# Prove the triangle identity

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• Jun 24th 2012, 06:06 AM
NitroNbg
Prove the triangle identity
Hi everybody, I need your help again. I have to prove the following identity:
(b-2*cos(c)*a)/(a*sin(c))+(c-2*cos(a)*b)/(b*sin(a))+(a-2*cos(b)*c)/(c*sin(b))=0
For every triangle (a,b,c represents length of sides and opposite angles)
I know I should use laws of sine and cosine but I couldn't complete it.
Also sorry for my post, I'm on my tablet so I didn't use equation editor.
$(b-2*a*cos(gama))/(a*sin(gama))+(c-2*b*cos(alpha))/(b*sin(alpha))+(a-2*cos(beta))/(c*sin(beta))$
Alpha - opposite of a
Beta - opposite of b
Gama - opposite of c
• Jun 24th 2012, 06:25 AM
HallsofIvy
Re: Prove the triangle identity
It would help if you would distinguish between "A", the measure of an angle, and "a", the length of the side opposite that angle.