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Math Help - sin(-)-6cos(-)+3tan(2)

  1. #1
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    Question sin(-)-6cos(-)+3tan(2)

    Edit: Added some tex tags
    This is a question on a review I am doing, my answer doesn't match up with the answer key and I am looking for some input on what I am doing wrong.

    If cos\theta = \frac{1}{6} with \theta in quadrant IV find:

    sin(-\theta)-6cos(-\theta)+3tan(2\theta)

    I solved for sin\theta = \frac{-\sqrt{35}}{6} cos\theta = \frac{1}{6} and  tan\theta = -\sqrt{35}

    Using negative angle identities and the double angle identity I end up with:

    -sin\theta-6cos\theta+3(\frac{2tan\theta}{1-tan^2\theta})

    Plugging in my values gives me:

    \frac{\sqrt{35}}{6} - \frac{6}{6} + 3[\frac{2(-\sqrt{35})}{1-35}]

    Simplified to:

    \frac{35(\sqrt{35}) -102}{102}

    The answer key gives: \frac{42+25\sqrt{35}}{42}

    Not sure where I am messing up =\ any help is much appreciated.
    Last edited by hermeticcharm; February 26th 2013 at 09:57 AM.
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: sin(-)-6cos(-)+3tan(2)

    I believe your answer is correct:

    Given  \cos \phi = \frac 1 6 in the fourth quadrant then  \phi = -1.4 radians = -80.45 degrees.

     \sin(- \phi) - 6 \cos( - \phi) + 3 \tan (2 \phi) = 0.986 - 1+ 1.044 = 1.030

    Your answer matches:

    \frac { 35 \sqrt {35} - 102}{102} = 1.030

    but their answer is:

     \frac {42+ 25 \sqrt{35}}{42} = 4.521

    **EDIT
    I think I may have found where whoever came up with the answer key made mistakes: if (a) they used  + 6\cos(\- \phi) instead of  -6 \cos (- \phi), and also (b) instead of simplifiying the  \tan (2 \phi) term to  \frac { 3 \sqrt{35}}{17} as you have it they made it  \frac { 3 \sqrt {35}}{7}, that would explain it.
    Last edited by ebaines; February 26th 2013 at 10:27 AM.
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  3. #3
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    Re: sin(-)-6cos(-)+3tan(2)

    Thanks for the quick reply ebaines! That is what I was starting to think. I always assume my side is where the problem is but I am slowly learning to not always trust the given answers.
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