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Math Help - Prove

  1. #1
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    Prove

    Given triangle ABC prove that c\sin\frac{A-B}{2}=(a-b)\cos\frac{C}{2}
    I know that \cos\frac{C}{2}=\sin\frac{A+B}{2}
    but after this i'm lost.
    Thanks
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  2. #2
    MHF Contributor
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    Hello arze
    Quote Originally Posted by arze View Post
    Given triangle ABC prove that c\sin\frac{A-B}{2}=(a-b)\cos\frac{C}{2}
    I know that \cos\frac{C}{2}=\sin\frac{A+B}{2}
    but after this i'm lost.
    Thanks
    The relationship you've quoted is correct, but in fact you need to use a similar one; one that begins:
    \sin\tfrac12C = ... ...(1)
    I'll start the proof for you. See whether you can finish it.
    \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}

    \Rightarrow a -b = \frac{c}{\sin C}(\sin A - \sin B)
    =\frac{2c\cos\tfrac12(A+B)\sin\tfrac12(A-B)}{2\sin\tfrac12C\cos\tfrac12C}

    = ... ?
    Can you complete the relationship I started in (1), and so complete the proof?

    Grandad
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