This simplifies to $ rac2(2\cos heta + 2\cos\phi)2i(\sin heta - \sin\phi) = rac2(\cos heta + \cos\phi)i(\sin heta - \sin\phi) $, which is purely imaginary. However, the original expression simplifies directly: - AMAZONAWS

📅 March 6, 2026 👤 scraface

Mar 06, 2026

$This simplifies to $ rac{2(2\cos heta + 2\cos\phi)}{2i(\sin heta - \sin\phi)} = rac{2(\cos heta + \cos\phi)}{i(\sin heta - \sin\phi)} $, which is purely imaginary. However, the original expression simplifies directly:$

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