Question

If $-\pi \le x\le \pi$, $-\pi \le y\le \pi$ and $\cos x+\cos y=2$ then $\cos (x-y)$

Answer 1

As we know the range of $cos\theta$ is $[-1,1]$ so the maximum value of $cos\theta$ is $1$

now, here $cosx+cosy=2$ can only be getting by its maximum value

so , $x=y=0$ because $\cos 0=1$

so, $cos(x-y)=cos(0-0)=\cos 0=1$

so, $\cos (x-y)=1$