1. Zadatak

Prvo rješenje.

Primijetimo da je

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

Primjenom poučka o kosinusu na trokut $ABC$ dobivamo

$$\cos\alpha=\frac{|AB|^2+|AC|^2-|BC|^2}{2|AC|\cdot|AB|} =\frac{16+25-49}{2\cdot5\cdot4} =-\frac{1}{5}.$$

Stoga je

$$\sin^6\frac{\alpha}{2}+\cos^6\frac{\alpha}{2} =\frac{1}{4}+\frac{3}{4}\cdot\frac{1}{25} =\frac{7}{25}.$$

Drugo rješenje.

Kao u prvom rješenju, dobivamo $\cos\alpha=-\frac{1}{5}$.

Budući da je $\cos^2\frac{\alpha}{2}=\frac{1-\cos\alpha}{2}$ i $\sin^2\frac{\alpha}{2}=\frac{1+\cos\alpha}{2}$, vrijedi $\cos^2\frac{\alpha}{2}=\frac{3}{5}$ i $\sin^2\frac{\alpha}{2}=\frac{2}{5}$.

Slijedi

$$\sin^6\frac{\alpha}{2}+\cos^6\frac{\alpha}{2} =\frac{27}{125}+\frac{8}{125} =\frac{7}{25}.$$