$\begin{align} & \text{If }A,B,C\text{ are angles of a triangle show that } \\ & sin2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C \\\end{align}$
$\begin{align} & \sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C \\ & \sin 2A+\sin 2B+\sin 2C=2\sin \frac{2A+2B}{2}\cos \frac{2A-2B}{2}+\sin 2C \\ & \sin 2A+\sin 2B+\sin 2C=2\sin (A+B)\cos (A-B)+\sin 2C \\ & \text{but }A+B=\pi -C \\ & \sin 2A+\sin 2B+\sin 2C=2\sin (\pi -C)\cos (A-B)+\sin 2C \\ & \sin 2A+\sin 2B+\sin 2C=2\sin (C)\cos (A-B)+\sin 2C \\ & \sin 2A+\sin 2B+\sin 2C=2\sin C\cos (A-B)+2\sin C\cos C \\ & \sin 2A+\sin 2B+\sin 2C=2\sin C\left[ \cos (A-B)+\cos C \right] \\ & \sin 2A+\sin 2B+\sin 2C=2\sin C\left[ \cos (A-B)+\cos [\pi -(A+B)] \right] \\ & \sin 2A+\sin 2B+\sin 2C=2\sin C\left[ \cos (A-B)-\cos (A+B) \right] \\ & \sin 2A+\sin 2B+\sin 2C=2\sin C\left[ -2\sin \tfrac{A-B+A+B}{2}\sin \tfrac{(A-B)-(A+B)}{2} \right] \\ & \sin 2A+\sin 2B+\sin 2C=2\sin C\left[ -2\sin A\sin (-B) \right] \\ & \sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C\text{ }\left| note:\sin (-\theta )=-\sin \theta \right. \\\end{align}$