SOLUTION: if alpha +beta + gamma = 0 , then prove that sin2α+sin2β+sin2γ=2(sinα+sinβ+sinγ)(1+cosα+cosβ+cosγ)
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Question 1005148: if alpha +beta + gamma = 0 , then prove that sin2α+sin2β+sin2γ=2(sinα+sinβ+sinγ)(1+cosα+cosβ+cosγ)
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
By expanding the right-hand side (RHS) we can invoke a lot of trig identities:
 + \sin 2\alpha + \sin 2\beta + \sin 2\gamma + 2(\sin(\alpha + \beta) + \sin(\alpha + \gamma) + \sin(\beta + \gamma)))
using the fact that  = \sin x \cos y + \sin y \cos x)
.
Since 
, we have that  = \sin(-\gamma) = -\sin \gamma)
, etc., so we can simplify as follows:
 + \sin 2\alpha + \sin 2\beta + \sin 2\gamma - 2(\sin \alpha - \sin \beta - \sin \gamma))
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