\section*{Problem 4 - 051224}

Determine all real numbers \(x, y\) for which the equation

\[
\sin (x+y)=\sin x+\sin y
\]

is satisfied.
}

The following trigonometric relationships apply:

\[
\begin{array}{r}
\sin 2 \alpha=2 \sin \alpha \cos \alpha \\
\sin \alpha+\sin \beta=2 \sin \frac{\alpha+\beta}{2} \cdot \cos \frac{\alpha-\beta}{2} \\
\cos \alpha-\cos \beta=-2 \cdot \sin \frac{\alpha+\beta}{2} \cdot \sin \frac{\alpha-\beta}{2}
\end{array}
\]

Thus, for the given equation with (1) and (2):

\[
\begin{array}{r}
\sin (x+y)=\sin x+\sin y \\
2 \sin \frac{x+y}{2} \cos \frac{x+y}{2}=2 \sin \frac{x+y}{2} \cdot \cos \frac{x-y}{2}
\end{array}
\]

Case 1: \(\sin \frac{x+y}{2}=0 \Rightarrow \frac{x+y}{2}=k \pi \Rightarrow x+y=2 k \pi ; k \in \mathbb{Z}\)

Case 2: \(\sin \frac{x+y}{2} \neq 0\) in (4) and with (3):

\[
\begin{aligned}
\cos \frac{x+y}{2}-\cos \frac{x-y}{2} & =0 \\
-2 \cdot \sin \frac{x}{2} \cdot \sin \frac{y}{2} & =0
\end{aligned}
\]

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

Case 2a: \(\sin \frac{x}{2}=0 \Rightarrow x=2 k \pi, k \in \mathbb{Z}\)

Case 2b: \(\sin \frac{y}{2}=0 \Rightarrow y=2 k \pi, k \in \mathbb{Z}\)