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Prove That: Sin2 B = Sin2 a + Sin2 (A − B) − 2 Sin a Cos B Sin (A − B)

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Question

Prove that:
sin2 B = sin2 A + sin2 (A − B) − 2 sin A cos B sin (A − B)

Answer in Brief
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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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Chapter 7: Values of Trigonometric function at sum or difference of angles - Exercise 7.1 [Page 20]

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R.D. Sharma Mathematics [English] Class 11
Chapter 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.1 | Q 16.4 | Page 20

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