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Thread: Half Angle Formula problem

  1. #1
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    Half Angle Formula problem

    Given:

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


    with 90 < x < 180.

    Find:

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

    So I actually did this:

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

    $\displaystyle (6) - (1) \over 12 $

    And simplified to:

    $\displaystyle 1 \over 2 $ $\displaystyle \sqrt 15 $

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

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


    with 90 < x < 180.

    Find:

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

    So I actually did this:

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

    $\displaystyle (6) - (1) \over 12 $

    And simplified to:

    $\displaystyle 1 \over 2 $ $\displaystyle \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 $\displaystyle \cos (x) = - \frac{1}{6}$ where $\displaystyle 90^0 < x < 180^0$ and asked to find $\displaystyle \cos\left( \frac{1}{2} x\right)$.

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

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

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

    Now rationalise the denominator. The correct option becomes obvious.
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