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Proving trigo identity

  1. May 6, 2005 #1
    Prove (given that A+B+C=π):
    [tex]\sin^2 A + \sin^2 B - \sin^2 C = 2 \sin A \sin B \sin C[/tex]
    I got this far:
    [tex]\begin{matrix}\mbox{LHS} & = & \sin^2 A + \sin^2 B - \sin^2 C \\ \ & = & \sin^2 A + \left ( \sin B + \sin C \right ) \left ( \sin B - \sin C \right ) \\ \ & = & \sin^2 A + \left ( 2 \sin \frac{B+C}{2} \cos \frac{B-C}{2} \right) \left ( 2 \cos \frac{B+C}{2} \sin \frac{B-C}{2} \right ) \\ \ & = & \sin^2 A + \left (2 \sin \frac{A}{2} \cos \frac{B-C}{2} \right) \left ( 2 \cos \frac{-A}{2} \sin \frac{B-C}{2} \right ) \\ \ & = & \sin^2 A + \left ( -2 \sin \frac{A}{2} \cos \frac{A}{2} \right ) \left ( 2 \cos \frac{B-C}{2} \sin \frac{B-C}{2} \right ) \\ \ & = & \sin^2 A + \left ( -2 \sin A \right ) \left (2 \cos \frac{B-C}{2} \sin \frac{B-C}{2} \right ) \end{matrix}[/tex]

    Help please...

    TIA
     
    Last edited: May 6, 2005
  2. jcsd
  3. May 6, 2005 #2
    In its current form, it's false. Try a = b = c = pi/3.
     
  4. May 6, 2005 #3
    Never mind, I got it already... right hand side should be cos C instead of sin C.
     
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