How do you evaluate #cos ((3pi)/8)#?

1 Answer
May 8, 2016

#sqrt(2 - sqrt2)/2#

Explanation:

Property of complementary arcs -->
#cos ((3pi)/8) = cos (-pi/8 + pi/2) = sin (pi/8).#
Evaluate #sin (pi/8)# by applying the trig identity:
#cos 2a = 1 - 2sin^2 a.#
Trig table -->
#cos (pi/4) = sqrt2/2 = 1 - 2sin^2 (pi/8)#.
#2sin^2 (pi/8) = 1 - sqrtt2/2 = (2 - sqrt2)/2#
#sin^2 (pi/8) = (2 - sqrt20/4#
#sin (pi/8) = sqrt(2 - sqrt2)/2#
Since #pi/8# is in Quadrant I, then we select the positive value.
#cos ((3pi)/8) = sin (pi/8) = sqrt(2 - sqrt2)/2#