If alpha + beta - gamma = pi , then {sin ^2}a | vedclass

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Question

(A) Given $\alpha + \beta - \gamma = \pi ,$ so $\gamma = \alpha + \beta - \pi .$Consider the expression $E = \sin^2 \alpha + \sin^2 \beta - \sin^2 \ga

Vedclass · Accepted Answer

$2\sin \alpha \sin \beta \cos \gamma $

Vedclass · Answer

(A) Given $\alpha + \beta - \gamma = \pi ,$ so $\gamma = \alpha + \beta - \pi .$Consider the expression $E = \sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma .$Using the identity $\sin^2 A - \sin^2 B = \sin(A - B)\sin(A + B),$$E = \sin^2 \alpha + \sin(\beta - \gamma)\sin(\beta + \gamma).$Since $\beta - \gamma = \pi - \alpha,$ we have $\sin(\beta - \gamma) = \sin(\pi - \alpha) = \sin \alpha.$Also,$\beta + \gamma = \beta + (\alpha + \beta - \pi) = \alpha + 2\beta - \pi.$Alternatively,using $\gamma = \alpha + \beta - \pi,$$E = \sin^2 \alpha + \sin^2 \beta - \sin^2(\alpha + \beta - \pi) = \sin^2 \alpha + \sin^2 \beta - \sin^2(\alpha + \beta).$Using $\sin^2 A - \sin^2 B = \sin(A-B)\sin(A+B),$$E = \sin^2 \alpha - \sin(\alpha + \beta - \beta)\sin(\alpha + \beta + \beta) = \sin^2 \alpha - \sin \alpha \sin(\alpha + 2\beta).$$E = \sin \alpha [\sin \alpha - \sin(\alpha + 2\beta)].$Using $\sin C - \sin D = 2\cos(\frac{C+D}{2})\sin(\frac{C-D}{2}),$$E = \sin \alpha [2\cos(\alpha + \beta)\sin(-\beta)] = -2\sin \alpha \sin \beta \cos(\alpha + \beta).$Since $\alpha + \beta = \pi + \gamma,$ $\cos(\alpha + \beta) = \cos(\pi + \gamma) = -\cos \gamma.$Therefore,$E = -2\sin \alpha \sin \beta (-\cos \gamma) = 2\sin \alpha \sin \beta \cos \gamma.$

If $\alpha + \beta - \gamma = \pi ,$ then ${\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma = $

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