4. Zadatak

S obzirom da su $\alpha$ i $\beta$ kutovi trokuta, slijedi da su kutovi $\frac\alpha2$ i $\frac\beta2$ između $0^\circ$ i $90^\circ$.

Na tom intervalu funkcija sin raste, a funkcija cos pada.

Zato $\alpha<\beta$ povlači

$$\sin\left(\frac\alpha2\right)<\sin\left(\frac\beta2\right)\quad\text{i}\quad\cos\left(\frac\alpha2\right)>\cos\left(\frac\beta2\right),$$

a onda i

$$\sin^2\left(\frac\alpha2\right)<\sin^2\left(\frac\beta2\right)\quad\text{i}\quad\cos^3\left(\frac\alpha2\right)>\cos^3\left(\frac\beta2\right),$$

pa je

$$\sin^2\left(\frac\alpha2\right)\cos^3\left(\frac\beta2\right)<\sin^2\left(\frac\beta2\right)\cos^3\left(\frac\alpha2\right).$$

S druge strane, $\alpha>\beta$ povlači

$$\sin\left(\frac\alpha2\right)>\sin\left(\frac\beta2\right)\quad\text{i}\quad\cos\left(\frac\alpha2\right)<\cos\left(\frac\beta2\right),$$

a onda i

$$\sin^2\left(\frac\alpha2\right)>\sin^2\left(\frac\beta2\right)\quad\text{i}\quad\cos^3\left(\frac\alpha2\right)<\cos^3\left(\frac\beta2\right),$$

pa je

$$\sin^2\left(\frac\alpha2\right)\cos^3\left(\frac\beta2\right)>\sin^2\left(\frac\beta2\right)\cos^3\left(\frac\alpha2\right).$$

Zato je jednakost u zadatku moguća samo za $\alpha=\beta$.