Question

How do you use the point (-6,-8) on the terminal side of the angle to evaluate the six trigonometric functions?

Answer 1

`sin(x) = -4/5`
`csc(x) = -5/4`
`cos(x) = -3/5`
`sec(x) = -5/3`
`tan(x) = 4/3`
`cot(x) = 3/4`

Explanation:

let o represent length of the opposite side:

`o = -8`

let a represent the length of the adjacent side:

`a = -6`

The length of the hypotenuse is:

`h = sqrt((-6)^2 + (-8)^2)`

`h = 10`

`sin(x) = o/h = -8/10 = -4/5`
`csc(x) = 1/sin(x) = -5/4`
`cos(x) = a/h = -6/10 = -3/5`
`sec(x) = 1/cos(x) = -5/3`
`tan(x) = o/a = (-8)/-6 = 4/3`
`cot(x) = 1/tan(x) = 3/4`