Closure of the algebra generated by $\{\sin x,1\}$ in $C_{\mathbb{R}}([-\pi,\pi])$ 
let $\mathcal{A}\subset C_{\mathbb{R}}([-\pi,\pi])$ equipped with the $\sup$ norm be a sub-algebra generated by $\{\sin x,1\}$.
What is $\overline{\mathcal{A}}$?

This excerise is suppose to be an application of the Stone-Weierstrass theorem but I'm having trouble seeing what this algebra looks like. Any help would be much appreciated
 A: Consider a closed subalgebra $\mathcal L \subset C[-\pi,\pi]$ consisting of such functions $f$, that
$$ f(x) = f(\pi-x),\;\; f(-x)=f(x-\pi) \;\quad\; \forall x\in [0,\tfrac{\pi}{2}].$$
Roughly speaking, this is all symmetries of the sine on $[-\pi,\pi]$. For a function in $\mathcal L$, it's fully determined by the behavior on $[-\frac{\pi}{2},\frac{\pi}{2}]$. Surely $\mathcal A \subset \mathcal L$.
Define $\Phi\colon \mathcal L \rightarrow C[-\frac{\pi}{2},\frac{\pi}{2}]$ by function restriction, i.e.
$\Phi(f) = \left.f\right|_{[-\frac{\pi}{2},\frac{\pi}{2}]}$. It's not hard to see that $\Phi$ is isomorphism of algebras plus homeomorphism (as it preserves $\sup$-norm).
Algebra $\mathcal B = \Phi(\mathcal A) \subset C[-\frac{\pi}{2},\frac{\pi}{2}]$ is generated by $\{\sin x, 1\}$ and separates points, since the sine is strictly monotone on $[-\frac{\pi}{2},\frac{\pi}{2}]$. Hence $\overline{\mathcal B} = C[-\frac{\pi}{2},\frac{\pi}{2}]$ by the Stone—Weierstrass theorem. Applying $\Phi^{-1}$ to the latter equality, we get
$ \overline{\mathcal A} = \mathcal L $.
