How do you use # sin(pi/6)=1/2# to find cos(-pi/12)?

1 Answer
Nghi N.
Mar 2, 2016

#cos (-pi/12) = sqrt(2 + sqrt3)/2#

Explanation:

#cos (-pi/12) = cos (pi/12)#
#sin (pi/6) = 1/2# -->
#cos^2 (pi/6) = 1 - sin^2 (pi/6) = 1 - 1/4 = 3/4# -->
#cos (pi/6) = sqrt3/2.# (cos #(pi/6)# is positive)
Use the identity: #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# (since cos (pi/12) is positive)

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