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Question
Prove that: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]
Solution
\[LHS = \sin^2 24° - \sin^2 6° \]
\[ = \sin\left( 24° + 6° \right) \sin\left( 24° - 6° \right) \left[ \sin\left( A + B \right) \sin\left( A - B \right) = \sin^2 A - \sin^2 B \right]\]
\[ = \sin30° \sin18° \]
\[ = \frac{1}{2} \times \frac{\sqrt{5} - 1}{4} \left( \because \sin18° = \frac{\sqrt{5} - 1}{4} \right)\]
\[ = \frac{\sqrt{5} - 1}{8}\]
\[ = RHS\]
\[\text{ Hence proved } .\]
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