Determine the exact values of cos(13pi/12).....
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Originally Posted by polymerase Determine the exact values of cos(13pi/12)..... $\displaystyle \cos \left( \frac{13\pi}{12} \right) = \cos \left( \frac{13\pi}{12} - \pi \right) = -\cos \frac{\pi}{12}$ HINT: Use half-angle formula. EDIT: Mistake fixed.
Last edited by ThePerfectHacker; Aug 26th 2007 at 11:05 AM.
Originally Posted by polymerase Determine the exact values of cos(13pi/12)..... $\displaystyle \cos \left( \frac {13 \pi}{12} \right) = \cos \left( \frac {1}{2} \cdot \frac {13 \pi}{6} \right)$ Now use the half angle formula for cosine. Note that $\displaystyle \frac {13 \pi}{12}$ is in the third quadrant, so the answer is negative
Last edited by Jhevon; Aug 26th 2007 at 10:02 AM. Reason: silly mistake. i said the angle was in the wrong quadrant, fixed it
$\displaystyle \displaystyle\cos\frac{13\pi}{12}=\cos\left(\pi+\f rac{\pi}{12}\right)=-\cos\frac{\pi}{12}=-\cos\left(\frac{\pi}{3}-\frac{\pi}{4}\right)$ Now use $\displaystyle \cos(a-b)=\cos a\cos b+\sin a\sin b$
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