How do you find the six trigonometric functions of #(-2pi)/3# degrees?

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Answer 1 · 2015-05-25T22:37:50

Use the trig unit circle as proof.

#sin ((-2pi)/3) = - sin (pi/3) = -(sqrt3)/2#

#cos (-(2pi)/3) = - cos ((pi)/3) = -1/2#

#tan ((-2pi)/3) = (sin/cos) = sqrt3#

#cot ((-2pi)/3) = 1/(sqrt3) = (sqrt3)/3#

#sec = 1/cos # =

#csc = 1/sin # =