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Math Help - Double-Angle Identity

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    Double-Angle Identity

    Find \sin \left( {\theta} \right), \cos \left( {\theta} \right), and \tan \left( {\theta} \right) given that \cot \left( {2\theta} \right) = -5/12; 2θ is in Quadrant II.

    I'm pretty sure solving this involves the double-angle (or maybe a half-angle) identity for tangent (the inverse of cotangent), but I'm not sure how to get \cot \left( {2\theta} \right) = -5/12 in terms of tangent. Please just help me get this started. I can solve for \sin \left( {\theta} \right) and \cos \left( {\theta} \right) if it were written in terms in terms of tangent. Thanks.
    Last edited by lightningstab714; March 18th 2009 at 07:00 PM.
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  2. #2
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    Quote Originally Posted by lightningstab714 View Post
    Find \sin \left( {\theta} \right), \cos \left( {\theta} \right), and \tan \left( {\theta} \right) given that \cot \left( {2\theta} \right) = -5/12; 2θ is in Quadrant II.

    I'm pretty sure solving this involves the double-angle (or maybe a half-angle) identity for tangent (the inverse of cotangent), but I'm not sure how to get \cot \left( {2\theta} \right) = -5/12 in terms of tangent. Please just help me get this started. I can solve for \sin \left( {\theta} \right) and \cos \left( {\theta} \right) if it were written in terms in terms of tangent. Thanks.
    \cot(2t) = -\frac{5}{12}

    \tan(2t) = -\frac{12}{5}

    \tan(2t) = \frac{2\tan{t}}{1 - \tan^2{t}}

    -\frac{12}{5} = \frac{2\tan{t}}{1 - \tan^2{t}}

    6\tan^2{t} - 5\tan{t} - 6 = 0

    (3\tan{t} + 2)(2\tan{t} - 3) = 0

    since \frac{\pi}{2} < 2t < \pi

    \frac{\pi}{4} < t < \frac{\pi}{2}

    \tan{t} = \frac{3}{2}
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