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Question
Prove that:
sin2 B = sin2 A + sin2 (A − B) − 2 sin A cos B sin (A − B)
Solution
\[\text{ RHS }= \sin^2 A + \sin^2 \left( A - B \right) - 2\sin A \cos B \sin\left( A - B \right)\]
\[ = \sin^2 A + \sin\left( A - B \right) \left\{ \sin\left( A - B \right) - 2\sin A \cos B \right\}\]
\[ = \sin^2 A + \sin\left( A - B \right) \left( \sin A \cos B - \cos A \sin B - 2\sin A \cos B \right)\]
\[ = \sin^2 A + \sin\left( A - B \right) \left( - \sin A \cos B - \cos A \sin B \right)\]
\[ = \sin^2 A - \sin\left( A - B \right) \left( \sin A \cos B + \cos A \sin B \right)\]
\[ = \sin^2 A - \sin\left( A - B \right) \sin\left( A + B \right)\]
\[ = \sin^2 A - \left( \sin^2 A - \sin^2 B \right)\]
\[ = \sin^2 A - \sin^2 A + \sin^2 B\]
\[ = \sin^2 B\]
= LHS
Hence proved.
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