Question

The terminal side of `theta` in standard position contains (-8,-15), how do you find the exact values of the six trigonometric functions of `theta`?

Answer 1

Call t the arc (or angle). t lies in Quadrant 3
`tan t = y/x = -15/(-8) = 15/8`
`cos^2 t = 1/(1 + tan^2 t) = 1/(1 + 225/64) = 64/289`
`cos t = +- 8/17`
Since t lies in Quadrant 3, cos t is negative
`cos t = - 8/17`
`sin^2 t = 1 - cos^2 t = 1 - 64/289 = 225/289`
`sin t = +- 15/17`
Since t is in Quadrant 3, sin t is negative
`sin t = - 15/17`
`tan t = sin t/(cos t) = (-15/17)(-17/8) = 15/8`
`cot = 1/(tan) = 8/15`
`sec = 1/(cos) = - 17/8`
`csc t = 1/(sin) = - 17/15`