^\infty (-1)^{r-1}w_r,
$$
where
$$
w_r = v_r + 2v_{2r} + 4v_{4r} + \dots.
$$
(I got this from "Numerical Recipes in C" as referenced in
the linked paper above)
It states that
Since,$\ldots$ [the indices] increase tremendously rapidly,
as powers of 2, it often requires only a few terms to converge [$w_r$] to
extraordinary accuracy.
So, to utilize this, we need to convert $\operatorname{acos}()$ into an alternating series.
From wikipedia
we get that
$$
\operatorname{acos}(z) = \frac{\pi}{2} - \sum_{n=0}^\infty
\binom{2n}{n}\frac{z^{2n+1}}{4^n(2n+1)}
$$
Then we can use the transform above to get
$$
\begin{split}
\frac{\pi}{2} - \operatorname{acos}(z)
&= \sum_{n=0}^\infty
\binom{2n}{n}\frac{z^{2n+1}}{4^n(2n+1)}\\
&= 1 + \sum_{n=1}^\infty
\binom{2n}{n}\frac{z^{2n+1}}{4^n(2n+1)}\\
&= 1 + \sum_{r=1}^\infty (-1)^{r-1}w_r,
\end{split}
$$
where
$$
\begin{split}
w_r &= \sum_{k=0}^\infty2^k\binom{2^{k+1}r}{2^kr} \frac{z^{2^{k+1}r+1}}{4^{2^kr}\left(2^{k+1}r+1\right)}\\
&= \sum_{k=0}^\infty\binom{2^{k+1}r}{2^kr} \frac{2^kz^{2^{k+1}r+1}}{2^{2^{k+1}r}\left(2^{k+1}r+1\right)}.
\end{split}
$$
Let $\alpha(r) = 2^{k+1}r + 1$, then we can write
$$
w_r= \sum_{k=0}^\infty\binom{\alpha(r)-1}{\frac{(\alpha(r)-1)}{2}} \frac{z^{\alpha(r)}}{2^{\alpha(r)-k}\left(\alpha(r)\right)}.
$$
So, we have
$$
\begin{align*}
\frac{\pi}{2} - 1 - \operatorname{acos}(z) &=
\sum_{r=1}^\infty (-1)^{r-1}w_r &
w_r &= \sum_{k=0}^\infty\binom{\alpha(r)-1}{(\alpha(r)-1)/2} \frac{z^{\alpha(r)}}{2^{\alpha(r)-k}\left(\alpha(r)\right)}.
\end{align*}
$$
So, we want to accelerate the calculation of the sum in the left
equation.
Now, the second algorithm in Cohen et. al is given as
Algorithm 2
- Let $Q_n(X) = \sum_{k=0}^nb_kX^k$
- Set $d = Q_n(-1)$
- Set $c = -d$
- Set $s = 0$
- For $k=0\:\text{to}\: n-1,$
$c = b_k - c$
$s + c\cdot a_k$
- Output $s/d$.
They specify in the paper that you should let $Q_n = B_n = P_n^{(m)}/((n-m)(m+1)!2^m)$. It was here that I had to stop.
I did find another explanation of the Cohen et al. acceleration, although it's explanation is not very in depth. You can find it (along with a postscript version) here. Good luck.
