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

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

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Answer 1 · 2016-04-06T23:43:23

#sqrt(2 + sqrt3)/2#

Explanation:

Apply the trig 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# --> #cos (pi/12)# is positive.

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

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