How do you evaluate #cos[ (5pi) / 8 ]#?

1 Answer
Nghi N.
Jan 8, 2016

#cos ((5pi)/8) = - (sqrt(2 - sqrt2))/2#

Explanation:

#cos ((5pi)/8) = cos ((-3pi)/8 + pi) = -cos ((3pi)/8)#
Find #cos ((3pi)/8)#. Call cos x = #cos ((3pi)/8)#
#cos 2x = cos ((3pi)/4) = -sqrt2/2#
Apply the trig identity: #cos 2x = 2cos^2 x - 1#. We get:
#2cos^2 x = 1 - sqrt2/2 = (2 - sqrt2)/2#
#cos^2 x = (2 - sqrt2)/4#
#cos x = cos ((3pi)/8) = +- (sqrt(2 - sqrt2))/2#
Finally,
#cos ((5pi)/8) = - cos ((3pi)/8) = - (sqrt(2 - sqrt2)/2)#

Check by calculator.
#cos ((5pi)/8) = cos 112^@5 = - 0.38#
#-(sqrt(2 - sqrt2)/2) = 0.768/2 = - 0.38#. OK

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