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Question

(B) Let the expression be $E = \sin (\beta + \gamma - \alpha ) + \sin (\gamma + \alpha - \beta ) + \sin (\alpha + \beta - \gamma ) - \sin (\alpha + \b

Vedclass · Accepted Answer

$4\sin \alpha \sin \beta \sin \gamma $

Vedclass · Answer

(B) Let the expression be $E = \sin (\beta + \gamma - \alpha ) + \sin (\gamma + \alpha - \beta ) + \sin (\alpha + \beta - \gamma ) - \sin (\alpha + \beta + \gamma )$.Using the formula $\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$:First,combine the first two terms:$\sin (\beta + \gamma - \alpha ) + \sin (\gamma + \alpha - \beta ) = 2 \sin \gamma \cos (\alpha - \beta) = 2 \sin \gamma \cos (\beta - \alpha)$.Next,combine the last two terms:$\sin (\alpha + \beta - \gamma ) - \sin (\alpha + \beta + \gamma ) = 2 \cos (\alpha + \beta) \sin (-\gamma) = -2 \sin \gamma \cos (\alpha + \beta)$.Now,combine these results:$E = 2 \sin \gamma [\cos (\beta - \alpha) - \cos (\alpha + \beta)]$.Using the identity $\cos (A-B) - \cos (A+B) = 2 \sin A \sin B$:$E = 2 \sin \gamma [2 \sin \alpha \sin \beta] = 4 \sin \alpha \sin \beta \sin \gamma$.

$\sin (\beta + \gamma - \alpha ) + \sin (\gamma + \alpha - \beta ) + \sin (\alpha + \beta - \gamma ) - \sin (\alpha + \beta + \gamma ) = $

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