3. (2 points) A poor student wrote the following incorrect formulas for the sine and cosine of a sum: $\sin (\alpha+\beta)=$ $\sin \alpha+\sin \beta$ and $\cos (\alpha+\beta)=\cos \alpha+\cos \beta$. In his defense, he said that for some $\alpha$ and $\beta$ his formulas are still correct. Find all such pairs $(\alpha, \beta)$.
Answer: $\alpha= \pm \frac{\pi}{3}+2 \pi m, m \in \mathbb{Z}$ and $\beta=\mp \frac{\pi}{3}+2 \pi l, l \in \mathbb{Z}$.

Solution:

$$
\left\{\begin{array}{l}
\sin (\alpha+\beta)=\sin \alpha+\sin \beta \\
\cos (\alpha+\beta)=\cos \alpha+\cos \beta
\end{array}\right.
$$

From the first equality, it follows that $2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha+\beta}{2}\right)=2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)$

Hence,

$$
\left[\begin{array} { l } 
{ \operatorname { sin } ( \frac { \alpha + \beta } { 2 } ) = 0 } \\
{ \operatorname { cos } ( \frac { \alpha + \beta } { 2 } ) = \operatorname { cos } ( \frac { \alpha - \beta } { 2 } ) }
\end{array} \Leftrightarrow \left[\begin{array}{l}
\alpha+\beta=2 \pi k, k \in \mathbb{Z} \\
\frac{\alpha+\beta}{2}= \pm \frac{\alpha-\beta}{2}+2 \pi n, n \in \mathbb{Z}
\end{array}\right.\right.
$$

In the first case, $\cos \alpha=\cos \beta$. Thus, the "cosine of the sum" formula becomes $1=2 \cos \alpha$, from which $\alpha= \pm \frac{\pi}{3}+2 \pi m, m \in \mathbb{Z}$ and, accordingly, $\beta=\mp \frac{\pi}{3}+2 \pi l, l \in \mathbb{Z}$.

In the second case, we get either $\beta=2 \pi n, n \in \mathbb{Z}$ or $\alpha=2 \pi n, n \in \mathbb{Z}$. Then the second equality becomes $\cos \alpha=1+\cos \alpha$ or a similar equality for $\beta$, which is impossible. Therefore, only the first case remains.