How do you use the point #(1,-sqrt3)# on the terminal side of the angle to evaluate the six trigonometric functions?

1 Answer
Nghi N
Jul 8, 2017

Call t the angle (arc)
#tan t = y/x = - sqrt3/1 = - sqrt3#.
t is in Quadrant 4.
Trig table and unit circle give -->
#t = (2pi)/3 + pi = (5pi)/3#
#sin t = sin (5pi)/3 = sin (2pi - pi/3) = sin (-pi/3) = = - sin (pi/3) = - sqrt3/2#
#cos t = cos (- pi/3) = cos pi/3 = 1/2#
#cot t = 1/(tan) = - 1/sart3 = - sqrt3/3#
#sec t = 1/(cos) = 2#
#csc t = 1/(sin) = - 2/sqrt3 = - (2sqrt3)/3#

Related questions
See all questions in Trigonometric Functions of Any Angle
Impact of this question
2584 views around the world
You can reuse this answer
Creative Commons License