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This is Chapter 15 Question 31 in Spivak:

a) Show sin is not a rational function.

By definition of a rational function, a rational function cannot be $0$ at infinite points unless it is $0$ everywhere. Obviously, sin have infinite points that are 0 and infinite points that are not zero, thus not a rational function.

b) Show that there do not exist rational functions $f_0, \ldots, f_{n-1}$ such that $(\sin x)^n + f_{n-1}(x)(\sin x)^{n-1} + \ldots + f_0({x}) = 0$ for all x

First choose $x = 2k\pi$, so $f_0(x) = 0$ for $x = 2k\pi$. Since $f_0$ is rational $\implies f_0(x) = 0$ for all x. Thus can write $\sin x[(\sin x)^{n-1} + f_{n-1}(\sin x)^{n-2} ... +f_1(x)] = 0$

Question: The second factor is $0$ for all $x \neq 2k\pi$ How does this imply that it is 0 for all x? And how does this lead to the result?

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Which do you call the second factor. – Student Feb 17 '13 at 20:21
The right one, where the left is sinx – mathnoob Feb 17 '13 at 20:22
Yes, use continuity. If a continuous $f(x)\not= 0$, then $f\not=0$ for a whole ball around x, which you know is not true. – Student Feb 17 '13 at 20:31

Hint: Use continuity of $\sin$ and rational functions.

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