How do you simplify #cos²(pi/12)-sin²(pi/12)#?

1 Answer
Nghi N.
Jul 19, 2015

Find #cos^2 (pi/12) - sin^2 (pi/12)#
Ans:# (sqrt3)/2#

Explanation:

Call #cos (pi/12) = cos t# --> #cos 2t = cos pi/6 = sqrt3/2#
Call #sin (pi/12) = sin t#
Use the trig identity: #cos 2t = 2cos^2t - 1# and #cos 2t = 1 - 2sin^2 t#
#cos 2t = sqrt3/2 = 2cos^2 t - 1 -> cos^2 t = (2 + sqrt3)/4#

#cos 2t = sqrt3/2 = 1 - 2sin^2 t -> sin^2 t = (2 - sqrt3)/4#

#cos^2 t - sin^2 t = (2 + sqrt3 - 2 + sqrt3)/4 = (2sqrt3)/4 = sqrt3/2#

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