The approximation sin(x)≃16(π−x)x5π2−4(π−x)x(0≤x≤π) 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)≃a(π−x)x5π2−b(π−x)x I so computed Φ(a,b)=∫π0(sin(x)−a(π−x)x5π2−b(π−x)x)2dx 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 Φ : 2.98×10−6 only decreased to 2.17×10−6. 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)≃π2−4x2π2+x2(−π2≤x≤π2) which is amazing too !

**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(π−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)=x(π−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=π/6,π/2,5π/6.

We want f(π/6)=f(5π/6)=1/2=5π2/36p(π/6)=5π2/36p(5π/6)

And we conclude that p(π/6)=p(5π/6)=5π2/18

We do the same at x=π/2 to conclude that p(π/2)=π2/4.

The only parabola through those points is

p(x)=116(5π2−4x(π−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.

**Attribution***Source : Link , Question Author : Claude Leibovici , Answer Author : BeaumontTaz*