$\sin x$ does not satisfy this quadratic equation

Prove that $\sin x$ is not a rational function using the fact that it is not of the form $p(x)/q(x)$ where $p$ and $q$ are polynomials. Then, by using the above proof, prove that $\sin x$ does not satisfy a "quadratic equation" of the form: $$(\sin x)^2 f_2(x) + (\sin x)f_1(x) + f_0(x) = 0,$$ where $f_0, f_1, f_2$ are rational functions.

I know that a rational function cannot be zero at infinitely many points unless it is $0$ everywhere, but how does one use this information to formulate $p(x)/q(x)$ argument? If anybody could please help.

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The zeros of the rational function $\,\frac{p(x)}{q(x)}\,$ are the roots of the polynomial $\,p(x)\,$... – DonAntonio Feb 21 '13 at 23:20

For the second part, by multiplying the equation with a suitable polynomial, we may assume that $f_0$, $f_1$, $f_2$ are polynomials. We have: $$\forall n \in \Bbb Z : \sin^2(\pi n) f_2(\pi n) + \sin(\pi n) f_1(\pi n) + f_0(\pi n) = 0$$

Therefore: $$f_0(\pi n) = 0$$

Since a non-constant polynomial function can only have a finite number of zeros, $f_0$ must be $0$ everywhere. The equation reduces to: $$\sin^2(x) f_2(x) + \sin(x) f_1(x) = 0$$

Assume $x \ne \pi n$ and factor out $\sin(x)$ to get:

$$\forall x \ne \pi n : \sin(x) f_2(x) + f_1(x) = 0$$

Which means that: $$\sin(x) = -\frac{f_1(x)}{f_2(x)}$$

But $-f_1/f_2$ is also a rational function. By the continuity of $\sin$ and $-f_1/f_2$, equality must hold everywhere, which contradicts the fact that $\sin$ is not a rational function.

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Minor typo: $f_0$ became $f_2$. – 1015 Feb 21 '13 at 23:36
@julien Fixed now. Thanks! – Ayman Hourieh Feb 21 '13 at 23:38
Typo in the 2nd last equation: $\sin(x) (f_2(x) + f_1(x)) = 0$ – Américo Tavares Feb 21 '13 at 23:43
@AméricoTavares I don't see it. It should be $\sin(x) f_2(x) + f_1(x) = 0$, not $\sin(x) (f_2(x) + f_1(x)) = 0$ as you suggest. – Ayman Hourieh Feb 21 '13 at 23:46
By the way, if you spot a typo, feel free to fix. I don't mind. :) – Ayman Hourieh Feb 21 '13 at 23:47

In fact, the curve $y = \sin x$ is not equal to (or contained in) any algebraic curve $f(x,y) = 0$, where $f(x,y)$ is any polynomial in two variables with real coefficients. Indeed, the algebraic curve $f(x,y) = 0$ meets the $x$-axis in only finitely many points, but $y = \sin x$ meets the $x$-axis in infinitely many points.

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