135 Let $\sin x+\sin y=1$, then the range of $\cos x+\cos y$ is
$135[-\sqrt{3}, \sqrt{3}]$. Let $\cos x+\cos y=t$, squaring both sides of this equation and $\sin x+\sin y=1$ and then adding them yields
$$
2(\cos x \cos y+\sin x \sin y)=t^{2}-1,
$$

which simplifies to
$$
2 \cos (x-y)=t^{2}-1 .
$$

Given $-1 \leqslant \cos (x-y) \leqslant 1$, we have
$$
t^{2} \leqslant 3,
$$

thus $-\sqrt{3} \leqslant t \leqslant \sqrt{3}$.