I'm having trouble getting behind the process of applying the theory to calculating specific conditional expectations. I know that this might be a simple question, but I can't see the simplicity. Let $$U \sim \mathrm{Unif}([- \frac 1 2, \frac 1 2]), \, X := \cos(U), \,Y := \sin(U),$$ the task is to compute $ \mathbb E[X|Y] $ and $ \mathbb E[Y|X]$.
I am already struggling to check measurability of $X$ with respect to $Y$ and vice versa; after some hard thoughts and non-rigorous intuition, I know that $X$ is $\sigma(Y)$-measurable, since you can "basically reconstruct" the value of $X$, given a realisation of $Y$ (how to rigorously prove this is beyond me). Thus we have $$ \mathbb E[X|Y]=X. $$ I am pretty sure that, on the other hand, $Y$ is not $\sigma(X)$-measurable, since - given a realisation of $X$ - one can only "reconstruct" the value of $|Y|$, and not $Y$ itself (again, still failing to understand why exactly).
Question is now, how to evaluate $\mathbb E[Y|X]$?
Since $\sin$ is bijective on $I\equiv[-1/2,+1/2]$, $$ \mathbb{E}\left[X\mid Y\right]=\cos(\arcsin Y)=\sqrt{1-Y^2}. $$ Since $\cos$ is even on $I$, $$ \mathbb{E}\left[Y\mid X\right]=\frac{1}{2}\sin(\arccos X)+\frac{1}{2}\sin(-\arccos X)=0 $$ where the last equality follows from $\sin(-x)=-\sin x$.