1. Sep 23, 2012 #1

   anthonych414

   

   1. The problem statement, all variables and given/known data
   
   if A+B+C=π, prove that asin(B-C)+bsin(C-A)+csin(A-B)=0
   
   2. Relevant equations
   
   sin(x-y)=sinxcosy-cosxsiny
   
   3. The attempt at a solution
   
   I tried to expand but to no avail. Any help is appreciated.

    

   anthonych414, Sep 23, 2012
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3. Sep 23, 2012 #2

   NasuSama

   

   Here are some advices:
   
   →Take A + B + C = π and solve for one of the variables. Then, substitute C, B or A for asin(B-C)+bsin(C-A)+csin(A-B)=0.
   →Since you get the shift by π in sine expressions, you need to use these formulas:
   
   sin(x - π) = -sin(x) and sin(-x + π) = sin(x) [Make sure that when get the expression like sin(A + B - π), we have -sin(A + B)]
   
   Oh! This link might help you! http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Symmetry.2C_shifts.2C_and_periodicity
   
   Good luck, and let me know if you have comments or problems. :D

    

   NasuSama, Sep 23, 2012

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