Before explaining the situation I have read this other question which deals with the problem and shows a solution.
The problem is as follows:
$\textrm{Prove:}$
$\sin 2\omega+\sin 2\phi+\sin 2\psi= 4 \sin\omega \sin\phi \sin\psi$
$\textrm{Given this condition:}$
$$\omega +\phi+ \psi= \pi$$
However I tried to solve it in a different way, without appending the double angles and summing them together at the end, as it has been suggested by one user to the OP in the preceding question.
My attempt differs from the OP as he has not mentioned an effort into solve the problem, to which it is to be exposed below. In request of proof reading.
Therefore, in my attempt to solve this problem I did what I felt obvious and that was to take the sine function to the whole condition as shown below:
$$\sin \left (\omega +\phi+ \psi \right )= \sin \left ( \pi \right )$$
Grouping:
$$\sin \left (\omega + \left(\phi+ \psi\right) \right )= \sin \left ( \pi \right )$$
$$\sin \left (\omega + \left(\phi+ \psi\right) \right )= 0 $$
$$\sin \left(\omega\right) \cos \left(\phi+ \psi\right) + \cos \left(\omega\right) \sin \left(\phi+ \psi\right) = 0 $$
$$\sin \omega \left(\cos \phi \cos \psi - \sin \phi \sin \psi\right) + \cos \omega \left(\sin \phi \cos \psi + \cos \phi \sin \psi \right) = 0 $$
$$\sin \omega \cos \phi \cos \psi - \sin \omega \sin \phi \sin \psi + \cos \omega \sin \phi \cos \psi + \cos \omega \cos \phi \sin \psi = 0$$
$$\sin \omega \cos \phi \cos \psi + \cos \omega \sin \phi \cos \psi + \cos \omega \cos \phi \sin \psi = \sin \omega \sin \phi \sin \psi$$
$$\sin \omega \left(\cos \phi \cos \psi\right) + \cos \omega \left(\sin \phi \cos \psi + \cos \phi \sin \psi\right) = \sin \omega \sin \phi \sin \psi$$
Using product to sum identity:
$\sin \omega \left(\frac{1}{2}(\cos (\phi + \psi) + \cos \left(\phi -\psi\right))\right) + \cos \omega \left(\frac{1}{2} \left(\sin \phi \cos \psi + \cos \phi \sin \psi\right)\right) = \sin \omega \sin \phi \sin \psi$
$\sin \omega \left(\frac{1}{2}(\cos (\phi + \psi) + \cos \left(\phi -\psi\right))\right) $ $+ \cos \omega \left(\frac{1}{2} \sin (\phi + \psi) + \sin (\phi-\psi) + \sin(\phi+\psi) - \sin(\phi-\psi)\right)= \sin \omega \sin \phi \sin \psi$
$\sin \omega \left(\frac{1}{2}(\cos (\phi + \psi) + \cos \left(\phi -\psi\right))\right)$ $ + \cos \omega \left(\frac{1}{2} \left(\sin (\phi + \psi) + \sin (\phi-\psi) + \sin(\phi+\psi) - \sin(\phi-\psi)\right)\right)= \sin \omega \sin \phi \sin \psi$
Simplifying similar terms:
$\sin \omega \left(\frac{1}{2}(\cos (\phi + \psi) + \cos \left(\phi -\psi\right))\right) + \cos \omega \left(\frac{1}{2} \left( 2 \sin (\phi + \psi) \right)\right)= \sin \omega \sin \phi \sin \psi$
Then multiplying all by $4$
$$2 \sin \omega \left((\cos (\phi + \psi) + \cos \left(\phi -\psi\right))\right) + 2 \cos \omega \left( \left( 2 \sin (\phi + \psi) \right)\right)= 4 \sin \omega \sin \phi \sin \psi$$
By returning to the initial condition: $\omega +\phi + \psi =\pi$
$$2 \sin \omega (\cos (\pi-\psi)+\cos(\phi-\psi))+2 \cos \omega (2 \sin(\pi-\psi))= 4 \sin \omega \sin \phi \sin \psi$$
$$2 \sin (-\cos \psi + \cos (\phi - \psi)) + 2 \cos \omega ( 2 \sin \psi) = 4 \sin \omega \sin \phi \sin \psi$$
And that's how far I went, I feel I'm almost there but I don't know if I am in the right path or maybe not. Can somebody instruct me where exactly can I relate the double angle of omega, phi and psi?. Why is my method not working?
what did I missed or what should I do to my steps to prove this identity?.
