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Question
Prove that sin2 (n + 1) A − sin2 nA = sin (2n + 1) A sin A.
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Solution
\[\text{ LHS }= \sin^2 \left( n + 1 \right)A - \sin^2 nA\]
\[ = \sin\left[ \left( n + 1 \right)A + nA \right] \sin\left[ \left( n + 1 \right)A - n A \right] [\text{ Using the formula }\sin^2 X - \sin^2 Y = \sin\left( X + Y \right) \sin\left( X - Y \right)\]
\[\text{ and taking }X = \left( n + 1 \right)A \text{ and }Y = n A \]
\[ = \sin\left[ \left( n + 1 + n \right)A \right] \sin \left[ \left( n + 1 - n \right)A \right]\]
\[ = \sin\left( 2n + 1 \right)A \sin A\]
= RHS
Hence proved .
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