cos alpha sin (beta-gamma)+cos beta sin (gamm | vedclass

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Question

(A) Expand each term using the formula $\sin(x-y) = \sin x \cos y - \cos x \sin y$:$\cos \alpha (\sin \beta \cos \gamma - \cos \beta \sin \gamma) + \c

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(A) Expand each term using the formula $\sin(x-y) = \sin x \cos y - \cos x \sin y$:$\cos \alpha (\sin \beta \cos \gamma - \cos \beta \sin \gamma) + \cos \beta (\sin \gamma \cos \alpha - \cos \gamma \sin \alpha) + \cos \gamma (\sin \alpha \cos \beta - \cos \alpha \sin \beta)$$= \cos \alpha \sin \beta \cos \gamma - \cos \alpha \cos \beta \sin \gamma + \cos \beta \sin \gamma \cos \alpha - \cos \beta \cos \gamma \sin \alpha + \cos \gamma \sin \alpha \cos \beta - \cos \gamma \cos \alpha \sin \beta$Observe that all terms cancel out:$(\cos \alpha \sin \beta \cos \gamma - \cos \gamma \cos \alpha \sin \beta) + (-\cos \alpha \cos \beta \sin \gamma + \cos \beta \sin \gamma \cos \alpha) + (-\cos \beta \cos \gamma \sin \alpha + \cos \gamma \sin \alpha \cos \beta) = 0 + 0 + 0 = 0$

$\cos \alpha \sin (\beta-\gamma)+\cos \beta \sin (\gamma-\alpha)+\cos \gamma \sin (\alpha-\beta)$ is equal to

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