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

1 Answer
Nghi N.
May 10, 2016

#- sqrt(2 - sqrt3)/2#

Explanation:

Use the trig identity: #cos 2a = 2cos^2 a - 1#
#cos 2a = cos ((14pi)/12) = cos ((7pi)/6) = cos (pi/6 + pi) = #
#-cos (pi/6) = - sqrt3/2#
#cos ((7pi)/6) = -sqrt3/2 = 2cos^2 ((7pi)/12) - 1#
#2cos^2 ((7pi)/12) = 1 - sqrt3/2 = (2 - sqrt3)/2#
#cos^2 ((7pi)/12) = (2 - sqrt3)/4#
#cos ((7pi)/12) = +- sqrt(2 - sqrt3)/2#
Since #(7pi)/12# is in Quadrant II, its cos is negative -->
#cos ((7pi)/12) = - sqrt(2 - sqrt3)/2#

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