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How do you evaluate #cos^2(pi/8) - sin^2(pi/8)#?

1 Answer

Apr 7, 2016

#sqrt2/2#

Explanation:

Apply the trig identity: #cos 2a = cos^2 a - sin^2 a#
#cos^2 (pi/8) - sin^2 (pi/8) = cos (pi/4)#
Trig table --> #cos (pi/4) = sqrt2/2#

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