5. If $\sin x+\sin y=\frac{\sqrt{2}}{2}, \cos x+\cos y=$ $\frac{\sqrt{6}}{2}$, then $\sin (x+y)$ equals $(\quad)$.
(A) $\frac{\sqrt{2}}{2}$
(B) $\frac{\sqrt{3}}{2}$
(C) $\frac{\sqrt{6}}{2}$
(D) 1
5. B,

Square the two expressions and add them, we get
$$
\begin{array}{l}
\left(\sin ^{2} x+\cos ^{2} x\right)+\left(\sin ^{2} y+\cos ^{2} y\right)+ \\
2(\cos x \cdot \cos y+\sin x \cdot \sin y)=2 .
\end{array}
$$

Therefore, $\cos (x-y)=0$.
Multiply the two expressions, we get
$\left(\sin x^{\prime} \cos y+\sin y^{\prime} \cos x\right)+\left(\sin x^{\prime} \cos x+\sin y^{\prime} \cos y\right)=\frac{\sqrt{3}}{2}$.
Therefore, $\sin (x+y)+\sin (x+y) \cdot \cos (x-y)=\frac{\sqrt{3}}{2}$. Hence $\sin (x+y)=\frac{\sqrt{3}}{2}$.