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In Triangle Abc, Prove the Following: a 2 Sin ( B − C ) = ( B 2 − C 2 ) Sin a

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Question

In triangle ABC, prove the following: 

\[a^2 \sin \left( B - C \right) = \left( b^2 - c^2 \right) \sin A\]

 

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Solution

Let 

\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\] 

Consider the RHS of the equation

\[a^2 \sin \left( B - C \right) = \left( b^2 - c^2 \right) \sin A\]

\[RHS = k^2 \sin A\left( \sin^2 B - \sin^2 C \right) \]
\[ = k^2 \sin A\left[ \sin\left( B + C \right)\sin\left( B - C \right) \right] \left[ \because \sin^2 B - \sin^2 C = \sin\left( B + C \right)\sin\left( B - C \right) \right]\]
\[ = k^2 \sin A\left[ \sin\left( \pi - A \right)\sin\left( B - C \right) \right] \left[ \because A + B + C = \pi \right]\]
\[ = k^2 \sin A\left[ \sin\left( A \right)\sin\left( B - C \right) \right]\]
\[ = k^2 \sin^2 A\sin\left( B - C \right)\]
\[ = a^2 \sin\left( B - C \right) = LHS \left[ \because a = k\sin A \right]\]
\[\text{ Hence proved } .\]

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Sine and Cosine Formulae and Their Applications
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