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

[Monoxdifly]

If \(\displaystyle \alpha+\beta+\gamma=180°\), prove that \(\displaystyle 2sin\alpha+2sin\beta+2sin\gamma=4sin\alpha sin\beta sin\gamma\)!
All I knew is that \(\displaystyle sin(\beta+\gamma)=sin(180°-\alpha)=sin\alpha\), but I think it doesn't help in this case.

 

[skeeter]

If \(\displaystyle \alpha+\beta+\gamma=180°\), prove that \(\displaystyle \color{red}2sin\alpha+2sin\beta+2sin\gamma=4sin\alpha sin\beta sin\gamma\)!
All I knew is that \(\displaystyle sin(\beta+\gamma)=sin(180°-\alpha)=sin\alpha\), but I think it doesn't help in this case.

check that equation again, because it's not an identity

 

[Monoxdifly]

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.

 

[Greg]

If $x+y+z=\pi$ then

\(\displaystyle \sin2x+\sin2y+\sin2z=4\sin x\sin y\sin z\)

 

[Greg]

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