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#### Monoxdifly

##### Well-known member

- Aug 6, 2015

- 271

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

- Thread starter Monoxdifly
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- Thread starter
- #1

- Aug 6, 2015

- 271

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

- Mar 1, 2012

- 644

check that equation again, because it's not an identityIf \(\displaystyle \alpha+\beta+\gamma=180°\), 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.

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- Aug 6, 2015

- 271

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- Feb 5, 2013

- 1,382

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

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

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

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- #5

- Feb 5, 2013

- 1,382

&=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*}\)