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Question
If cos A + sin B = m and sin A + cos B = n, prove that 2 sin (A + B) = m2 + n2 − 2.
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Solution
\[\text{ RHS }= m^2 + n^2 - 2\]
\[ = \left( \cos A + \sin B \right)^2 + \left( \sin A + \cos B \right)^2 - 2\]
\[ = \cos^2 A + \sin^2 B + 2\cos A\sin B + \sin^2 A + \cos {}^2 B + 2\sin A\cos B - 2\]
\[ = 1 + 1 + 2\cos A\sin B + 2\sin A\cos B - 2\]
\[ = 2\left( \cos A\sin B + \sin A \cos B \right)\]
\[ = 2\sin\left( A + B \right)\]
= LHS
Hence proved.
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