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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
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up vote 3 down vote accepted

For the second part, 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 rational 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
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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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