Given that $$\cos x + \cos y = 1$$, the range of $$\sin x - \sin y$$ is

Given,

$$\cos x + \cos y = 1$$

Squaring both sides, $$\cos^2x+\cos^2y+2\cos x\cos y=1$$

or, $$1-\sin^2x+1-\sin^2y+2\cos x\cos y=1$$

or, $$1+2\cos x\cos y=\sin^2x+\sin^2y$$

Adding $$-2\sin x\sin y$$ on both sides,

$$1+2\cos x\cos y-2\sin x\sin y=\sin^2x+\sin^2y-2\sin x\sin y$$

or, $$1+2\cos\left(x+y\right)=\left(\sin x-\sin y\right)^2$$

Now, the maximum value of $$\cos\left(x+y\right)$$ can be 1

So, the maximum value of $$\left(\sin x-\sin y\right)^2$$ will be $$1+2\times\ 1=3$$

So, the maximum and minimum possible values of  $$\sin x-\sin y$$ will be $$\sqrt{\ 3}$$ and $$-\sqrt{\ 3}$$ respectively

So, the range of $$\sin x-\sin y$$ is $$[-\surd{3}, \surd{3}]$$