3. (2 points) A poor student wrote the following incorrect formulas for the sine and cosine of the difference: $\sin (\alpha-\beta)=$ $\sin \alpha-\sin \beta$ 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: $\alpha=2 \pi n, n \in \mathbb{Z}$ 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 sum" formula turns into $1=0$, which is impossible.

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 turns into $\cos \alpha=\cos \alpha-1$, which is impossible, or $\cos \beta=1-\cos \beta$, from which $\beta=\mp \frac{\pi}{3}+2 \pi l, l \in \mathbb{Z}$.