If $A+B+C=\pi$ and $\frac{\sin 2 A+\sin 2 B+\sin 2 C}{\sin A+\sin B+\sin C}=\lambda \sin \left(\frac

Question

If $A+B+C=\pi$ and $\frac{\sin 2 A+\sin 2 B+\sin 2 C}{\sin A+\sin B+\sin C}=\lambda \sin \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right)$, then the value of $\lambda$ must be

Professor PC Thomas Classes · Answer

$ \begin{aligned} & \sin 2 A+\sin 2 B+\sin 2 C=4 \sin A \sin B \sin C=32 \sin A / 2 \sin B / 2 \sin C / 2 \ & \quad \cos A / 2 \cos B / 2 \cos C / 2 	ext { and } \sin A+\sin B+\sin C=4 \cos A / 2 \cos B / 2 \cos C / 2 \ & \quad 	ext { (from conditional identities) } \ & \quad 	herefore \frac{\sin 2 A+\sin 2 B+\sin 2 C}{\sin A+\sin B+\sin C}=8 \sin \left(\frac{A}{2}ight) \sin \left(\frac{B}{2}ight) \sin \left(\frac{C}{2}ight) \end{aligned} $ \ on comparing, we get $\lambda=8$