TITLE: Expressing $\sin(2x)$ as a polynomial of $\sin{x}$
QUESTION [13 upvotes]: Using trignometric identities ( double angle forumlas) one can see that $\cos{2x} = 2 \: \cos^{2}{x} - 1$ can be expressed as a polynomial of $\cos{x}$, where $p(\cos{x})=2 \: \cos^{2}{x}-1$. Then its natural to ask the same question for the sine function. 

Can we express $\sin{2x}$ as a polynomial of $\sin{x}$?

REPLY [20 votes]: Here's a slightly different way of proving that $\sin nx$ is not
a polynomial in $\sin x$ when $n$ is even. Indeed $\sin nx$
cannot be written as $f(\sin x)$ for any function $f$.
To prove this all one has to do is to write down two numbers $a$
and $b$ such that $\sin a=\sin b$ but $\sin na\ne\sin nb$.
Let $a=\pi/(2n)$ and $b=\pi-a$. Then $\sin a=\sin b$. But
$na=\pi/2$ and $nb=n\pi-\pi/2$ so that $\sin na=1$ and $\sin nb=-1$
(recalling that $n$ is even).