Given,

$\cos x+\cos y=p .........\left(1\right)$

And $\sin x+\sin y=q .......\left(2\right)$

Now squaring and adding both the equation we get,

${\cos }^{2}x+{\sin }^{2}x+{\cos }^{2}y+{\sin }^{2}y+2\cos x\cos y+2\sin x\sin y=p^2+q^2$

$\Rightarrow 1+1+2\left(\cos x\cos y+\sin x\sin y\right)=p^2+q^2$

$\Rightarrow 1+\cos \left(x-y\right)=\frac{p^2+q^2}{2} ......\left(3\right)$

Now we know that $1+\cos 2\theta =2{\cos }^2\frac{\theta }{2}$

Now using the above formula in equation $\left(3\right)$ we get,

$2{\cos }^2\left(\frac{x-y}{2}\right)=\frac{p^2+q^2}{2}$

$\Rightarrow {\cos }^2\left(\frac{x-y}{2}\right)=\frac{p^2+q^2}{4}$

$\Rightarrow \cos \left(\frac{x-y}{2}\right)=\pm \frac{\sqrt{p^2+q^2}}{2}$