square $\sin x+\cos y=1$ $\sin^{2}x+2\sin x\cos y+\cos^{2}y=1$ square $\sin y+\cos x=-1$ $\sin^{2}y+2\sin y\cos x+\cos^{2}x=1$ add the two and simplify $\sin (x+y)=0$ subtract the two and simplify $\sin^{2}x-\cos^{2}x+\cos^{2}y-\sin^{2}y+2\sin x\cos y-2\sin y\cos x=0$ $-\cos2x+\cos2y+2(\sin(x-y))=0$(1) multiply two original equations together $\sin x\sin y+\cos x\cos y+\sin x\cos y+\sin y\cos x=1$ $\cos(x-y)+\sin(x+y)=1$ $\cos(x-y)=1$ so $\sin(x-y)=0$ then (1) becomes $\cos2y-\cos2x=0$ $\cos2y=\cos2x$ AWD

Let $z = \frac{\pi}{2}-y$. Then $\sin x+\sin z = 1$ $\cos x+\cos z =-1$ And we want to prove $\cos 2x = \cos (\pi-2z) =-\cos (2z)$ Well, let $u = e^{i x}, v = e^{i z}$. Then $u+v = i-1$ Of course, geometrically, we can only have two cases (there are only two right triangles with side lengths $1, 1, \sqrt{2}$ such that the hypotenuse is the line from $(0, 0)$ to $(-1, 1)$): $u = i, v =-1$ Or $u =-1, v = i$ Both of which satisfy our condition. QED.

square $\sin x+\cos y=1$ $\sin^{2}x+2\sin x\cos y+\cos^{2}y=1$ square $\sin y+\cos x=-1$ $\sin^{2}y+2\sin y\cos x+\cos^{2}x=1$ add the two and simplify $\sin (x+y)=0$ subtract the two and simplify $\sin^{2}x-\cos^{2}x+\cos^{2}y-\sin^{2}y+2\sin x\cos y-2\sin y\cos x=0$ $-\cos2x+\cos2y+2(\sin(x-y))=0$(1) multiply two original equations together $\sin x\sin y+\cos x\cos y+\sin x\cos y+\sin y\cos x=1$ $\cos(x-y)+\sin(x+y)=1$ $\cos(x-y)=1$ so $\sin(x-y)=0$ then (1) becomes $\cos2y-\cos2x=0$ $\cos2y=\cos2x$ AWD

From $\sin(x+y) = 0$, you immediately get $x+y = k\pi$, so $y = k\pi-x$. Therefore, \[\cos 2y = \cos(2k\pi-2x) = \cos(-2x) = \cos(2x)\]