Identity for $4\sin^2\alpha\sin^2\beta\sin^2\gamma$ expressed as product of terms $(\pm\sin\alpha\pm\sin\beta\pm\sin\gamma)$

Question

It is well known (?) that if $\alpha+\beta+\gamma=\pi$ then $4\sin\alpha\sin\beta\sin\gamma = \sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)$ (I think I've seen it in some late-19th-century books, and I read somewhere on the internet (therefore it's true!! right?) that it has repeatedly appeared on the joint entrance examination of the Indian Institutes of Technology).

It seems very probable that this similar identity is in the literature somewhere, and I wonder where: $$ \begin{align} & {}\qquad \text{If }\alpha+\beta+\gamma=\pi\text{ then }4\sin^2\alpha\;\sin^2\beta\;\sin^2\gamma \\ & = (\sin\alpha+\sin\beta+\sin\gamma)(\sin\alpha+\sin\beta-\sin\gamma)(\sin\alpha-\sin\beta+\sin\gamma)(-\sin\alpha+\sin\beta+\sin\gamma). \end{align} $$

Answer 1

It can be done using Heron's formula (*) as stated in the comments, but notice that this is simply the constraint (or the Zariski closure of the constraint) on the sines of three angles to be sines of a triangle. I don't think there is a lower-degree polynomial expressing the same condition. This raises the question of whether a conceptual solution exists, avoiding the details of classical geometry.

[(*) Start from Heron's formula (squared), replace Area by $abc/4R$, replace $a,b,c$ by $2R\sin \alpha, 2R \sin \beta, 2R \sin \gamma$. Factors of $R$ will disappear from the final result, leaving the formula on sines.

The "well known" formula at the top of the question is a similar restatement of the fact that the area of a triangle is the sum of the small triangles formed by joining vertices to the center of the circumscribed circle.]