1. Zadatak

Primijenimo na desnoj strani dane jednakosti formulu za polovični kut

$$\begin{aligned} &(b+c)^2\cdot\sin^2\frac{\alpha}{2}+(b-c)^2\cdot\cos^2\frac{\alpha}{2}\\ &=(b+c)^2\cdot\frac{1-\cos\alpha}{2}+(b-c)^2\cdot\frac{1+\cos\alpha}{2}\\ &=\frac12\left[(b+c)^2-(b+c)^2\cos\alpha+(b-c)^2+(b-c)^2\cos\alpha\right]\\ &=\frac12\left[2b^2+2c^2-\cos\alpha\left(b^2+2bc+c^2-b^2+2bc-c^2\right)\right]\\ &=\frac12\left[2b^2+2c^2-4bc\cos\alpha\right]=b^2+c^2-2bc\cdot\cos\alpha. \end{aligned}$$

Kako je u danom trokutu prema kosinusovom poučku $b^2+c^2-2bc\cos\alpha=a^2$, tvrdnja je dokazana.