sin 50^circ - sin 70^circ + sin 10^circ = | vedclass

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(B) Using the trigonometric identity $\sin C - \sin D = 2 \cos \left( \frac{C+D}{2} \right) \sin \left( \frac{C-D}{2} \right)$,we have:$\sin 50^\circ

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(B) Using the trigonometric identity $\sin C - \sin D = 2 \cos \left( \frac{C+D}{2} ight) \sin \left( \frac{C-D}{2} ight)$,we have:$\sin 50^\circ - \sin 70^\circ + \sin 10^\circ$$= 2 \cos \left( \frac{50^\circ + 70^\circ}{2} ight) \sin \left( \frac{50^\circ - 70^\circ}{2} ight) + \sin 10^\circ$$= 2 \cos(60^\circ) \sin(-10^\circ) + \sin 10^\circ$Since $\sin(-	heta) = -\sin 	heta$ and $\cos 60^\circ = 1/2$:$= 2 \left( \frac{1}{2} ight) (-\sin 10^\circ) + \sin 10^\circ$$= -\sin 10^\circ + \sin 10^\circ = 0.$

$\sin 50^\circ - \sin 70^\circ + \sin 10^\circ = $

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