What is the conditional expectation of $\mathbb{E}(X\mid\sin(X))$ if $X$ is uniformly distributed on $[0,\pi]$?
Intuitively I expect that it is constant and equal to $\frac{\pi}{2}$, since the Borel set generated by $\sin(X)$ is symmetric around $\frac{\pi}{2}$.
Is that true? What would be the formal proof?
On
$\frac{\pi}{2}$ is correct, yes.
Proof: We want to use the following theorem: $$\mathbb{E}(X|\mathcal{F}) = Y \Leftrightarrow \forall G \in \mathcal{G}: \int_G Y \, d\mathbb{P} = \int_G X \, d\mathbb{P} \qquad (\ast)$$
where $G$ is a generator of the $\sigma$-Algebra $\mathcal{F}$. In this case we can choose $\mathcal{G} = \{\{\sin X \leq y\}; y \in \mathbb{R}\}$. Let $G = \{\sin X \leq y\} \in \mathcal{G}$, $y \in [0,1]$, then
$$\int_G X \, d\mathbb{P} = \frac{1}{\pi} \int_0^\pi x \cdot 1_{(-\infty,y]}(\sin x) \, dx = \frac{1}{\pi} \cdot \left( \int_0^{\arcsin y} x \,dx + \int_{\pi-\arcsin y}^{\pi} x \, dx \right) \\ = \ldots = \arcsin y = \ldots = \frac{1}{\pi} \cdot \int_0^\pi \underbrace{\frac{\pi}{2}}_{=:Y} 1_{(-\infty,y]}(x) \, dx = \int_G \frac{\pi}{2}$$
Clearly the equality holds also for $y \in \mathbb{R} \backslash [0,1]$. From $(\ast)$ follows $\mathbb{E}(X|\sin X)=\frac{\pi}{2}$.
Yes it is.
A formal proof would be to show that $\mathbb E(Xu(\sin X))=\frac\pi2\mathbb E(u(\sin X))$ for every bounded measurable function $u$.
Recall that in full generality, $\mathbb E(Y\mid Z)$ may be defined as the only $Z$-measurable random variable $T$ such that $\mathbb E(Yu(Z))=\mathbb E(Tu(Z))$ for every bounded measurable function $u$. Use this for $Y=X$, $Z=\sin X$ and $T=\frac\pi2$.
A painless way to show that $\mathbb E(Xu(\sin X))=\frac\pi2\mathbb E(u(\sin X))$ is to start with the fact that $\pi-X$ and $X$ are identically distributed hence $(\pi-X,\sin(\pi-X))=(\pi-X,\sin X)$ and $(X,\sin X)$ are identically distributed. Thus, $(*)=\mathbb E(Xu(\sin X))$ solves the identity $$ (*)=\mathbb E((\pi-X)u(\sin(X)))=\pi\mathbb E(u(\sin X))-(*), $$ and the proof is complete.