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Question
In ∆ABC, prove the following:
\[a^2 = \left( b + c \right)^2 - 4 bc \cos^2 \frac{A}{2}\]
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Solution
\[RHS = \left( b + c \right)^2 - 4bc \cos^2 \frac{A}{2}\]
\[ = b^2 + c^2 + 2bc - 4bc\left( \frac{1 + \cos A}{2} \right)\]
\[ = b^2 + c^2 + 2bc - 2bc\left( 1 + \cos A \right)\]
\[ = b^2 + c^2 + 2bc\left( 1 - 1 - \cos A \right)\]
\[ = b^2 + c^2 - 2bc\cos A\]
\[ = b^2 + c^2 - 2bc\left( \frac{b^2 + c^2 - a^2}{2bc} \right) \left( \because \cos A = \frac{b^2 + c^2 - a^2}{2bc} \right)\]
\[ = b^2 + c^2 - b^2 - c^2 + a^2 \]
\[ = a^2 = LHS\]
Hence proved.
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