How do you find the exact value of #Cos ((5pi)/12)+cos((2pi)/3-pi/4)#?

1 Answer
Apr 5, 2016

#sqrt(2 - sqrt3)#

Explanation:

#S = cos ((5pi)/12) + cos ((2pi)/3 - pi/4)#
#cos ((2pi)/3 - pi/4) = cos ((8pi)/12 - ((3pi)/12) = cos ((5pi)/12)#
Trig unit circle and property of complementary arcs give -->
#cos ((5pi)/12) = cos (pi/2 - pi/12) = sin (pi/12)#
Therefor,
#S = 2sin (pi/12)#
Evaluate #sin (pi/12)# by the trig identity: #cos 2a = 1 - 2sin^2 a#
#cos (pi/6) = sqrt3/2 = 1 - 2sin^2 (pi/12)#
#2sin^2 (pi/12) = 1 - sqrt3/2 = (2 - sqrt3)/2#
#sin^2 (pi/12) = (2 - sqrt3)/4#
#sin (pi/12) = sqrt(2 - sqrt3)/2# --> #sin (pi)/12# is positive
#S = 2sin (pi/12) = sqrt((2 - sqrt3)#