1. Zadatak

$$\begin{aligned} &\sin^2\left(x-\frac{2016\pi}{3}\right)+\sin^2\left(x-\frac{2017\pi}{3}\right)+\sin^2\left(x-\frac{2018\pi}{3}\right) \\ &=\sin^2(x-672\pi)+\sin^2\left(x-\frac{\pi}{3}-672\pi\right)+\sin^2\left(x-\frac{2\pi}{3}-672\pi\right) \\ &=\sin^2x+\sin^2\left(x-\frac{\pi}{3}\right)+\sin^2\left(x-\frac{2\pi}{3}\right) \\ &=\sin^2x+\left(\sin x\cos\frac{\pi}{3}-\cos x\sin\frac{\pi}{3}\right)^2+\left(\sin x\cos\frac{2\pi}{3}-\cos x\sin\frac{2\pi}{3}\right)^2 \\ &=\sin^2x+\left(\sin x\cdot\frac12-\cos x\cdot\frac{\sqrt3}{2}\right)^2+\left(\sin x\cdot\frac12+\cos x\cdot\frac{\sqrt3}{2}\right)^2 \\ &=\sin^2x+\frac14\sin^2x-\frac{\sqrt3}{2}\sin x\cos x+\frac34\cos^2x+\frac14\sin^2x+\frac{\sqrt3}{2}\sin x\cos x+\frac34\cos^2x \\ &=\frac32\sin^2x+\frac32\cos^2x=\frac32. \end{aligned}$$

Alternativno rješenje.

Možemo koristiti formulu za dvostruki kut:

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

Preostaje dokazati činjenicu da je

$$\cos(2x)+\cos\left(2x-\frac{2\pi}{3}\right)+\cos\left(2x-\frac{4\pi}{3}\right)=0.$$