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Math Help - Complex number - De Moivre

  1. #1
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    Complex number - De Moivre

    After proving that:

    <br />
tan(3x) = \frac{3tanx - tan^3(x)}{1-3tan^2(x)}<br />

    ...how would i go about HENCE solving:

    <br />
1 - 3t^2 = 3t - t^3<br />

    Even just a nudge in the right direction would be of much help - I can see a link but just can't bring the two together!
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  2. #2
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    1-3t^2=3t-t^3

    \implies t^3-3t^2-3t+1=0

    The rational root theorem suggests t=\pm1 as possible roots. Try dividing this polynomial by t-1 or t+1 to see if either is a factor. (One of them will be.)

    Edit: I realize now this isn't the intended way to solve the polynomial. Follow Archie Meade's method below. (But leaving this up in case you were curious about another way to solve such a problem.)
    Last edited by drumist; June 2nd 2010 at 02:02 PM.
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  3. #3
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    Quote Originally Posted by aceband View Post
    After proving that:

    <br />
tan(3x) = \frac{3tanx - tan^3(x)}{1-3tan^2(x)}<br />

    ...how would i go about HENCE solving:

    <br />
1 - 3t^2 = 3t - t^3<br />

    Even just a nudge in the right direction would be of much help - I can see a link but just can't bring the two together!
    \frac{3t-t^3}{1-3t^2}=1

    tan(3x)=\frac{3tanx-tan^3x}{1-3tan^2x}

    Use the substitution t=tanx

    tan(3x)=1\ \Rightarrow\ 3x=\frac{{\pi}}{4},\ \frac{5{\pi}}{4},\ \frac{9{\pi}}{4}

    x=\frac{{\pi}}{12},\ \frac{5{\pi}}{12},\ \frac{9{\pi}}{12}

    Then t=tanx
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