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