Prove that:
$\sin ^{2} \frac{2 \pi}{5}-\sin ^{2-} \frac{\pi}{3}=\frac{\sqrt{5}-1}{8}$
$\frac{2 \pi}{5}=72^{\circ}, \frac{\pi}{2}=60^{\circ}$
$\mathrm{LHS}=\sin ^{2} 72^{\circ}-\sin ^{2} 60^{\circ}$
$=\sin ^{2}\left(90^{\circ}-18^{\circ}\right)-\frac{3}{4}$
$=\cos ^{2} 18^{\circ}-\frac{3}{4} \quad\left(\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right)$
$=\left(\frac{\sqrt{10+2 \sqrt{5}}}{4}\right)^{2}-\frac{3}{4} \quad\left(\because \cos 18^{\circ}=\frac{\sqrt{10+2 \sqrt{5}}}{4}\right)$
$=\frac{10+2 \sqrt{5}}{16}-\frac{3}{4}$
$=\frac{10+2 \sqrt{5}-12}{16}$
$=\frac{2 \sqrt{5}-2}{16}$
$=\frac{\sqrt{5}-1}{8}$
= RHS
Hence proved.
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