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

1 Answer
Oct 14, 2016

#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#