Sine not a Rational Function Spivak 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?

since we defined all the functions here to be rational (other than sin(x)), we see that f0(x) has infinite roots since there are infinite points at which 2kpi must equal zero. Therefore the only rational function f0(x) could possibly be is the function f0(x)=0, so we can get rid of it and now refactor the equation. the rest is done similarly.
One can show by induction that the equation can be factored, and when derived always gives you an equation involving sin[..] and cos[..], now set x=2kpi and we see that by a similar argument fn(x) must be equal to 0 for all n. Therefore the only function that satisfies the implicit thing is fn(x)=0 for all n and x. We are done.
