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#### Monoxdifly

##### Well-known member
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

##### Well-known member
MHB Math Helper
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-known member
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

##### Perseverance
Staff member
If $x+y+z=\pi$ then

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

#### Greg

##### Perseverance
Staff member
\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*}