Start from the identity \(\cos(x-y)=\cos x\cos y+\sin x\sin y\).

Compute \(p^2+q^2\):

\[ p^2+q^2 =(\sin x+\sin y)^2 + (\cos x+\cos y)^2 \] \[ =(\sin^2x+\sin^2y + 2\sin x\sin y) +(\cos^2x+\cos^2y + 2\cos x\cos y) \] \[ =( \sin^2x+\cos^2x ) + ( \sin^2y+\cos^2y ) + 2(\sin x\sin y+\cos x\cos y) \] \[ =1+1+2\cos(x-y) =2+2\cos(x-y). \]

Solve for \(\cos(x-y)\):

\[ \cos(x-y)=\frac{p^2+q^2-2}{2}. \]

Result: \(\displaystyle \boxed{\cos(x-y)=\tfrac{p^2+q^2-2}{2}}\).