Question

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

Answer 1

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`