If $\sin x + \sin y = 1$ and $\cos x + \cos y = 0$, solve for $x$ and $y$

There is an interesting trick: you may couple the two equations by writing $$ e^{ix}+e^{iy} = i \tag{1}$$ hence $e^{ix}$ and $e^{iy}$, that are two points on the unit circle, are simmetric with respect to the imaginary axis. By imposing that their sum has unit norm, we clearly get $\{x,y\}=\left\{\frac{\pi}{6},\frac{5\pi}{6}\right\}$:

There are so many different ways to solve it the real question is which way.

What reaches out to grab me is:

$\cos x + \cos y = 0$

$\cos x = - \cos y$ which means either $y = \pi - x$ (within a period of $2\pi$) or $y = x + \pi$ (within a period of $2\pi$).

If $y = x + \pi$ then $\sin y = - \sin x$ and $\sin y + \sin x = 0 \ne 1$ which is impossible.

If $y = x - \pi$ then $\sin y = \sin x$ and $\sin y + \sin x = 2 \sin x$. If this is so (and it's our only option) then $\sin x = 1/2$ which means $x = \{\pi/6, 5\pi/6\}$.

So $(x,y) = (\pi/6, 5\pi/6)$ or $(x,y)= (5\pi/6, \pi/6)$ (within periods of $2\pi$)

HINT

Squaring both equations you get $$ \sin^2 x + \sin^2 y + 2\sin x \sin y = 1\\ \cos^2 x + \cos^2 y + 2\cos x \cos y = 0 $$ Now add them together to get $$ 2 + 2 \sin x \sin y + 2 \cos x \cos y = 1 $$ or in other words $$ \frac{-1}{2} = \cos x \cos y + \sin x \sin y = \cos (x-y) $$