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Math Help - trig identity..

  1. #1
    Senior Member polymerase's Avatar
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    trig identity..

    Determine the exact values of cos(13pi/12).....
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    Quote Originally Posted by polymerase View Post
    Determine the exact values of cos(13pi/12).....
    \cos \left( \frac{13\pi}{12} \right) = \cos \left( \frac{13\pi}{12} - \pi \right) = -\cos \frac{\pi}{12}
    HINT: Use half-angle formula.

    EDIT: Mistake fixed.
    Last edited by ThePerfectHacker; August 26th 2007 at 11:05 AM.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by polymerase View Post
    Determine the exact values of cos(13pi/12).....
    \cos \left( \frac {13 \pi}{12} \right) = \cos \left( \frac {1}{2} \cdot \frac {13 \pi}{6} \right)

    Now use the half angle formula for cosine. Note that \frac {13 \pi}{12} is in the third quadrant, so the answer is negative
    Last edited by Jhevon; August 26th 2007 at 10:02 AM. Reason: silly mistake. i said the angle was in the wrong quadrant, fixed it
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  4. #4
    MHF Contributor red_dog's Avatar
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    \displaystyle\cos\frac{13\pi}{12}=\cos\left(\pi+\f  rac{\pi}{12}\right)=-\cos\frac{\pi}{12}=-\cos\left(\frac{\pi}{3}-\frac{\pi}{4}\right)
    Now use \cos(a-b)=\cos a\cos b+\sin a\sin b
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