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Question
Prove that: \[\left( \cos \alpha + \cos \beta^2 \right) + \left( \sin \alpha + \sin \beta \right)^2 = 4 \cos^2 \left( \frac{\alpha - \beta}{2} \right)\]
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Solution
\[LHS = \left( cos \alpha + cos\beta \right)^2 + \left( sin\alpha + sin\beta \right)^2 \]
\[ = \cos^2 \alpha + \cos^2 \beta + 2cos\alpha cos\beta + \sin^2 \alpha + \sin^2 \beta + 2sin\alpha sin\beta\]
\[ = ( \cos^2 \alpha + \sin^2 \alpha) + ( \cos^2 \beta + \sin^2 \beta) + 2\left( cos\alpha cos\beta + sin\alpha sin\beta \right)\]
\[ = 1 + 1 + 2\cos(\alpha - \beta)\]
\[ = 2\left\{ 1 + \cos(\alpha - \beta) \right\}\]
\[ = 2\left\{ 2 \cos^2 \left( \frac{\alpha - \beta}{2} \right) \right\}\]
\[ = 4 \cos^2 \left( \frac{\alpha - \beta}{2} \right) = RHS\]
\[\text{ Hence proved } .\]
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