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Trigonometry Proof

  1. Sep 23, 2012 #1
    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.
     
  2. jcsd
  3. Sep 23, 2012 #2
    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
     
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