Question

How do you find the exact values of the six trigonometric function of `theta` if the terminal side of `theta` in the standard position contains the point (5,-8)?

Answer 1

The point (x = 5, y = -8), on the terminal side of the angle, lies in
Quadrant 4.
Call t the angle (or arc), we get:
`tan t = y/x = -8/5`
`cos^2 t = 1/(1 + tan^2 t) + 1/(1 + 64/25) = 25/89`
`cos t = 5/sqrt89` (because t lies in Q. 4)
`sin^2 t = 1 - cos^2 t = 1 - 25/89 = 64/89`
`sin t = - 8/sqrt89` (because t lies in Q.4)
`tan t = - 8/5`
`cot = 1/(tan) = - 5/8`
`sec t = 1/(cos) = sqrt89/5`
`csc t = 1/(sin) = - sqrt89/8`