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Thread: What's wrong with my proof?

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

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

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

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

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

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

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

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

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

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

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

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

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