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Math Help - Prove the triangle identity

  1. #1
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    Beograd
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    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
    Last edited by NitroNbg; June 24th 2012 at 08:14 AM.
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  2. #2
    MHF Contributor

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    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.
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