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Trigometry Identity

  1. Apr 7, 2007 #1
    Question Statement
    If A+B+C = 180, prove that cos (A+B-C) + cos (B+C-A) + cos (C+A-B) = 1+4 cosAcosBcosC

    My Attempt
    If A+B+C=180,
    Then A+B-C=180-2C
    cos (A+B-c)=cos(180-2C)

    (After some substitution and caculation)

    cos (A+B-C) = -cos 2C

    Similarily, I obtain the same expression for cos (B+C-A), cos (C+A-B).

    But that does not get me to the desire result.

    Is my attempt wrong from the beginning?
     
  2. jcsd
  3. Apr 7, 2007 #2
    No. Youre right. There is another trigo identity that you might find useful:
    cos2a + cos2b + cos2c=-1-4cosa cosb cosc.
     
  4. Apr 7, 2007 #3

    VietDao29

    User Avatar
    Homework Helper

    Have you covered the Sum to Product formulae?
    [tex]\cos \alpha + \cos \beta = 2 \cos \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right)[/tex]
    [tex]\cos \alpha - \cos \beta = -2 \sin \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right)[/tex]
    [tex]\sin \alpha + \sin \beta = 2 \sin \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right)[/tex]
    [tex]\sin \alpha - \sin \beta = 2 \cos \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right)[/tex]

    So you've shown the LHS to be equal to:
    - (cos (2A) + cos (2B) + cos(2C))
    Using the identites above, we have:
    [tex]- (\cos (2A) + \cos (2B) + \cos(2C)) = - (2 \cos(A + B) \cos(A - B) + \cos (2C)) = -(2 \cos(\pi - C) \cos (A - B) + 2 \cos ^ 2 C - 1)[/tex]
    [tex]= 1 - (- 2 \cos C \cos (A - B) + 2 \cos ^ 2 C) = ...[/tex]
    Can you go from here? :)
     
    Last edited: Apr 7, 2007
  5. Apr 7, 2007 #4

    Gib Z

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    Homework Helper

    Heres a Trig Identity that you will find useful :

    [tex]\mbox{If A+B+C = 180, then} \cos (A+B-C) +\cos (B+C-A) + \cos (C+A-B) = 1 + 4 \cos A \cos B \cos C[/tex].

    LOL
     
  6. Apr 9, 2007 #5
    Thanks a lot :)
     
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