Question

How do you use the point (-1,5) on the terminal side of the angle to evaluate the six trigonometric functions?

Answer 1

Call t the angle that is located in Quadrant 2, with
`tan t = y/x = 5/-1 = -5`
`cos ^2 t = 1/(1 + tan^2 t) = 1/(1 + 25) = 1/26`
`cos t = - 1/(sqrt26) = - sqrt26/26`
(because t is in Quadrant 2, cos t < 0)
`sin^2 t = 1/(1 + cot^2 t) = 1/(1 + 1/25) = 25/26`
`sin t = + 5/sqrt26` (t in Q. 2 --> sin t > 0)
`tan t = sin/(cos) = (5/(sqrt26))((- sqrt26)/1) = - 5`OK
`sec t = 1/(cos) = - sqrt26`
`csc t = 1/(sin) = sqrt26/5`