How would you find the exact value of the six trigonometric function of 5pi/3?

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How would you find the exact value of the six trigonometric function of 5pi/3?

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Answer 1 · 2016-07-17T04:21:40

arc $(5pi)/3$ is in Quadrant IV.
$sin ((5pi)/3) = sin ((-pi/3 + (6pi)/3) = sin (-pi/3 + 2pi) =$
$= sin (-pi/3) = - sin (pi/3) = - sqrt3/2$
$cos ((5pi)/3) = cos (-pi/3 + 2pi) = cos (-pi/3) = cos (pi/3) = 1/2$
$tan ((5pi)/3) = (sin)/(cos) = -(sqrt3/2)(2/1) = -sqrt3$
$cot ((5pi)/3) = 1/(tan) = - 1/sqrt3 = - sqrt3/3$
$sec ((5pi)/3) = 1/(cos) = 2$
$csc ((5pi)/3) = 1/(sin) = -2/sqrt3 = - (2sqrt3)/3$

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How would you find the exact value of the six trigonometric function of 5pi/3?

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