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📅 July 6, 2026 👤 scraface

Jul 06, 2026

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Using the identity $ \cos(90^\circ - x) = \sin x $, but here we use:
\cos(2lpha - 90^\circ) = \sin(2lpha),
since $ \cos(90^\circ - x) = \sin x $, but sign depends on identity:
\cos(2lpha - 90^\circ) = \cos(90^\circ - 2lpha) = \sin(2lpha) \quad...
\cos(2lpha - 90^\circ) = \cos(90^\circ - 2lpha) = \sin(2lpha) \quad...
Better:
\cos(2lpha - 90^\circ) = \cos(90^\circ - 2lpha) = \sin(2lpha) \quad...
Actually, $ \cos(90^\circ - x) = \sin x $, so $ \cos(2lpha - 90^\circ...
\sin(2lpha) = \cos(2lpha - 90^\circ) = rac{1}{3}.
So $ \sin(2lpha) = rac{1}{3} $.

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