The problem is as follows:

$\textrm{Prove:}$

$\sin 2\omega+\sin 2\phi+\sin 2\psi= 4 \sin\omega \sin\phi \sin\psi$

$\textrm{Given this condition:}$

$$\omega +\phi+ \psi= \pi$$

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:

$$\sin \omega \left(\frac{1}{2}(\cos (\phi + \psi) + \cos \left(\phi -\psi\right))\right) + \cos \omega \left(\left(\sin \phi \cos \psi + \cos \phi \sin \psi\right)\right) = \sin \omega \sin \phi \sin \psi$$

$$\sin \omega \left(\frac{1}{2}(\cos (\phi + \psi) + \cos \left(\phi -\psi\right))\right) + \cos \omega \left(\frac{1}{2} \left(\sin (\phi + \psi) + \sin (\phi-\psi) + \sin(\phi+\psi) - \sin(\phi-\psi)\right)\right)= \sin \omega \sin \phi \sin \psi$$

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 $4$

$$2 \sin \omega \left((\cos (\phi + \psi) + \cos \left(\phi -\psi\right))\right) + 2 \cos \omega \left( \left( 2 \sin (\phi + \psi) \right)\right)= 4 \sin \omega \sin \phi \sin \psi$$

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 what is it missing or what should I do to prove this identity?.

$\textrm{Prove:}$

$\sin 2\omega+\sin 2\phi+\sin 2\psi= 4 \sin\omega \sin\phi \sin\psi$

$\textrm{Given this condition:}$

$$\omega +\phi+ \psi= \pi$$

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:

$$\sin \omega \left(\frac{1}{2}(\cos (\phi + \psi) + \cos \left(\phi -\psi\right))\right) + \cos \omega \left(\left(\sin \phi \cos \psi + \cos \phi \sin \psi\right)\right) = \sin \omega \sin \phi \sin \psi$$

$$\sin \omega \left(\frac{1}{2}(\cos (\phi + \psi) + \cos \left(\phi -\psi\right))\right) + \cos \omega \left(\frac{1}{2} \left(\sin (\phi + \psi) + \sin (\phi-\psi) + \sin(\phi+\psi) - \sin(\phi-\psi)\right)\right)= \sin \omega \sin \phi \sin \psi$$

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 $4$

$$2 \sin \omega \left((\cos (\phi + \psi) + \cos \left(\phi -\psi\right))\right) + 2 \cos \omega \left( \left( 2 \sin (\phi + \psi) \right)\right)= 4 \sin \omega \sin \phi \sin \psi$$

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 what is it missing or what should I do to prove this identity?.

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