2. Zadatak

Izraz $\cos^2\alpha+\cos^2\beta+\cos^2\gamma-1$ može se transformirati u umnožak kosinusa na neki od sljedećih načina: Prvi način:

$$\begin{aligned} 0&=\cos^2\alpha+\cos^2\beta+\cos^2\gamma-1=\cos^2\alpha+\cos^2\beta-\sin^2\gamma\\ &=\cos^2\alpha+\cos^2\beta-\sin^2[\pi-(\alpha+\beta)]\\ &=\cos^2\alpha+\cos^2\beta-\sin^2(\alpha+\beta)\\ &=\cos^2\alpha+\cos^2\beta-(\sin\alpha\cos\beta+\cos\alpha\sin\beta)^2\\ &=\cos^2\alpha+\cos^2\beta-\sin^2\alpha\cos^2\beta-\cos^2\alpha\sin^2\beta-2\sin\alpha\cos\beta\cos\alpha\sin\beta\\ &=\cos^2\alpha(1-\sin^2\beta)+\cos^2\beta(1-\sin^2\alpha)-2\sin\alpha\cos\alpha\cos\beta\sin\beta\\ &=2\cos^2\alpha\cos^2\beta-2\sin\alpha\cos\alpha\cos\beta\sin\beta\\ &=2\cos\alpha\cos\beta(\cos\alpha\cos\beta-\sin\alpha\sin\beta)\\ &=2\cos\alpha\cos\beta\cos(\alpha+\beta)\\ &=-2\cos\alpha\cos\beta\cos\gamma \end{aligned}$$

Drugi način:

$$\begin{aligned} 0&=\cos^2\alpha+\cos^2\beta+\cos^2\gamma-1\\ &=\frac{1+\cos2\alpha}{2}+\frac{1+\cos2\beta}{2}+\cos^2\gamma-1\\ &=\frac{\cos2\alpha+\cos2\beta}{2}+\cos^2\gamma\\ &=\cos(\alpha+\beta)\cos(\alpha-\beta)+\cos^2\gamma\\ &=\cos(\pi-\gamma)\cos(\alpha-\beta)+\cos^2\gamma\\ &=-\cos\gamma\cos(\alpha-\beta)+\cos^2\gamma\\ &=-\cos\gamma(\cos(\alpha-\beta)-\cos\gamma)\\ &=-\cos\gamma(\cos(\alpha-\beta)+\cos(\alpha+\beta))\\ &=-2\cos\alpha\cos\beta\cos\gamma \end{aligned}$$

Sad zaključujemo da neki od kosinusa $\cos\alpha$, $\cos\beta$, $\cos\gamma$ mora biti jednak nuli, pa je taj kut pravi.