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Complex Numbers

Prove that: ...

Question

Prove that: ${sin}^{2} 42^{o} - {cos}^{2} 78^{o} = \frac{(\sqrt{5} + 1)}{8}$

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Solution

LHS $= - ({cos}^{2} 78^{o} - {sin}^{2} 42^{o}) = - cos (78^{o} + 42^{o}) cos (78^{o} - 42^{o}) = - cos 120^{o} cos 36^{o} = - (- \frac{1}{2}) \frac{(\sqrt{5} + 1)}{4} = (\sqrt{5} + 1) / 8$

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