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

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 !

• Am I the only one who's more interested in the analytical solution of the integral? – UserX Oct 16 '14 at 11:38
• How about replacing $5$ as well? – lhf Oct 16 '14 at 11:40
• Yeah, Indians cooked up some pretty good pi back in that era. – Daniel R Hicks Oct 16 '14 at 17:33
• Hm, it's very pity that this is not really a question. Couldn't you just add some question to your post so that this beautiful thing does not get closed and deleted for formal reasons? – Hagen von Eitzen Oct 16 '14 at 18:15
• @HagenvonEitzen. Thanks for pushing me ! I added. Feel free to add in the post. – Claude Leibovici Oct 16 '14 at 18:23

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 $$\pm1$$

• Let me give one : $\cos(x) \approx \frac{5 \pi ^2}{x^2+\pi ^2}-4$. I don't know if it is new or not (I built it in the same spirit as the one for the sine). Is that any good ? ($0\leq x\leq \pi/2$). – Claude Leibovici Oct 16 '14 at 15:02
• @ClaudeLeibovici: It's brilliant! Far simpler and more accurate! :-$)$ – Lucian Oct 16 '14 at 15:07
• Amazing, isn't ? – Claude Leibovici Oct 16 '14 at 15:08
• @ClaudeLeibovici: Probably based on the fact that $\dfrac1{1+x^2}\approx1-x^2$, since $x^4\approx0$. – Lucian Oct 16 '14 at 15:10
• @Lucian. Just use $\cos(x) \approx \frac{5 \pi ^2}{x^2+\pi ^2}-4$ and make $x=\frac{\pi y}{2}$ to get $\cos\bigg(\dfrac\pi2 y\bigg)\simeq \frac{4 \left(1-y^2\right)}{y^2+4}$ – Claude Leibovici Oct 17 '14 at 7:19

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.

• That was wonderful – Fezvez Oct 17 '14 at 6:44
• @BeaumontTaz. Inspired by your answer, I selected three points at $x=\alpha,\frac{\pi}{2},\pi-\alpha$ from which the coefficients can be computed as functions of $\alpha$. I set $\alpha=\frac{n\pi}{8!}$ to maintain rationality and I searched for $n$ such that the integral in the post be minimum. I found $n=7017$ instead of $n=6720$ (corresponding to $\frac{\pi}{6}$). The value of the integral changed from $2.97776\times 10^{-6}$ to $2.94776\times 10^{-6}$ which is just ridiculous. One more proof of how clever is the choice of $\frac{\pi}{6}$ ! – Claude Leibovici Oct 18 '14 at 5:10
• @BeaumontTaz. I did not want to have an irrational value for the optimum $\alpha$. That's it ! Cheers :-) – Claude Leibovici Oct 18 '14 at 11:15
• @BeaumontTaz amazing answer. What is the intuition behind scaling of $x(\pi-x)$ by dividing with $p(x)$, and not by some multiplicative function ? – kaka Oct 22 '14 at 19:58
• @kaka. My intuition behind dividing by $p(x)$ was that it was the already established form of the approximation. It was using the knowledge that we know what we're going for. Kinda cheating. However, I did just spend some time solving for a $q(x)$ such that $g(x) = q(x)x(\pi-x)$ and it doesn't turn out nearly as appealing, not to mention the fact that it's a fourth order polynomial, now. However, I will mention that it is a more accurate approximation having $\Phi = 1.118\times10^{-6}$. I will post the both $q(x)$ and $g(x)$ in a proceeding comment. – BeaumontTaz Oct 22 '14 at 20:57

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 $\frac{4}{\pi}$.

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

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

• Very interesting ! Thanks & cheers :-) – Claude Leibovici Oct 16 '14 at 12:16
• This last sum is just fascinating too ! Thank you so much. This opens new windows in my mind !! – Claude Leibovici Oct 16 '14 at 13:58
• Maybe $\frac{5}{16} = 0.3125 \approx \frac{\pi}{10}$ ? $\frac{1}{\pi} \approx 0.3183$. – jpmc26 Oct 17 '14 at 0:17
• @jpmc26, the fractions arise naturally from trying to fit a parabola $p(x)$ to the points $\left(\frac{\pi}{6},\frac{5\pi^2}{18}\right), \left(\frac{\pi}{2},\frac{\pi^2}{4}\right), \left(\frac{5\pi}{6},\frac{5\pi^2}{18}\right)$ which would make $$f(x)=\frac{x(\pi-x)}{p(x)}=\sin{(x)}$$ at $x=\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6}$. I flushed it out a little more in my answer. Also, this wiki page and this answer might help. – BeaumontTaz Oct 17 '14 at 6:32
• @jpmc26 Also, the slope of our approximation at $x=0$ is $\frac{16}{5\pi}$ and the slope of $\sin$ at $x=0$ is $1$. So basically $\frac{16}{5\pi}\approx 1$ which means that $\frac{16}{5}\approx \pi$ You seem to be right or at least it's a coincidence that $\frac{5}{16}\approx \frac{1}{\pi}$. – BeaumontTaz Oct 17 '14 at 6:40