 Solution

The triangular support can be written $0\le x\le \pi /2$, $\pi /2-x\le y\le \pi /2$. The density is nonnegative there, and the normalization of a probability distribution requiresHence $\boxed{A=2}$. Multiplying the joint probability density by the required function cancels $\sin x$ on the interior. Its expectation isThe singularity at $x=0$ lies at a support vertex of zero probability; the displayed integral establishes integrability. Assigning any value to the random function at that single vertex does not change the expectation.