#cos ((3pi)/4 - pi/3) = cos ((9pi - 4pi)/12) = cos ((5pi)/12)#
Find #cos ((5pi)/12)# by using trig identity: #2cos^2 a = 1 + cos 2a#
In this case, trig table gives: #cos 2a = cos ((10pi)/12) = cos (5pi/6) = - sqrt3/2#
Call #cos ((5pi)/12) = cos t#, we get: #2cos^2 t = 1 - sqrt3/2 = (2 - sqrt3)/2# #cos^2 t = (2 - sqrt3)/4# #cos t = +- sqrt(2 - sqrt3)/2#
Since #(5pi)/12# is in Quadrant 1, its cos is positive --> #cos ((5pi)/12) = cos t = sqrt(2 - sqrt3)/2#