Sine Approximation of Bhaskara An Indian mathematician, Bhaskara I, gave the following amazing approximation of the sine (I checked the graph and some values, and the approximation is truly impressive.)
$$\sin x \approx \frac{{16x\left( {\pi  - x} \right)}}{{5{\pi ^2} - 4x\left( {\pi  - x} \right)}}$$
for $(0,\pi)$
Here's an image. Cyan for the sine and blue for the approximation. 

¿Is there any way of proving such rational approximation? ¿Is there any theory similar to Taylor's or Power Series for rational approximations? 
 A: *

*Here is an article which is written by Shailesh Shirali. Unfotunately my university doesn't have access to it.

*Here is one more article.
A: This is very close to a Padé approximant, and in this case the formula is simple enough that it's easy to derive.  Firstly, we know that $\sin(x)$ is $0$ at $x=0, x=\pi$; this suggests recasting in terms of the variable $y=x(\pi-x)$.  What we're after is a first-order rational approximation $\sin(x) = f(y) = \frac{ay+b}{cy+d}$; since we know that $f(y) = 0$ at $y=0$ (i.e., as $x$ approaches $0$ or $\pi$) then the constant term in the numerator is $0$, and after dividing out the approximation takes the form $\frac{y}{a+by}$.
Now, we certainly want our approximation to give $\sin(\pi/2) = f(\pi^2/4) = 1$; this means $\displaystyle{\frac{\pi^2/4}{a+b\pi^2/4}} = 1$, or $4a+b\pi^2 = \pi^2$, or $a=\frac{1-b}{4}\pi^2$.  The other relation between $a$ and $b$ presumably comes from trying to match the derivative at $0$, $\left.d(\sin(x))/dx\right|_{x=0} = 1$; the condition for this this can easily be written out in terms of $df/dy$ at $y=0$.  I'll spare the arithmetic (unless someone's really curious), but the result works out to be that $a=\pi$; this would give $b=(\pi-4)/\pi$ and the overall approximant $f(y) = \frac{\pi y}{\pi^2+(\pi-4)y}$, but instead the formula uses a second approximation by setting $a=5\pi^2/16$, which (thanks to the first relation) gives a rational value of $b$ (and in fact, the 'nice' value $1/4$).  This approximation is equivalent to saying that $5\pi^2/16\approx\pi$, or in other words that $\pi\approx 16/5 = 3.2$; it means a slight error in the slope of the approximation at $x=0$, but that's a fair tradeoff for the ease of calculation gained.
A: Writing $x = \pi/2 + \pi t$, the approximation becomes $\cos(\pi t) \approx \frac{1-4t^2}{1+t^2} = 1 - 5 t^2 + O(t^4)$.  In fact $\cos(\pi t) = 1 - \frac{\pi^2}{2} t^2 + O(t^4)$, but $\pi^2/2 \approx 4.9348$ is not far from $5$. In terms of uniform approximation to $\cos(\pi t)$ for $t \in [-1/2, 1/2]$, $\frac{1 - 4 t^2}{1+1.0043 t^2}$ would be somewhat better.
A: That's a pretty neat visualization of how can you intiutively build, proove and understand the approximation.

The full proof with video and explanation.
