11.49 For what values of $\alpha$ and $\beta$ is the equality $\sin \alpha + \sin \beta = \sin (\alpha + \beta)$ possible?
11.49 We have

\[
\begin{aligned}
& 2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}-2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha+\beta}{2}=0 \\
& \sin \frac{\alpha+\beta}{2}\left(\cos \frac{\alpha-\beta}{2}-\cos \frac{\alpha+\beta}{2}\right)=0 \\
& -2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha}{2} \sin \frac{\beta}{2}=0
\end{aligned}
\]

The product is zero if at least one of the factors is zero:

1) \(\sin \frac{\alpha+\beta}{2}=0 ; \frac{\alpha+\beta}{2}=\pi n ; \alpha+\beta=2 \pi n\);

\[
\sin \frac{\alpha}{2}=0 ; \frac{\alpha}{2}=\pi n ; \alpha=2 \pi n, \beta-\text{any number; }
\]

3) \(\sin \frac{\beta}{2}=0 ; \frac{\beta}{2}=\pi n ; \beta=2 \pi n, \alpha\) - any number.

Answer: \(\alpha+\beta=2 \pi n ; \alpha=2 \pi n, \beta \in \mathbb{R} ; \beta=2 \pi n, \alpha \in \mathbb{R}(n \in \mathbb{Z})\).