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?
