Prove that:

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.

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