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Math Help - Please help me with this proof

  1. #1
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    Please help me with this proof



    Prove the identity:
    (hint: use the laws of cosine)

    cos A
    = b + c - a
    ..a.........2abc

    Using the above fact, now prove:
    (Hint: what would cos B and cos C look like?)
    ....................b.........c

    cos A + cos B + cos C = a + b + c
    ..a.......b.......c........2abc

    thanks in advance.
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  2. #2
    Senior Member DivideBy0's Avatar
    Joined
    Mar 2007
    From
    Melbourne, Australia
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    Quote Originally Posted by adhesive View Post

    Prove the identity:
    (hint: use the laws of cosine)

    cos A = b + c - a
    ..a.........2abc

    Using the above fact, now prove:
    (Hint: what would cos B and cos C look like?)
    ....................b.........c

    cos A + cos B + cos C = a + b + c
    ..a.......b.......c........2abc

    thanks in advance.
    The Law of cosines is a^2 = b^2+c^2-2bc\cos A

    Re-arranging: -2bc\cos A = a^2-b^2-c^2

    Multiplying both sides by -1: 2bc\cos{A}=b^2+c^2-a^2

    Dividing both sides by 2abc: \frac{\cos{A}}{a}=\frac{b^2+c^2-a^2}{2abc}

    Similarly, by changing around variables, you can get:

    \frac{\cos B}{b}=\frac{a^2+c^2-b^2}{2abc}

    \frac{\cos C}{c} = \frac{a^2+b^2-c^2}{2abc}

    So if you simply add them all up:

    \frac{\cos A}{a}+\frac{\cos B}{b} + \frac{\cos C}{c} = \frac{b^2+c^2-a^2}{2abc}+\frac{a^2+c^2-b^2}{2abc}+\frac{a^2+b^2-c^2}{2abc}

    =\frac{a^2+b^2+c^2}{2abc}
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  3. #3
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    Fantastic, thank you very much!
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