Elementary proof of Bhaskara I's approximation: $\sin\theta=\frac{4\theta(180-\theta)}{40500-\theta(180-\theta)}$

Question

This site has seen umpteen questions about efficient ways to calculate the sine of an angle. But a remarkable formula was given in ~CE 615 by the Indian mathematician Bhaskara I in his Mahabhaskariya. The formula goes as follows: $$sin\,\theta=\frac{4\theta(180-\theta)}{40500-\theta(180-\theta)}$$ The remarkable things about this formula:

1. Apparently derived using only geometry because it occurs in the sections on geometry in Mahabhaskariya and other books that follow.
2. Maximum error of 0.92% occurs at $\theta=10^\circ$.
3. A simpler formula than this cannot be found without introducing higher order polynomials.

The challenge to all interested folks is: Imagine yourself armed with only pen and paper. How can you derive the above formula by using geometry and elementary mathematics? We ask this question since Bhaskara I does not give in his work how he arrived at this formula nor do his successors! Wikipedia does give some relations https://en.wikipedia.org/wiki/Bhaskara_I%27s_sine_approximation_formula based on Prof. R.C. Gupta's paper, but can anyone better it?

https://web.archive.org/web/20120316083451/http://www.new.dli.ernet.in/rawdataupload/upload/insa/INSA_1/20005af0_121.pdf

Anyone who can give any decent (meaningful, not silly.. sorry unable to quantify what decent or meaningful means, but each poster can have his or her own quality check for their answer before posting) answer to this gets a chocolate. As an aside, can a simpler than this formula be derived for sine?

Answer 1

The book "Dead Reckoning: Calculating Without Instruments" show this formula: $$\sin\theta\approx \frac{\theta}{10000}\left[174.4-\frac{\theta(\theta+1)}{120}\right]$$ With $0^\circ\leq\theta\leq54^\circ$.

If $54^\circ<\theta\leq90^\circ$ then the formula is: $$\sin\theta\approx 1-\frac{(90-\theta)(91-\theta)}{7000}$$

Answer 2

COMMENT.- Extraordinary formula. It is better taking $x$ in radians and formulate it in terms of cosinus. One get $$\cos x=\frac{\pi^2-4x^2}{\pi^2+x^2}$$ it is practically coincident to $\cos x$ over the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$.

A possible way to get it is to put $\cos x=\dfrac{a+bx^2}{c+dx^2}$ and giving some fitting values to $x$. For example for $x=0$ we get at once $a=c$, for $x=\dfrac{\pi}{3}$ we have the value $\dfrac12$ and so on with fine careful obviously, it seems we can arrive to the exposed approximating formula.