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

The approximation sin(x)16(πx)x5π24(πx)x(0xπ) 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(πx)x5π2b(πx)x I so computed Φ(a,b)=π0(sin(x)a(πx)x5π2b(πx)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 : 2.98×106 only decreased to 2.17×106. 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)π24x2π2+x2(π2xπ2) 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(πx)

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(πx)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(5π24x(πx))

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

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