The approximation $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ was proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician.
I wondered how much this could be improved using our computers and so I tried (very immodestly) to see if we could do better using $$\sin(x) \simeq \frac{a (\pi -x) x}{5 \pi ^2-b (\pi -x) x}$$ I so computed $$\Phi(a,b)=\int_0^{\pi} \left(\sin (x)-\frac{a (\pi -x) x}{5 \pi ^2-b (\pi -x)x}\right)^2 dx$$ the analytical expression of which not being added to the post. Settings the derivatives equal to $0$ and solving for $a$ and $b$, I arrived to $a=15.9815,b=4.03344$ so close to the original approximation !
What is interesting is to compare the values of $\Phi$ : $2.98 \times 10^{-6}$ only decreased to $2.17 \times 10^{-6}$. Then, no improvement and loss of attractive coefficients.
Now, since this is a matter of etiquette on this site, I ask a simple question:
with all the tools and machines we have in our hands, could any of our community propose something as simple (or almost) for basic trigonometric functions ?
In the discussions, I mentioned one I made (it is probable that I reinvented the wheel) in the same spirit $$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad (-\frac \pi 2 \leq x\leq\frac \pi 2)$$ which is amazing too !
Best Answer
One simple way to derive this is to come up with a parabola approximation. Just getting the roots correct we have
$$f(x)=x(\pi-x)$$
Then, we need to scale it (to get the heights correct). And we are gonna do that by dividing by another parabola $p(x)$
$$f(x)=\frac{x(\pi-x)}{p(x)}$$
Let's fix this at three points (thus defining a parabola). Easy rational points would be when $\sin$ is $1/2$ or $1$. So we fix it at $x=\pi/6,\pi/2,5\pi/6$.
We want $$f(\pi/6)=f(5\pi/6)=1/2=\frac{5\pi^2/36}{p(\pi/6)}=\frac{5\pi^2/36}{p(5\pi/6)}$$ And we conclude that $p(\pi/6)=p(5\pi/6)=5\pi^2/18$
We do the same at $x=\pi/2$ to conclude that $p(\pi/2)=\pi^2/4$.
The only parabola through those points is
$$p(x)=\frac{1}{16}(5\pi^2-4x(\pi-x))$$
And thus we have the original approximation.
In the spirit of answering the question: This method could be applied for most trig functions on some small symmetric bound.