# trig identity..

• Aug 26th 2007, 09:32 AM
polymerase
trig identity..
Determine the exact values of cos(13pi/12).....
• Aug 26th 2007, 09:43 AM
ThePerfectHacker
Quote:

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.
• Aug 26th 2007, 09:44 AM
Jhevon
Quote:

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
• Aug 26th 2007, 12:21 PM
red_dog
$\displaystyle \displaystyle\cos\frac{13\pi}{12}=\cos\left(\pi+\f rac{\pi}{12}\right)=-\cos\frac{\pi}{12}=-\cos\left(\frac{\pi}{3}-\frac{\pi}{4}\right)$
Now use $\displaystyle \cos(a-b)=\cos a\cos b+\sin a\sin b$