Question:
Prove that:
$\sin ^{2} 24^{\circ}-\sin ^{2} 6^{\circ}=\frac{\sqrt{5}-1}{8}$
Solution:
LHS $=\sin ^{2} 24^{\circ}-\sin ^{2} 6^{\circ}$
$=\sin \left(24^{\circ}+6^{\circ}\right) \sin \left(24^{\circ}-6^{\circ}\right) \quad\left[\sin (\mathrm{A}+\mathrm{B}) \sin (\mathrm{A}-\mathrm{B})=\sin ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~B}\right]$
$=\sin 30^{\circ} \sin 18^{\circ}$
$=\frac{1}{2} \times \frac{\sqrt{5}-1}{4} \quad\left(\because \sin 18^{\circ}=\frac{\sqrt{5}-1}{4}\right)$
$=\frac{\sqrt{5}-1}{8}$
= RHS
Hence proved.