How do you evaluate #Cos(pi/12) + Cos((5pi)/12)#?

1 Answer
Nghi N.
Apr 1, 2016

#(sqrt(2 + sqrt3)/2) + (sqrt(2 - sqrt3)/2)#

Explanation:

#S = cos (pi/12) + cos ((5pi)/12)#
#cos ((5pi)/12) = cos (-pi/2 + pi/2) = sin (pi/12).#
Reminder: #cos (pi/2 - a) = sin a# --> Complementary arcs.
Evaluate #sin (pi/12#) and #cos (pi/12)# by using the trig identity:
#cos 2a = 2cos^2 a - 1 = 1 - 2sin^2 a#.
a. #cos (pi/6) = sqrt3/2 = 2cos^2 (pi/12) - 1#
#2cos^2 (pi/12) = 1 + sqrt3/2 = (2 + sqrt3)/2#
#cos^2 (pi/12) = (2 + sqrt3)/4#
#cos (pi/12) = (sqrt(2 + sqrt3)/2) # --> #cos (pi/12)# is positive.
b. #cos (pi/6) = sqrt3/2 = 1 - 2sin^2 (pi/12)#
#2sin^2 (pi/12) = 1 - sqrt3/2 = (2 - sqrt3)/2#
#sin (pi/12) = (sqrt(2 - sqrt3)/2)# --> #sin (pi/12)# is positive.

#S = (sqrt(2 + sqrt3)/2) + (sqrt(2 - sqrt3)/2)#

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