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

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Discussion Overview

The discussion centers around proving the equation $$2\sin\alpha + 2\sin\beta + 2\sin\gamma = 4\sin\alpha\sin\beta\sin\gamma$$ under the condition that $$\alpha + \beta + \gamma = 180°$$. Participants explore various approaches and provide insights related to trigonometric identities and relationships.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant notes that $$\sin(\beta+\gamma) = \sin(180° - \alpha) = \sin\alpha$$ but expresses uncertainty about its usefulness in proving the equation.
  • Another participant questions the validity of the equation, stating it is not an identity and suggests checking it again.
  • A third participant mentions that a student attempted to verify the equation using angles of 60° but found that it did not hold, leading to the conclusion that it could not be proved.
  • One participant introduces a related equation involving $$\sin2x + \sin2y + \sin2z$$ and provides a mathematical derivation that leads to a different identity, which may or may not relate to the original equation.

Areas of Agreement / Disagreement

Participants express disagreement regarding the validity of the original equation, with some questioning its identity status and others providing alternative equations. The discussion remains unresolved as no consensus is reached on the proof or validity of the original statement.

Contextual Notes

There are indications of missing assumptions or dependencies on specific angle values, as well as unresolved mathematical steps in the discussions presented.

Monoxdifly
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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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