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प्रश्न
Prove that: \[\sin^2 42° - \cos^2 78 = \frac{\sqrt{5} + 1}{8}\]
उत्तर
\[LHS = \sin^2 42° - \cos^2 78° \]
\[ = \sin^2 \left( 90° - 48° \right) - \cos^2 \left( 90° - 12° \right)\]
\[ = \cos^2 48° - \sin^2 12° \]
\[ = \cos\left( 48° + 12° \right) \cos\left( 48° - 12° \right) \left[ \cos\left( A + B \right) \cos\left( A - B \right) = \cos^2 A - \sin^2 \right]\]
\[ = \cos60° \cos36° \]
\[ = \frac{1}{2} \times \frac{\sqrt{5} + 1}{4} \left( \because \cos36° = \frac{\sqrt{5} + 1}{4} \right)\]
\[ = \frac{\sqrt{5} + 1}{8}\]
\[ = RHS\]
\[\text{ Hence proved } .\]
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