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Math Help - Concerning one step in Trig substitution

  1. #1
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    Concerning one step in Trig substitution

    How does sqrt(9-9(sin theta)^2 = sqrt(9(cos theta)^2) ?

    Thanks!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by C.C. View Post
    How does sqrt(9-9(sin theta)^2 = sqrt(9(cos theta)^2) ?

    Thanks!
    You have some mismatched delimiters as WolframAlpha would say. I believe the question at hand is 9-9\sin^2(\theta)=9\left(1-\sin^2(\theta)\right)=9\cos^2(\theta)
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  3. #3
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     \sqrt{9 - 9 sin^2(\theta) } = \sqrt{9 * (1 - sin^2(\theta) }

    Recall that

     cos^2 (\theta) + sin^2(\theta) = 1

    Manipulating,

     cos^2(\theta) = 1 - sin^2 (\theta)

    Using this property, we see that

     \sqrt{9 * ( 1 - sin^2(\theta)} = \sqrt{9 * cos^2(\theta)}

    Which neatly reduces to

     +/- 3 * cos(\theta)
    Last edited by Math Major; December 1st 2009 at 03:59 PM. Reason: Corrected
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  4. #4
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    Quote Originally Posted by Math Major View Post
    [tex]  \sqrt{9 * ( 1 - sin^2(\theta)} = \sqrt{9 * cos^2(\theta)}

    Which neatly reduces to

     3 * cos(\theta)
    No it does not. That is false.
    What about \theta=\frac{3\pi}{4}.
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