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Math Help - trig proof

  1. #1
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    trig proof

    In tirangle ABC, angle C is  90^o . Show that

     sin\frac{1}{2}A = \sqrt{\frac{c-b}{2c}}

    I think I am suppose to be using the half angle formula;  sin^{2}\frac{A}{2} = \frac{1+cosA}{2}

    I did draw a right angle triangle, labelling the hypotenuse c and got;  sinA = \frac{b}{c}

      cosA = \frac{a}{c}

    Using the half angle formula, wouldn't I just sub in what cosA = a\c, but this does not work.

    Can some point me to the right direction?

    Thanks.
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  2. #2
    MHF Contributor red_dog's Avatar
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    You have made two mistakes:

    I think I am suppose to be using the half angle formula;
    and
    .

    In fact we have \sin^2\frac{A}{2}=\frac{1-\cos A}{2}

    and \cos A=\frac{b}{c}
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  3. #3
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    oh right, thanks.


    Can you also tell me why my labelling of the right angled triangle is wrong?

    I labelled the other two angles A and B and the sides opposite them a , b .
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  4. #4
    MHF Contributor red_dog's Avatar
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    Your labelling is right.

    But \cos A=\frac{adjacente \ side}{hypotenuse}=\frac{b}{c}
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