The inexplicable approach of an Indian mathematician for Cosecant

Question

I am reading about the old Indian mathematician approximation $$\csc (z)\simeq \frac{z^4+\pi ^2 z^2+2 \pi ^4}{2 \pi ^4 z-2 \pi ^2 z^3}$$ reminiscent of Bhaskara's.
I tried to use Taylor series and I got something similar to a Padé approximation $$\csc (z)\simeq \frac{\frac{11 z^4}{5880}+\frac{3 z^2}{49}+1}{z-\frac{31 z^3}{294}}$$ but the Indian formula is a fairly superior approximation.
How is this possible?

Answer 1

The Indian approximation encodes a large portion of the Weierstrass product for the sine function: $$ \sin(z) = z\left(1-\frac{z^2}{\pi^2}\right)\prod_{n\geq 2}\left(1-\frac{z^2}{n^2\pi^2}\right) \tag{1}$$ The problem of approximating $\csc(z)$ boils down to producing approximations for $$ g(z)=\frac{z}{\sin z}\left(1-\frac{z^2}{\pi^2}\right)=\prod_{n\geq 2}\left(1-\frac{z^2}{n^2 \pi^2}\right)^{-1}\tag{2}$$ which is an even and approximately quadratic function, ranging from $1$ to $2$ on the interval $(0,\pi)$.
Actually $$ g(z) \approx 1+\left(\zeta(2)-1\right)\left(\frac{z}{\pi}\right)^2+(2-\zeta(2))\left(\frac{z}{\pi}\right)^4\tag{3} $$ leads to an approximation similar to the Indian one $$ \csc(z)\approx\frac{1+\left(\zeta(2)-1\right)\left(\frac{z}{\pi}\right)^2+(2-\zeta(2))\left(\frac{z}{\pi}\right)^4}{z\left(1-\frac{z^2}{\pi^2}\right)}.\tag{4}$$ and more accurate if $z$ is close to $0$. $$ \csc(z)\approx\frac{1+\left(\zeta(2)-1\right)\left(\frac{z}{\pi}\right)^2+(\frac{7}{2}-2\zeta(2))\left(\frac{z}{\pi}\right)^4+(\zeta(2)-\frac{3}{2})\left(\frac{z}{\pi}\right)^6}{z\left(1-\frac{z^2}{\pi^2}\right)}\tag{5}$$ is more accurate on the whole interval $(0,\pi)$. Folklore: the evaluation of $(5)$ at $z=\frac{\pi}{6}$ leads to $$ \pi\approx \frac{3}{35}\left(432-\sqrt{156301}\right) $$ with an approximation error of $\approx 3$ parts in one hundred thousands.