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Math Help - Trig Identities Question

  1. #1
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    Trig Identities Question

    I'm trying to find cos \alpha. Earlier I determined using tan \alpha = \frac{1}{cot \alpha} that tan \alpha = \frac{1}{5}. This is correct, according to my book.

    Knowing this, can't I determine that because tan \alpha = \frac{sin \alpha}{cos \alpha} that cos \alpha = 5. However, this logic seems incorrect. The answer my book is looking for is \frac{5\sqrt{26}}{26}. Can someone help me understand why my way doesn't work?
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    Re: Trig Identities Question

    Quote Originally Posted by fogownz View Post
    I'm trying to find cos \alpha. Earlier I determined using tan \alpha = \frac{1}{cot \alpha} that tan \alpha = \frac{1}{5}. This is correct, according to my book.

    Knowing this, can't I determine that because tan \alpha = \frac{sin \alpha}{cos \alpha} that cos \alpha = 5. However, this logic seems incorrect. The answer my book is looking for is \frac{5\sqrt{26}}{26}. Can someone help me understand why my way doesn't work?

    because \cos{\alpha} \ne 5 ...

    -1 \le \cos(anything) \le 1

    using the identity ...

    \tan^2{\alpha} + 1 = \sec^2{\alpha}

    \frac{1}{25} + 1 = \sec^2{\alpha}

    \frac{26}{25} = \sec^2{\alpha}

    \frac{25}{26} = \cos^2{\alpha}

    \pm \frac{5}{\sqrt{26}} = \pm \frac{5\sqrt{26}}{26} = \cos{\alpha}


    you could also find \cos{\alpha} by sketching a reference triangle in quads I and/or IV
    Last edited by skeeter; November 15th 2012 at 12:41 PM.
    Thanks from fogownz
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