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Math Help - What's wrong with my proof?

  1. #1
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    What's wrong with my proof?

    Prove \cos A + \cos B + \cos C -1=4\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} when A+B+C=180^\circ

    \cos A + \cos B + \cos C -1= 2\cos \frac{A+B}{2}\cos \frac{A-B}{2}-2\sin^2\frac{C}{2}

    =2\sin\frac {C}{2}\cos \frac {A-B}{2}-2\sin^2 \frac{C}{2}

    2\sin\frac {C}{2} ( \cos \frac{A-B}{2}-\sin \frac {C}{2} )

    2\sin\frac {C}{2} ( \cos \frac{A-B}{2}-\cos \frac{A+B}{2})

    2\sin \frac{C}{2}(-2\sin\frac {A}{2}\sin \frac {B}{2})

    -4\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}

    I had the same steps up towards 2\sin\frac {C}{2} ( \cos \frac{A-B}{2}-\cos \frac{A+B}{2}), but then the negative sign just disappeared from the next step. Isn't the formula for \cos A-\cos B=-2\sin \frac{A+B}{2}\sin \frac{A-B}{2}?
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  2. #2
    Senior Member eumyang's Avatar
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    From here:
    = 2\sin\frac {C}{2} ( \cos \frac{A-B}{2}-\cos \frac{A+B}{2})

    to here:
    = 2\sin \frac{C}{2}(-2\sin\frac {A}{2}\sin \frac {B}{2})

    is the problem, I think. That final angle of B/2 should be negative:
    = 2\sin \frac{C}{2}\left(-2\sin\frac {A}{2}\sin \left( -\frac {B}{2} \right) \right)

    And of course, sine being an odd function, you bring the negative out:
    = 2\sin \frac{C}{2}\left(2\sin\frac {A}{2}\sin \frac {B}{2}\right)

    = 4\sin \frac{A}{2} \sin\frac {B}{2} \sin \frac {C}{2}
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