Question #2af88

1 Answer
Feb 13, 2018

#cos t = - 5/6#
#cos 2t = 2cos^2 t - 1 = 2(25/36) - 1 = 14/36 = 7/18#
#sin^2 2t = 1 - cos^2 2t = 1 - 49/324 = 275/324 #
#sin 2t = +- (5sqrt11)/18#
Since t is in Quadrant 3, 2t is in Quadrant 1, so, sin 2t is positive
#sin 2t = (5sqrt11)/18#
Use half angle formula -->
#sin (t/2) = +- sqrt((1 - cos t)/2)#
#sin (t/2) = +- sqrt((1 + 5/6)/2) = +- sqrt11/sqrt12 #
Because t is in Quadrant 3, then, #t/2# is in Quadrant 2, so,
#sin (t/2)# is positive
#sin (t/2) = sqrt11/(2sqrt3)#
Use trig identity:
#cos (t/2) = +- sqrt((1 + cos t)/2)#
#cos (t/2) = +- sqrt((1 - 5/6)/2) = +- 1/sqrt12 = +- 1/(2sqrt3)#
Since #t/2# is in Quadrant 2, #cos (t/2)# is negative
#cos (t/2) = - sqrt3/6#