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Math Help - Trig Problem

  1. #1
    Newbie
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    Oct 2009
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    Trig Problem

    Looking for someone to verify my work ....

    If \sin\theta=\frac{1}{3} and \frac{\pi}{2}<\theta<\pi then what is \cos\theta?

    Solution

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

    \frac{1}{3}^2+\cos^2\theta=1

    \cos^2\theta=1-\frac{1}{9}

    \cos^2\theta=\frac{8}{9}

    \cos\theta=\frac{\sqrt{8}}{3}

    \cos\theta=\frac{2\sqrt{2}}{3}



    Now because \frac{\pi}{2}<\theta<\pi then my answer should read \cos\theta=\frac{-2\sqrt{2}}{3}



    I assume that I have completed everything properly.
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  2. #2
    Senior Member I-Think's Avatar
    Joined
    Apr 2009
    Posts
    288
    Yes my good sir, you are correct.

    And just to help you along, here's an alternative method to do the question.

    If sin\theta=\frac{1}{3}, then you may construct the right-angled \Delta with that ratio.

    |\
    | \
    1 | \ 3
    | \
    |____\
    \frac{2\sqrt{2}}{3}

    So cos\theta=\frac{2\sqrt{2}}{3}}.

    But since the desired range is \frac{\pi}{2}<\theta<\pi
    use the identity cos(180-\theta)=-cos\theta, to get the desired result

    Forgive my very shoddy triangle please. It's almost 11p.m.
    Last edited by I-Think; November 24th 2009 at 05:55 PM.
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