Expressing $\sin(2x)$ as a polynomial of $\sin{x}$ 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 $\sin{x}$?

 A: 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).
A: Just for the sake of completeness, here is a proof that you cannot express it that way.
(For a general proof, see the bottom of this answer).
Suppose we could express $\sin 2x$ as a polynomial in $\sin x$.
i.e.
$\sin x \cos x  = P(\sin x)$ for some polynomial $P$.
Since $\sin 0 = 0$, $x$ is a factor of $P(x)$. Hence we get
$\cos x = Q(\sin x)$ for a polynomial $Q$.
Put $x = \frac{\pi}{2}$ and $x=\frac{3\pi}{2}$ we see that $1,-1$ are roots of $Q(x)$.
Thus we see that for 0 < t < 1,
$\sqrt{1-t^2} = (t-1)(t+1)H(t)$ (because $Q(t) = (t-1)(t+1)H(t)$ for some polynomial $H$)
By squaring we see that the degree of the left hand side polynomial is $2$ while the degree of the right hand side polynomial is $>2$. This is not possible, as then their difference, which is not identically zero will have infinite roots.
Thus it is not possible to express $\sin 2x$ as a polynomial in $\sin x$.

General Case:
Here is a proof that in general $ \sin (2nx)$ cannot be written as a polynomial in $ \sin x$ based on the identity provided by J.M:
$$\frac{\sin(2nx)}{2\cos(x)}=(-1)^{\lfloor n-1/2\rfloor}\sum_{j=0}^{\lfloor n-1/2\rfloor}{(-1)^j \sin((2j+1)x)}$$
First notice that $\sin ((2m+1)x)$ can be written as a polynomial in $\sin x$ with integer coefficients (follows easily from the DeMoivre's identity).
Now suppose $ \sin (2nx)$ could be written as a polynomial $P$. Consider the $r^{th}$ derivative of that polynomial at $0$. This must be a rational number, as the $r^{th}$ derivative of $ \sin (2nx)$ at $0$ is rational. Thus the coefficients of $P$ must be rational.
Thus using the above identity we must have that $\cos x$ is of the form $\frac{P(\sin x)}{Q(\sin x)}$ where $P$ and $Q$ are polynomials with rational coefficients. Thus we must have that
$$\sqrt{1-t^2} = \frac{P(t)}{Q(t)}$$ 
$$ 0 < t < 1$$
Putting $t=\frac{1}{2}$ gives us a rational number on the RHS while an irrational number on the LHS.
Hence it is not possible to represent $ \sin (2nx)$ as a polynomial in $ \sin x$.
A: No, you can't get any simpler than $2\sin\;x\;\cos\;x$. In general, only sines of odd angle multiples ($\sin((2k+1)x)$) are expressible as polynomials in $\sin(x)$, because any powers of cosines that may appear are raised to even exponents, and thus the identity $\sin^2(x)=1-\cos^2(x)$ applies, to turn it into an expression entirely made up of powers of sines.
