Prove that:

$\frac{2 \pi}{5}=72^{\circ}, \frac{\pi}{2}=60^{\circ}$

$\mathrm{LHS}=\sin ^{2} 72^{\circ}-\sin ^{2} 60^{\circ}$

$=\sin ^{2}\left(90^{\circ}-18^{\circ}\right)-\frac{3}{4}$

$=\cos ^{2} 18^{\circ}-\frac{3}{4} \quad\left(\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right)$

$=\left(\frac{\sqrt{10+2 \sqrt{5}}}{4}\right)^{2}-\frac{3}{4} \quad\left(\because \cos 18^{\circ}=\frac{\sqrt{10+2 \sqrt{5}}}{4}\right)$

$=\frac{10+2 \sqrt{5}}{16}-\frac{3}{4}$

$=\frac{10+2 \sqrt{5}-12}{16}$

$=\frac{2 \sqrt{5}-2}{16}$

$=\frac{\sqrt{5}-1}{8}$

= RHS

Hence proved.