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An Indian mathematician, Bhaskara I, gave the following amazing approximation of the sine (I checked the graph and some values, and the approximation is truly impressive.)

$$\sin x \approx \frac{{16x\left( {\pi - x} \right)}}{{5{\pi ^2} - 4x\left( {\pi - x} \right)}}$$

for $(0,\pi)$

Here's an image. Cyan for the sine and blue for the approximation. enter image description here

¿Is there any way of proving such rational approximation? ¿Is there any theory similar to Taylor's or Power Series for rational approximations?

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    Off topic: May be you could say Indian Mathematician rather than a Hindu-Mathematician which attaches a religion to Mathematics. (Note: I am Hindu and not against any religion!)2012-02-06
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    @KannappanSampath: Edited accordingly :)2012-02-06
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    @KannappanSampath That is product of my own ignorance. I translated from my language, which uses "Hindú" for "Indian" ("Hinduista" is "Hindu", whereas "Indio" rather means native indian, and would be used for the people in cowboy movies).2012-02-06
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    There is the theory of [Padé approximants](http://en.wikipedia.org/wiki/Pad%C3%A9_approximant), but I don't think this particular function is one.2012-02-06
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    @Peter: I have changed the title. Hope it doesn't affect you.2012-02-06
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    You can compute the Taylor series of the difference between this function and sine around $\frac{\pi}{2}$. I think the leading term is $\left(\frac{1}{2}-\frac{5}{\pi^2}\right) x^2 \approx -0.0066 x^2$. (At least, that's what I got by converting the approximation into an approximation for cos and then [expanding the difference](http://www.wolframalpha.com/input/?i=(pi^2-4*x^2)/(pi^2%2Bx^2)-cos(x)).2012-02-06
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    @Chandrasekhar Was it too subjective?2012-02-06
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    @Peter: Well, i thought including Bhaskara's name in the question would be better because, if some unknown person wants to search something attributed to Bhaskara he may see this question in future :)2012-02-06

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  • Here is an article which is written by Shailesh Shirali. Unfotunately my university doesn't have access to it.

  • Here is one more article.

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This is very close to a Padé approximant, and in this case the formula is simple enough that it's easy to derive. Firstly, we know that $\sin(x)$ is $0$ at $x=0, x=\pi$; this suggests recasting in terms of the variable $y=x(\pi-x)$. What we're after is a first-order rational approximation $\sin(x) = f(y) = \frac{ay+b}{cy+d}$; since we know that $f(y) = 0$ at $y=0$ (i.e., as $x$ approaches $0$ or $\pi$) then the constant term in the numerator is $0$, and after dividing out the approximation takes the form $\frac{y}{a+by}$.

Now, we certainly want our approximation to give $\sin(\pi/2) = f(\pi^2/4) = 1$; this means $\displaystyle{\frac{\pi^2/4}{a+b\pi^2/4}} = 1$, or $4a+b\pi^2 = \pi^2$, or $a=\frac{1-b}{4}\pi^2$. The other relation between $a$ and $b$ presumably comes from trying to match the derivative at $0$, $\left.d(\sin(x))/dx\right|_{x=0} = 1$; the condition for this this can easily be written out in terms of $df/dy$ at $y=0$. I'll spare the arithmetic (unless someone's really curious), but the result works out to be that $a=\pi$; this would give $b=(\pi-4)/\pi$ and the overall approximant $f(y) = \frac{\pi y}{\pi^2+(\pi-4)y}$, but instead the formula uses a second approximation by setting $a=5\pi^2/16$, which (thanks to the first relation) gives a rational value of $b$ (and in fact, the 'nice' value $1/4$). This approximation is equivalent to saying that $5\pi^2/16\approx\pi$, or in other words that $\pi\approx 16/5 = 3.2$; it means a slight error in the slope of the approximation at $x=0$, but that's a fair tradeoff for the ease of calculation gained.

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    Thank you for the answer. Upvoted. I chose the other since he provided me a full article with an explanation.2012-02-06
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    Wholly understandable! (Note that the article gives a different derivation of the approximation $5\pi^2/16$ for $a$; it satisfies the relation that $\sin(\pi/6) = 1/2$, which keeps $b$ and $a/\pi^2$ rational at all steps).2012-02-06
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    (Since that's the only other relation of the form $\sin(p\cdot\pi)=q$ for rational $p,q$ this makes quite a bit of sense.)2012-02-06
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    How did you get the article?2012-02-06
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    @Peter I don't have JSTOR access, but the second link (the one directly to a PDF) in Chandrasekhar's reply worked for me; that PDF is the article I was referring to.2012-02-06
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    Analogous to taking $\pi \approx 16/5$ to get the correct gradient at $x=0$, you can take $\pi \approx \sqrt{10}$ to get the correct second differential at $x=\pi/2$.2012-02-06
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    As it turns out, the derivation given here is similar to what might be done for what is called a "two-point Padé approximant".2012-03-24
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    @PedroTamaroff isn't it against site policy to submit answers that are just links? Made worse in this case by paywall and dead links. (The possibility of which is of course the reason for the policy). I think you should accept Steven's answer instead.2018-08-07
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Writing $x = \pi/2 + \pi t$, the approximation becomes $\cos(\pi t) \approx \frac{1-4t^2}{1+t^2} = 1 - 5 t^2 + O(t^4)$. In fact $\cos(\pi t) = 1 - \frac{\pi^2}{2} t^2 + O(t^4)$, but $\pi^2/2 \approx 4.9348$ is not far from $5$. In terms of uniform approximation to $\cos(\pi t)$ for $t \in [-1/2, 1/2]$, $\frac{1 - 4 t^2}{1+1.0043 t^2}$ would be somewhat better.

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    Actually, putting $\cos \frac{\pi x}{2} $ gives what in the article is stated as the "Bhaskara function":$$\cos \frac{\pi x}{2} \approx 4 \frac{1-x^2}{4+x^2}$$ which is an outstoundingly good approximation of that function. Thanks!2012-02-10