If A+B+C=frac{pi}{3},then sin left(frac{pi-6A | vedclass

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(D) Given,$A+B+C=\frac{\pi}{3}$.We need to evaluate $S = \sin \left(\frac{\pi-6A}{6}\right)+\sin \left(\frac{\pi-6B}{6}\right)+\sin C$.Using the formu

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$4 \cos \left(\frac{\pi-6A}{12}\right) \cos \left(\frac{\pi-6B}{12}\right) \sin \frac{C}{2}$

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(D) Given,$A+B+C=\frac{\pi}{3}$.We need to evaluate $S = \sin \left(\frac{\pi-6A}{6}\right)+\sin \left(\frac{\pi-6B}{6}\right)+\sin C$.Using the formula $\sin X + \sin Y = 2 \sin \left(\frac{X+Y}{2}\right) \cos \left(\frac{X-Y}{2}\right)$:$S = 2 \sin \left(\frac{\frac{\pi-6A}{6} + \frac{\pi-6B}{6}}{2}\right) \cos \left(\frac{\frac{\pi-6A}{6} - \frac{\pi-6B}{6}}{2}\right) + \sin C$$S = 2 \sin \left(\frac{2\pi - 6(A+B)}{12}\right) \cos \left(\frac{6(B-A)}{12}\right) + \sin C$Since $A+B = \frac{\pi}{3} - C$,then $6(A+B) = 2\pi - 6C$.$S = 2 \sin \left(\frac{2\pi - (2\pi - 6C)}{12}\right) \cos \left(\frac{B-A}{2}\right) + \sin C$$S = 2 \sin \left(\frac{6C}{12}\right) \cos \left(\frac{B-A}{2}\right) + 2 \sin \frac{C}{2} \cos \frac{C}{2}$$S = 2 \sin \frac{C}{2} \left[ \cos \left(\frac{B-A}{2}\right) + \cos \frac{C}{2} \right]$Since $C = \frac{\pi}{3} - (A+B)$,$\frac{C}{2} = \frac{\pi}{6} - \frac{A+B}{2}$.$S = 2 \sin \frac{C}{2} \left[ \cos \left(\frac{B-A}{2}\right) + \cos \left(\frac{\pi}{6} - \frac{A+B}{2}\right) \right]$Using $\cos X + \cos Y = 2 \cos \left(\frac{X+Y}{2}\right) \cos \left(\frac{X-Y}{2}\right)$:$S = 2 \sin \frac{C}{2} \left[ 2 \cos \left(\frac{\frac{B-A}{2} + \frac{\pi}{6} - \frac{A+B}{2}}{2}\right) \cos \left(\frac{\frac{B-A}{2} - \frac{\pi}{6} + \frac{A+B}{2}}{2}\right) \right]$$S = 4 \sin \frac{C}{2} \cos \left(\frac{\pi - 6A}{12}\right) \cos \left(\frac{6B - \pi}{12}\right)$Since $\cos(-x) = \cos(x)$,$\cos \left(\frac{6B - \pi}{12}\right) = \cos \left(\frac{\pi - 6B}{12}\right)$.Thus,$S = 4 \cos \left(\frac{\pi-6A}{12}\right) \cos \left(\frac{\pi-6B}{12}\right) \sin \frac{C}{2}$.

If $A+B+C=\frac{\pi}{3}$,then $\sin \left(\frac{\pi-6A}{6}\right)+\sin \left(\frac{\pi-6B}{6}\right)+\sin C=$

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