1. Zadatak

Nakon kvadriranja zadanih jednakosti dobivamo

$$\sin^2x-2\sin x\sin y+\sin^2y=\frac{1008}{1009},$$ $$\cos^2x-2\cos x\cos y+\cos^2y=1.$$

Zbrojimo dobivene izraze:

$$\sin^2x+\cos^2x -2(\cos x\cos y+\sin x\sin y) +\sin^2y+\cos^2y =1+\frac{1008}{1009}.$$

Prema tome,

$$1-2\cos(x-y)+1=1+\frac{1008}{1009},$$

pa je konačno

$$\boxed{\cos(x-y)=\frac1{2018}}.$$