If $1+\cos \left(x-y\right)=0$ , then

$\cos \left(x-y\right)=-1$

$x-y=\pi$

$x=y+\pi$

$\cos x=\cos \left(\pi +y\right)$

$\cos x=-\cos y$

$\cos x+\cos y=0$

$x=y+\pi$

$\sin x=\sin \left(\pi +y\right)$

$\sin x=-\sin y$

$\sin x+\sin y=0$