How do you evaluate cos ((7pi)/12)cos(7π12)?

1 Answer
Nghi N.
May 10, 2016

- sqrt(2 - sqrt3)/2−√2−√32

Explanation:

Use the trig identity: cos 2a = 2cos^2 a - 1cos2a=2cos2a−1
cos 2a = cos ((14pi)/12) = cos ((7pi)/6) = cos (pi/6 + pi) = cos2a=cos(14π12)=cos(7π6)=cos(π6+π)=
-cos (pi/6) = - sqrt3/2−cos(π6)=−√32
cos ((7pi)/6) = -sqrt3/2 = 2cos^2 ((7pi)/12) - 1cos(7π6)=−√32=2cos2(7π12)−1
2cos^2 ((7pi)/12) = 1 - sqrt3/2 = (2 - sqrt3)/22cos2(7π12)=1−√32=2−√32
cos^2 ((7pi)/12) = (2 - sqrt3)/4cos2(7π12)=2−√34
cos ((7pi)/12) = +- sqrt(2 - sqrt3)/2cos(7π12)=±√2−√32
Since (7pi)/127π12 is in Quadrant II, its cos is negative -->
cos ((7pi)/12) = - sqrt(2 - sqrt3)/2cos(7π12)=−√2−√32

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