7・15 The necessary and sufficient condition for $\sin \alpha+\sin \beta=\sin (\alpha+\beta)$ is
(A) $\alpha+\beta=0$.
(B) $\alpha=\beta=2 k \pi(k \in Z)$.
(C) At least one of $\alpha, \beta$ is $2 k \pi(k \in Z)$.
(D) At least one of $\alpha, \beta, \alpha+\beta$ is $2 k \pi(k \in Z)$.
(High School Mathematics Competition in Huanggang Region, Hubei Province, China, 1989)
[Solution 1] If $\sin \alpha + \sin \beta = \sin (\alpha + \beta)$ holds,
and $\sin \alpha + \sin \beta = 2 \cos \frac{\alpha - \beta}{2} \sin \frac{\alpha + \beta}{2} = \sin (\alpha + \beta)$, then we have
$$
\alpha + \beta = 2 k \pi \quad (k \in \mathbb{Z}).
$$

Also, $\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta = \sin \alpha + \sin \beta$, then we have $\alpha = 2 k \pi$ or $\beta = 2 k \pi \quad (k \in \mathbb{Z})$.

That is, if the given equation holds, then at least one of $\alpha, \beta, \alpha + \beta$ is $2 k \pi \quad (k \in \mathbb{Z})$.
Conversely, if at least one of $\alpha, \beta, \alpha + \beta$ is $2 k \pi \quad (k \in \mathbb{Z})$,
if $\alpha = 2 k \pi \quad (k \in \mathbb{Z})$, then $\sin \alpha = 0$, and $\sin (\alpha + \beta) = \sin \beta$, the given equation holds;
Similarly, if $\beta = 2 k \pi \quad (k \in \mathbb{Z})$, it also holds;
If $\alpha + \beta = 2 k \pi \quad (k \in \mathbb{Z})$, then $\sin \beta = \sin (2 k \pi - \alpha) = -\sin \alpha$, i.e.,
$$
\sin \alpha + \sin \beta = 0,
$$

and $\sin (\alpha + \beta) = \sin 2 k \pi = 0$, then the given equation holds.
Therefore, the answer is $(D)$.
[Solution 2] $\sin \alpha + \sin \beta - \sin (\alpha + \beta) = 0$
$$
\begin{array}{l}
\Leftrightarrow 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} - 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha + \beta}{2} = 0 \\
\Leftrightarrow 2 \sin \frac{\alpha + \beta}{2} \left( \cos \frac{\alpha - \beta}{2} - \cos \frac{\alpha + \beta}{2} \right) = 0 \\
\Leftrightarrow 4 \sin \frac{\alpha + \beta}{2} \cdot \sin \frac{\alpha}{2} \sin \frac{\beta}{2} = 0 \\
\Leftrightarrow \sin \frac{\alpha + \beta}{2} = 0 \text{ or } \sin \frac{\alpha}{2} = 0 \text{ or } \sin \frac{\beta}{2} = 0 \\
\Leftrightarrow \alpha + \beta = 2 k \pi \text{ or } \alpha = 2 k \pi \text{ or } \beta = 2 k \pi \quad (k \in \mathbb{Z}) \\
\Leftrightarrow \alpha, \beta, \alpha + \beta \text{ at least one of them is } 2 k \pi \quad (k \in \mathbb{Z}).
\end{array}
$$

Therefore, the answer is $(D)$.