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Math Help - Natural log and trig exponents

  1. #1
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    Natural log and trig exponents

    I'm h
    Last edited by chris520; March 7th 2011 at 05:13 PM.
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  2. #2
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    Is this what you're trying to solve?

    \displaystyle \left(\frac{81}{256}\right)^{\sin^2{\theta}} + \left(\frac{81}{256}\right)^{\cos^2{\theta}} = \frac{75}{64}?

    If so...

    \displaystyle \left(\frac{81}{256}\right)^{1 - \cos^2{\theta}} + \left(\frac{81}{256}\right)^{\cos^2{\theta}} = \frac{75}{64}

    \displaystyle \frac{\frac{81}{256}}{\left(\frac{81}{256}\right)^  {\cos^2{\theta}}} + \left(\frac{81}{256}\right)^{\cos^2{\theta}} = \frac{75}{64}

    \displaystyle \frac{81}{256} + \left[\left(\frac{81}{256}\right)^{\cos^2{\theta}}\right]^2 = \frac{75}{64}\left(\frac{81}{256}\right)^{\cos^2{\  theta}}

    \displaystyle \left[\left(\frac{81}{256}\right)^{\cos^2{\theta}}\right]^2 - \frac{75}{64}\left(\frac{81}{256}\right)^{\cos^2{\  theta}} + \frac{81}{256} = 0.


    Now let \displaystyle x = \left(\frac{81}{256}\right)^{\cos^2{\theta}} so that we can write the equation as

    \displaystyle x^2 - \frac{75}{64}x + \frac{81}{256} = 0

    which is a quadratic equation you can solve for \displaystyle x, the solution of which you can use to solve for \displaystyle \theta.
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  3. #3
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    Last edited by chris520; March 7th 2011 at 05:14 PM.
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