How do you sketch the angle whose terminal side in standard position passes through #(-2,-3sqrt5)# and how do you find sin and cos?

1 Answer
May 26, 2018

#sin t = - 3sqrt5/7#
#cos t = - 2/7#

Explanation:

Call t the angle whose terminal side passes through
point #(-2, - 3sqrt5)# --> t's terminal side lies in Quadrant 3.
#tan t = y/x = (-3sqrt5)/(-2) = (3sqrt5)/2#
#cos^2 t = 1/(1 + tan^2 t) = 1/(1 + 45/4) = 4/49#
#cos t = - 2/7# (because t lies in Quadrant 3 --> cos t is negative)
#sin^2 t = 1 - cos^2 t = 1 - 4/49 = 45/49#
#sin t = - (3sqrt5)/7# (because t lies in Q.3 --> sin t is negative)