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.

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

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

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

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

$sec = 1/cos$ =

$csc = 1/sin$ =

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