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 !
One simple way to derive this is to come up with a parabola approximation. Just getting the roots correct we have
Then, we need to scale it (to get the heights correct). And we are gonna do that by dividing by another parabola 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
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.