Question

How would you find the exact value of the six trigonometric function of 330 degrees?

Answer 1

Use trig table of special arcs and unit circle -->
`sin 330 = sin (-30 + 360) = sin (-30) = - sin 30 = - 1/2`
`cos (330) = cos (-30 + 360) = cos (- 30) = cos 30 = sqrt3/2`
`tan 330 = sin/(cos) = (-1/2)(2/sqrt3) = - 1/sqrt3 = - sqrt3/3`
`cot 330 = 1/(tan) = -sqrt3`
`sec 330 = 1/(cos) = 2/sqrt3 = (2sqrt3)/3`
`csc 330 = 1/(sin) = - 2`