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

1 Answer
Jun 11, 2018

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 = -8/5tant=yx=85
cos^2 t = 1/(1 + tan^2 t) + 1/(1 + 64/25) = 25/89cos2t=11+tan2t+11+6425=2589
cos t = 5/sqrt89cost=589 (because t lies in Q. 4)
sin^2 t = 1 - cos^2 t = 1 - 25/89 = 64/89sin2t=1cos2t=12589=6489
sin t = - 8/sqrt89sint=889 (because t lies in Q.4)
tan t = - 8/5tant=85
cot = 1/(tan) = - 5/8cot=1tan=58
sec t = 1/(cos) = sqrt89/5sect=1cos=895
csc t = 1/(sin) = - sqrt89/8csct=1sin=898