Question

If $A+B+C=\pi$ then prove that sin 2A +sin 2B+sin 2C =4 sin A sin B sin C.

Answer 1

Consider the problem

$\sin x+\sin y=2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right),\sin 2A=2\mathrm{sinAcosA}$

And, $\cos x+\cos y=2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)$

$\begin{array}{c}\sin 2A+\sin 2B+\sin 2C=2\sin \left(\frac{2A+2B}{2}\right)\cos \left(\frac{2A-2B}{2}\right)+2\sin C.\cos C\\ =2\sin \left(A+B\right)\cos \left(A-B\right)+2\sin C.\cos C\\ =2\sin \left(\pi -C\right)\cos \left(A-B\right)+2\sin C.\cos C\\ =2\sin C\cos \left(A-B\right)+2\sin C.\cos C\\ =2\sin C\left(\cos \left(A-B\right)+\cos C\right)\\ =2\sin C\left(2\cos \left(\frac{A-B+C}{2}\right)\cos \left(\frac{A-B-C}{2}\right)\right)\\ =2\sin C\left(2\cos \left(\frac{\pi -B-B}{2}\right)\cos \left(\frac{A-\pi +A}{2}\right)\right)\\ =2\sin C\left(2\cos \left(\frac{\pi }{2}-B\right)\cos \left(A-\frac{\pi }{2}\right)\right)\\ =4\sin C\sin B\cos \left(-\left(\frac{\pi }{2}-A\right)\right)\\ =4\sin A\sin B\sin C\end{array}$