Question

How do you use the point (-12,-15) on the terminal side of the angle to evaluate the six trigonometric functions?

Answer 1

Call t the angle whose terminal side passes through point (-12/15).
t lies in Quadrant 3.
We have:
`tan t = y/x = -15/- 12 = 5/4`
`cos^2 t = 1/(1 + tan^2 t) = 1/(1 + 25/16) = 16/41`
`cos t = - 4/sqrt41` (t in Q. 3 --> cos t is negative)
`sin^2 t = 1 - cos^2 t = 1 - 16/41 = 25/41`
`sin t = - 5/sqrt41` (t lies in Q. 3 --> sin t is negative)
`tan t = 5/4`
`cot = 4/5`
`sec t = 1/(cos) = - sqrt41/4`
`csc t = - sqrt41/5`