Question 42, Given $\sin x+\sin y=\frac{1}{3}, \cos x-\cos y=\frac{1}{5}$, find the values of $\sin (x-y)$ and $\cos (x+y)$.
Question 42, Solution: According to the conditions, we have:
$$
\begin{array}{l}
\frac{34}{225}=\frac{1}{3^{2}}+\frac{1}{5^{2}} \\
=(\sin x+\sin y)^{2}+(\cos x-\cos y)^{2} \\
=\left(\sin ^{2} x+\cos ^{2} x\right)+\left(\sin ^{2} y+\cos ^{2} y\right)-2(\cos x \cos y-\sin x \sin y) \\
=2-2 \cos (x+y)
\end{array}
$$

Solving, we get $\cos (x+y)=\frac{208}{225}$
According to the conditions, we also have:
$$
\begin{array}{l}
\frac{1}{15}=(\sin x+\sin y)(\cos x-\cos y) \\
=(\sin x \cos x-\sin y \cos y)-(\sin x \cos y-\cos x \sin y) \\
=\frac{1}{2}(\sin 2 x-\sin 2 y)-\sin (x-y) \\
=\cos (x+y) \sin (x-y)-\sin (x-y) \\
=[\cos (x+y)-1] \sin (x-y)
\end{array}
$$

Therefore, $\sin (x-y)=\frac{1}{15[\cos (x+y)-1]}$. Since $\cos (x+y)=\frac{208}{225}$, we have $\sin (x-y)=\frac{1}{15\left(\frac{208}{225} - 1\right)}=$
$$
-\frac{15}{17} \text {. }
$$