How do I prove that sin is not defined implicitly by an algebraic equation? How do I prove that sin is not defined implicitly by an algebraic equation?
In essence, there does not exist rational functions $f_0,\ldots,f_{n-1}$ that satisfies
$$\sin^n(x)+f_{n-1}(x)\sin^{n-1}(x)+\cdots+f_0(x)=0$$
 A: No, that cannot be.
$f_0(x)$ would need to be $0$ whenever $x$ is a multiple of $\pi$, and since a nontrivial rational function has only finitely many zeroes, this means that $f_0$ is the constant zero function.
Then we can divide through by $\sin x$ at every point that is not a multiple of $\pi$ and get
$$ \sin^{n-1}(x) + f_{n-1}(x)\sin^{n-2}(x)+\cdots+f_1(x) = 0 $$
which, by continuity, must still hold at multiples of $\pi$.
Proceed by induction until $n=0$ and you get
$$ \sin^0(x) =0 $$
which is absurd.
A: If $f_0,...,f_{n-1}$ are all rational then there exist polynomials $g_0,...,g_{n-1}, g_n$ such that $$g_n(x) \sin^n(x)+g_{n-1}(x)\sin^{n-1}(x)+...+g_0(x)=0.$$ For $x=n\pi$, $n$ integer it must be true that $g_0(x)=0$. Because $g_0$ is just a polynomial, it can only have a finite number of roots, or it is identically $0$. Thus , $g_0(x) \equiv 0.$ Then one has $$g_n(x) \sin^n(x)+g_{n-1}(x)\sin^{n-1}(x)+...+g_1(x)\sin(x)=0$$ which for $x\ne n\pi$, is same as $$g_n(x) \sin^{n-1}(x)+g_{n-1}(x)\sin^{n-2}(x)+...+g_1(x)=0.$$ If $\alpha \ne n\pi$ is a root of above then $$g_n(x-\alpha) \sin^{n-1}(x-\alpha)+g_{n-1}(x-\alpha)\sin^{n-2}(x-\alpha)+...+g_1(x-\alpha)=0$$ is of the same form as the first equation and it follows $g_1(n\pi+x-\alpha)= 0$. By induction the result follows.
