How do you sketch the angle whose terminal side in standard position passes through #(sqrt5,-sqrt15)# and how do you find sin and cos?

1 Answer
Aug 25, 2016

#sin a = - sqrt3/2#
#cos a = 1/2#

Explanation:

Call M the point #M(sqrt5, -sqrt15)#
M is in Quadrant IV. On the trig unit circle, M is the terminal point of the arc a, that #tan a = - sqrt15/sqrt5 = - sqrt3#
Trig table gives --> #a = - pi/3 or a = (5pi)/3# (co -terminal arcs).

#sin a = sin (-pi/3) = - sin (pi/3) = - sqrt3/2#
#cos a = cos (-pi/3) = cos (pi/3) = 1/2#