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Trig Proof

  1. Dec 24, 2014 #1
    Prove for any triangle ABC that:
    [tex]cos(A)+cos(B)+cos(C)=1+4sin(\frac{1}{2}A)sin(\frac{1}{2}B)sin(\frac{1}{2}C)[/tex]

    I tried using the half-angle formula on the right side to get:

    [tex]1+4 \left(\frac{1-cos(A)}{2}\frac{1-cos(b)}{2}sin(\frac{1}{2}C)\right) {}[/tex]

    which simplifies to

    [tex]1+\bigg(1-cos(A)-cos(B)+cos(A)cos(B) \bigg)*sin(\frac{1}{2}C)[/tex]

    by the product to sum rule, this simplifies to

    [tex]1+\bigg(1-cos(A)-cos(B)+\frac{1}{2}\Big[cos(A+B)+cos(A-B)\Big]\bigg)*sin(\frac{1}{2}C)[/tex]

    I tried expanding the sin 1/2 C, but that just made things even more complicated. How should I approach this problem?
     
    Last edited: Dec 24, 2014
  2. jcsd
  3. Dec 24, 2014 #2
    Use product to sum formulas.

    I remember their being a shortcut for this problem where u rewrite the terms on the left allowing you to use double angle formulas (dont quote me on this). I'll try to give a shot.
     
  4. Dec 24, 2014 #3
    Forget the last part I said about product to sum. It just gives you back the original statement. I'll try to work on it later after work. Just had a 15 min break.
     
  5. Dec 24, 2014 #4

    NascentOxygen

    User Avatar

    Staff: Mentor

    Don't forget at some stage you are going to substitute C with 180° - (A+B)
    or C = ##\pi## - (A+B)

    whichever you prefer.
     
  6. Dec 24, 2014 #5

    NascentOxygen

    User Avatar

    Staff: Mentor

    If stymied, it's often worth trying a google search, such as: triangle "sin A + sin B + sin C"
     
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