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Question
Without using trigonometric tables, prove that:
(sin72° + cos18°)(sin72° − cos18°) = 0
Solution
LHS=(sin72°+cos18°)(sin72°−cos18°)
=(sin72°+cos18°)[cos(90°−72°)−cos18°]
=(sin72°+cos18°)(cos18°−cos18°)
=(sin72°+cos18°)(0)
=0
=RHS
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