How do you find the exact values of #costheta# and #sintheta# when #tantheta=-1/2#?

1 Answer
Dec 21, 2016

Answer:

#sin t = +- sqrt5/5#
#cos t = +- (2sqrt5)/5#

Explanation:

Use trig identity:
#cos^2 x = 1/(1 + tan^2 x)#
#cos^2 t = 1/(1 + 1/4) = 1/(5/4) = (4/5)#
#cos t = +- 2/sqrt5 = +- (2sqrt5)/5#
#sin ^2 t = 1 - cos^2 t = 1 - 4/5 = (5 - 4)/5 = 1/5#
#sin t = +- 1/sqrt5 = +- sqrt5/5#
Because tan #t = - 1/2#, then t is in Quadrant II, or in Quadrant IV.
sin t and cos t in these 2 Quadrants have opposite signs.
a. If t is in Quadrant II --> #sin t = sqrt5/5#, and #cos t = - (2sqrt5)/5#, and #tan t = - 1/2#
b. If t is in Quadrant IV --> #sin t = - sqrt5/5#, and #cos t = (2sqrt5)/5#, and #tan t = - 1/2#.