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Math Help - Some trig help please

  1. #1
    Junior Member rednest's Avatar
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    Cool Some trig help please

    Given that z=cos\theta + isin\theta, show that

     <br />
\frac {1}{1-z} = \frac {1}{2} + \frac {i}{2} cot \frac {\theta}{2}<br />
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  2. #2
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    Quote Originally Posted by rednest View Post
    Given that z=cos\theta + isin\theta, show that

     <br />
\frac {1}{1-z} = \frac {1}{2} + \frac {i}{2} cot \frac {\theta}{2}<br />
    Sub in c+si for z
    make the denominator real
    simplify

    You'll need a trig identity...

    \frac{\sin\theta}{1-\cos\theta}=\cot\left(\frac{\theta}{2}\right)
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  3. #3
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    Quote Originally Posted by rednest View Post
    Given that z=cos\theta + isin\theta, show that

     <br />
\frac {1}{1-z} = \frac {1}{2} + \frac {i}{2} cot \frac {\theta}{2}<br />
    \frac{1}{(1 - \cos \theta - i \sin \theta)} \times \frac{(1 - \cos \theta + i \sin \theta)}{(1 - \cos \theta + i \sin \theta)} = \frac{1 - \cos \theta + i \sin \theta}{(1 - \cos \theta)^2 + \sin^2 \theta}



    = \frac{1 - \cos \theta + i \sin \theta}{2(1 - \cos \theta)}.


    Seeing the real part of this is equal to 1/2 is trivial.

    Your job is to now prove that \frac{\sin \theta}{1 -  \cos \theta}= \cot \frac {\theta}{2} ......
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