How do you evaluate #\cos ( - \frac { 2\pi } { 3} ) + \sin ( - \frac { 4\pi } { 3} )#?

1 Answer
Nghi N
Jun 21, 2017

#(-1 + sqrt3)/2#

Explanation:

trig table gives:
#cos ((-2pi)/3) = cos ((2pi)/3) = - 1/2#
#sin ((-4pi)/3) = - sin ((4pi)/3) = - 2sin ((2pi)/3)cos ((2pi)/3) =#
#= - 2((sqrt3)/2).(-1/2) = sqrt3/2#
Finally,
#cos ((-2pi)/3) + sin ((-4pi)/3) = - 1/2 + sqrt3/2 = (-1 + sqrt3)/2#

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