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

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}

• Heron's formula + the law of sines? Commented Aug 24, 2011 at 18:24
• @Michael: I get $2m^2=a^2+b^2+c^2+4\sqrt{3s(s-a)(s-b)(s-c)}$ where $s=(a+b+c)/2$.
– robjohn
Commented Aug 24, 2011 at 23:25
• This would probably routine for a computer algebra package to verify by converting the trig functions into exponentials. Commented Aug 25, 2011 at 0:36
• @Ragib: It would be, but, as shown, these are all simple consequences of the Law of Sines, the Law of Cosines, and Heron's Formula. Why invoke a CA package? I find proof by computer a little less satisfying.
– robjohn
Commented Aug 25, 2011 at 2:52
• @rob, true, I probably overreacted to the apparent complexity of the problem. I just have that sort of reaction to problems like this, where I know there is a completely algorithmic procedure for solving it. In the same vein, I don't particularly enjoy integrating rational functions. Commented Aug 25, 2011 at 8:47

[(*) 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.