If $\alpha + \beta + \gamma = \pi$ prove that $\sin(2\alpha)+\sin(2\beta) + \sin(2\gamma) = 4\sin(\alpha)\sin(\beta)\sin(\gamma)$

Question

If $\alpha + \beta + \gamma = \pi$ prove that $\sin(2\alpha)+\sin(2\beta) + \sin(2\gamma) = 4\sin(\alpha)\sin(\beta)\sin(\gamma)$

I am really confused of how to utilize the information given.
Can someone please help.
Thanks.

Answer 1

Start with: $$\sin(2\alpha)+\sin(2\beta)=2\sin(\alpha+\beta)\cos(\alpha-\beta)=2\sin(\gamma)\cos(\alpha-\beta)$$

Then $$\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)=2\sin(\gamma)\cos(\alpha-\beta)+\sin(2\gamma)$$

$$\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)=2\sin(\gamma)\cos(\alpha-\beta)+2\sin(\gamma)\cos(\gamma)=$$ $$=2\sin(\gamma)(\cos(\alpha-\beta)+\cos(\gamma))=4\sin(\gamma)\cos\left(\frac{\alpha-\beta+\gamma}{2}\right)\cos\left(\frac{\alpha-\beta-\gamma}{2}\right)$$

Can you finish?