Compute E(sin(X)|X+Y) Okay, let $X,Y$ be independent random variables with the same exponential distribution. Compute $E(X|X+Y)$ and $E(\sin(X)|X+Y)$.
Solution:
$X+Y=E(X+Y|X+Y)=E(X|X+Y)+E(Y|X+Y)=2\cdot E(X|X+Y)$
Hence $E(X|X+Y)=\frac{X+Y}{2}$.
$\sin(\frac{X+Y}{2})=E(\sin(\frac{X+Y}{2})|X+Y)=E(\sin \frac{X}{2}\cos\frac{Y}{2}+\cos\frac{X}{2}\sin\frac{Y}{2}|X+Y)=2E(\sin\frac{X}{2}\cos\frac{Y}{2}|X+Y)=E(\sin(X)|X+Y)$
Hence
$E(\sin(X)|X+Y)=\sin(\frac{X+Y}{2})$.
Is that correct?

Set $Z=X+Y$.
First we'll show $X|Z=z\sim \mathcal{U}(0,z)$ for $z>0$ fixed. For $x\in(0,z)$ we have
$$f_{X|Z=z}(x|z)=\frac{f_{XY}(x,z-x)\sqrt{2}}{\int_0^zf_{XY}(x,z-x)\sqrt{2}dx}=\frac{1}{z}$$ Here we're using the fact that $f_{XY}(x,y)=\lambda^2e^{-\lambda(x+y)}$ for $(x,y)\in (0,\infty)^2$ and $f_{XY}(x,y)=0$ elsewhere. Therefore, $$E(X|Z=z)=\int_0^zxf_{X|Z=z}(x|z)dx=\frac{z}{2}$$ Moreover, $$E(\sin(X)|Z=z)=\int_0^z\sin(x)f_{X|Z=z}(x|z)dx=\frac{2}{z}\sin^2\Big(\frac{z}{2}\Big)$$ So we get $E(X|X+Y)=\frac{X+Y}{2}$ while $E(\sin(X)|X+Y)=\frac{2}{X+Y}\sin^2\Big(\frac{X+Y}{2}\Big)$.
Note: The problem arises in your calculation when you claim that $$2E\Big(\sin(X/2)\cos(Y/2)\Big|X+Y=z\Big)=E(\sin(X)|X+Y=z)$$
