What is #cos((-9pi)/4)#?

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Answer 1 · 2017-05-22T15:46:59

#cos(-9/4pi)= sqrt2/2#

#cos(-9/4pi) = cos((-2pi)-1/4pi)#

#=cos(-2pi)cos(1/4pi) + sin(-2pi)sin(1/4pi)#

#=(1)(sqrt2/2) + (0)(sqrt2/2)#

#= sqrt2/2#

Answer 2 · 2017-05-22T17:38:10

#sqrt2/2#

#cos ((-9pi)/4) = cos (-pi/4 - 2pi) = cos (-pi/4) = cos (pi/4)#
Trig table gives:
#cos ((-9pi)/4) = cos (pi/4) = sqrt2/2#