If Cos(t)=-7/8 with #pi<=t<=3pi/2#, how do you find the value of sin(t/2)?

1 Answer
Mar 24, 2016

Answer:

#sqrt15/4#

Explanation:

Apply the trig identity: #cos 2t = 1 - 2sin^2 t# -->
#cos t = -7/8 = 1 - 2sin^2 (t/2) #
#2sin^2 (t/2) = 1 + 7/8 = 15/8#
#sin^2 (t/2) = 15/16#
#sin (t/2) = +- sqrt15/4#
t is in Quadrant III, then, t/2 is in Quadrant II. Its sin is positive.
Therefor, #sin (t/2) = sqrt15/4#
Check by calculator
#cos t = -7/8 = --0.875# --> #t = 151^@04#
#t/2 = 75^@52# --> #sin (t/2) = sin 75.52 = 0.968 = sqrt15/4#. OK