1. Zadatak

Kvadriramo li danu jednakost dobivamo

$$\sin^2x-2\sin x\cos x+\cos^2x=\frac13,$$

odnosno $\sin x\cos x=\frac13$.

Izraz $\sin^6x+\cos^6x$ možemo zapisati u obliku umnoška

$$\begin{aligned} \sin^6x+\cos^6x &=(\sin^2x+\cos^2x)(\sin^4x-\sin^2x\cos^2x+\cos^4x)\\ &=\sin^4x-\sin^2x\cos^2x+\cos^4x\\ &=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x-\sin^2x\cos^2x\\ &=1-3\sin^2x\cos^2x. \end{aligned}$$

Koristeći $\sin x\cos x=\frac13$ konačno je

$$\sin^6x+\cos^6x=1-3(\sin x\cos x)^2=1-3\left(\frac13\right)^2=\frac23.$$