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Thread: very difficult trig question

  1. #1
    Member
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    very difficult trig question

    If

    $\displaystyle \frac{ax}{\cos\theta}+\frac{by}{\sin\theta}=a^2-b^2$

    and

    $\displaystyle \frac{ax\sin\theta}{{\cos}^2\theta}-\frac{by\cos\theta}{{\sin}^2\theta}=0$

    then prove that

    $\displaystyle {(ax)}^{2/3} + {(by)}^{2/3} = {(a^2 - b^2)}^{2/3} $
    Last edited by mr fantastic; Jul 24th 2009 at 04:45 AM.
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  2. #2
    MHF Contributor
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    Hi

    Multiply the second equation by $\displaystyle \frac{\sin \theta}{\cos \theta}$ and add it to the first one : $\displaystyle \frac{ax}{\cos^3 \theta}=a^2-b^2$

    Multiply the second equation by $\displaystyle \frac{\cos \theta}{\sin \theta}$ and subtract it from the first one : $\displaystyle \frac{by}{\sin^3 \theta}=a^2-b^2$

    $\displaystyle (ax)^{2/3} = (a^2 - b^2)^{2/3}\cos^2 \theta$
    $\displaystyle (by)^{2/3} = (a^2 - b^2)^{2/3}\sin^2 \theta$

    Then $\displaystyle (ax)^{2/3} + (by)^{2/3} = (a^2 - b^2)^{2/3} $
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