Solve the following system of equations:

$$
\begin{aligned}
& \cos x+\cos y=\cos (x+y) \\
& \sin x+\sin y=\sin (x+y)
\end{aligned}
$$
I. solution. Let us introduce the notations $u=\frac{x+y}{2}$ and $v=\frac{x-y}{2}$. With these, $x=u+v, y=u-v$, and our system of equations transforms into:

$$
\begin{aligned}
& 2 \cos u \cos v=2 \cos ^{2} u-1 \\
& 2 \sin u \cos v=2 \sin u \cos u
\end{aligned}
$$

From (1), $\cos u \neq \cos v$, since $2 \cos ^{2} u \neq 2 \cos ^{2} u-1$. Therefore, (2) holds if and only if $\sin u=0$, i.e.,

$$
u=k \pi
$$

for some integer $k$. Then $\cos u=(-1)^{k}$, and from (1), $\cos v=(-1)^{k} / 2$, so

$$
v= \pm \frac{\pi}{3}+k \pi+2 l \pi
$$

where $l$ is an integer. The equations (1) and (2) are satisfied by these and only these values of $u, v$, so the solutions to the original system of equations are

$$
x=u+v= \pm \frac{\pi}{3}+2(k+l) \pi, \quad y=u-v=\mp \frac{\pi}{3}+2 l \pi
$$

where $k$ and $l$ are arbitrary integers.

II. solution. Let the unit vector in the direction of the positive x-axis be denoted by $\mathbf{a}$, and the unit vector in the direction of the positive y-axis by $\mathbf{b}$. The unit vector in the direction of the angle $x+y$ is denoted by $\mathbf{c}$. In these notations, the system of equations in the problem is equivalent to the equation

$$
\mathbf{c}=\mathbf{a}+\mathbf{b}
$$

Furthermore, let the origin be denoted by $O$, and the endpoints of the vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ starting from $O$ be denoted by $A$, $B$, and $C$ respectively.

Since $\mathbf{a}$ and $\mathbf{b}$ are unit vectors, equation (3) implies that one of the following three cases must hold:

1. $\mathbf{a}$ and $\mathbf{b}$ are in the same direction, in which case $\mathbf{c}$ is twice their length;
2. $\mathbf{a}$ and $\mathbf{b}$ are in opposite directions, in which case $\mathbf{c}$ is the zero vector;
3. The quadrilateral $OACB$ is a parallelogram.

![](https://cdn.mathpix.com/cropped/2024_05_02_92508b6d11ed10cebdf7g-1.jpg?height=615&width=675&top_left_y=1427&top_left_x=714)

Since $\mathbf{c}$ is also a unit vector, only the third case can occur. In this case, since $AC \parallel OB$ and $BC \parallel OA$, the angles at $O$ also appear at $C$, and since $OA = OC = OB$, they can be copied to $A$ and $B$. However, in this case, they can only be $60^\circ$, and therefore

$$
\begin{aligned}
& x= \pm 60^\circ + n \cdot 360^\circ \\
& y=\mp 60^\circ + m \cdot 360^\circ
\end{aligned}
$$

where $n, m$ are arbitrary integers.