If sin A+sin B=sqrt{3}(cos B-cos A),then sin | vedclass

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(A) Given,$\sin A+\sin B=\sqrt{3}(\cos B-\cos A)$.Rearranging the terms,we get $\sin A+\sqrt{3}\cos A=\sqrt{3}\cos B-\sin B$.Dividing both sides by $2

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(A) Given,$\sin A+\sin B=\sqrt{3}(\cos B-\cos A)$.Rearranging the terms,we get $\sin A+\sqrt{3}\cos A=\sqrt{3}\cos B-\sin B$.Dividing both sides by $2$,we have $\frac{1}{2}\sin A+\frac{\sqrt{3}}{2}\cos A=\frac{\sqrt{3}}{2}\cos B-\frac{1}{2}\sin B$.This can be written as $\sin A \cos \frac{\pi}{3}+\cos A \sin \frac{\pi}{3}=\sin \frac{\pi}{3} \cos B-\cos \frac{\pi}{3} \sin B$.Using the formula $\sin(x+y)=\sin x \cos y+\cos x \sin y$ and $\sin(x-y)=\sin x \cos y-\cos x \sin y$,we get $\sin(A+\frac{\pi}{3})=\sin(\frac{\pi}{3}-B)$.Thus,$A+\frac{\pi}{3}=\frac{\pi}{3}-B$,which implies $A=-B$.Now,$\sin 3A+\sin 3B = \sin 3(-B)+\sin 3B = -\sin 3B+\sin 3B = 0$.

If $\sin A+\sin B=\sqrt{3}(\cos B-\cos A)$,then $\sin 3A+\sin 3B$ is equal to

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