MHB [ASK] Proving Trigonometry 2sinα+2sinβ+2sinγ=4sinαsinβsinγ

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The discussion centers on proving the equation 2sinα + 2sinβ + 2sinγ = 4sinαsinβsinγ under the condition that α + β + γ = 180°. Participants express skepticism about the validity of the equation, noting that testing specific angles like 60° yields inconsistent results. One contributor mentions that the equation does not hold as an identity, suggesting it cannot be proven. An alternative formulation involving sin2x, sin2y, and sin2z is presented, which leads to a valid conclusion under the same angle sum condition. The overall consensus indicates that the original equation is likely incorrect.
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If $$\alpha+\beta+\gamma=180°$$, prove that $$2sin\alpha+2sin\beta+2sin\gamma=4sin\alpha sin\beta sin\gamma$$!
All I knew is that $$sin(\beta+\gamma)=sin(180°-\alpha)=sin\alpha$$, but I think it doesn't help in this case.
 
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Monoxdifly said:
If $$\alpha+\beta+\gamma=180°$$, prove that $$\color{red}2sin\alpha+2sin\beta+2sin\gamma=4sin\alpha sin\beta sin\gamma$$!
All I knew is that $$sin(\beta+\gamma)=sin(180°-\alpha)=sin\alpha$$, but I think it doesn't help in this case.

check that equation again, because it's not an identity
 
Well, this was a question asked by a student I am tutoring. He did say that he tried all angles to be 60° and the equation didn't match, so I told him that it means that it couldn't be proved.
 
If $x+y+z=\pi$ then

$$\sin2x+\sin2y+\sin2z=4\sin x\sin y\sin z$$
 
$$\begin{align*}2(\sin2x+\sin2y+\sin2z)&=\sin x\cos(y-z)+\sin y\cos(x-z)+\sin z\cos(x-y) \\
&=6\sin x\sin y\sin z+\sin x\cos y\cos z+\sin y\cos x\cos z+\sin z\cos x\cos y \\
&=6\sin x\sin y\sin z+\sin x\cos x+\sin y\cos y+\sin z\cos z \\
\frac32(\sin2x+\sin2y+\sin2z)&=6\sin x\sin y\sin z \\
\sin2x+\sin2y+\sin2z&=4\sin x\sin y\sin z\end{align*}$$
 
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