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Math Help - Half Angle Formula problem

  1. #1
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    Half Angle Formula problem

    Given:

    (-1) \over (6) =  \cos x


    with 90 < x < 180.

    Find:

    \cos (x) =  ( -1 ) \over (2) X

    So I actually did this:

     \cos^2 (X)  =  (1) + (-1/6) \over 2

     (6) - (1)  \over 12

    And simplified to:

     1 \over 2  \sqrt 15

    But this is not an answer choice. I feel like I have missed a step. Thank you.
    Last edited by DaGr8Gatzby; November 30th 2009 at 06:23 PM.
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  2. #2
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    Quote Originally Posted by DaGr8Gatzby View Post
    Given:

    (-1) \over (10) =  \cos x


    with 90 < x < 180.

    Find:

    \cos (x) =  ( -1 ) \over (2) X

    So I actually did this:

     \cos^2 (X)  =  (1) + (-1/6) \over 2

     (6) - (1) \over 12

    And simplified to:

     1 \over 2  \sqrt 15

    But this is not an answer choice. I feel like I have missed a step. Thank you.
    This makes no sense.
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  3. #3
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    Edited: -1/6
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  4. #4
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    Quote Originally Posted by DaGr8Gatzby View Post
    Edited: -1/6
    It still makes no sense. What's X and what's meant to be done with X?
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  5. #5
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    Maybe a screenshot would help:

    http://img694.imageshack.us/img694/9...91130at824.png

    Still stumped on this one.
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  6. #6
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    Quote Originally Posted by DaGr8Gatzby View Post
    Maybe a screenshot would help:

    http://img694.imageshack.us/img694/9...91130at824.png

    Still stumped on this one.
    You are given \cos (x) = - \frac{1}{6} where 90^0 < x < 180^0 and asked to find \cos\left( \frac{1}{2} x\right).

    From the double angle formula: \cos (x) = 2 \cos^2 \left( \frac{1}{2} x\right) - 1.

    Substitute \cos (x) = - \frac{1}{6} and simplify: \cos^2\left( \frac{1}{2} x\right) = \frac{5}{12}.

    90^0 < x < 180^0 \Rightarrow 45^0 < \frac{1}{2} x < 90^0 and so the positive root is required: \cos \left( \frac{1}{2} x \right) = \frac{\sqrt{5}}{\sqrt{12}} = \frac{\sqrt{5}}{2 \sqrt{3}}.

    Now rationalise the denominator. The correct option becomes obvious.
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