A 1,400 years old approximation to the sine function by Mahabhaskariya of Bhaskara I

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.

This might be more explicable if you observe that it is the same thing as

$$ \csc(x) \approx -\frac{1}{4} + \frac{5 \pi}{16} \left( \frac{1}{x} + \frac{1}{\pi - x} \right) $$

The two summands in the parentheses are obvious if you want to get the poles of $\csc$ correct. If you wanted a good approximation of $\csc$ near the poles, then the coefficient out front should be $1$. But since we're approximating $\sin$, it's okay to get that wrong because anything near zero is near zero.

The extreme point is at $\csc(\pi/2) = 1$; in the approximation, this would become

$$ -\frac{1}{4} + \frac{5 \pi}{16} \left( \frac{2}{\pi} + \frac{2}{\pi} \right) = -\frac{1}{4} + \frac{5}{4} = 1$$

and so we see the appearance of the remaining copy of $\pi$ is to cancel out the other two $\pi$'s. All that's left is to tune the factor $\frac{5}{16}$ to something appropriate, and adjust the $-\frac{1}{4}$ to compensate. I'm not sure where the choice of $\frac{5}{16}$ comes from, although it's quite plausible it ought to be near $\frac{1}{\pi}$; maybe it was chosen just to be a small fraction whose denominator was divisible by $4$, so as to cancel the $4$ in $\frac{4}{\pi}$.

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As a bit of an aside, my comment about the poles suggests an infinite sum for $\csc(x)$ that I hadn't seen before:

$$ \csc(x) = \sum_{n=-\infty}^{+\infty} (-1)^n \frac{1}{x - \pi n} $$

While you're at it, try also $\cos\bigg(\dfrac\pi2x\bigg)\simeq\Big(1-x^2\Big)^\tfrac65$ and $\cos\bigg(\dfrac\pi2x\bigg)\simeq\Big(1-x^2\Big)^\tfrac76$.

But since the numerical evaluation of fractional powers is significantly more time-

consuming in terms of CPU, we can substantially improve this by using the binomial

series for $\Big(1-x^2\Big)^\tfrac15$, and experimentally adjusting the coefficient, finally arriving at

$\color{seagreen}{\cos\bigg(\dfrac\pi2x\bigg)\simeq\Big(1-x^2\Big)\bigg(1-\dfrac{x^2}{4.5}\bigg)}$, which yields an absolute error of about $\pm1$ ‰