#sin x = - 3/5#. First, find cos x.
#cos^2 x = 1 - sin^2 x = 1 - 9/25 = 16/25# --> #cos x = +- 4/5#
To Find sin (x/2), use trig identity:
#2sin^2 x = 1 - cos 2x#.
In this case: #2sin^2 (x/2) = 1 - cos x = 1 +- 4/5#
a. #2sin^2 (x/2) = 1 -(- 4/5) = 9/5#
#sin^2 (x/2) = 9/10# --> #sin (x/2) = +- 3/sqrt10 = +- (3sqrt10)/10#
sin x < 0, cos x < 0, x is in Quadrant 3, #x/2# is in quadrant 2, then,
#sin (x/2)# is positive.
#sin (x/2) = (3sqrt10)/10#
b. #2sin^2 (x/2) = 1 - 4/5 = 1/5#
#sin^2 (x/2) = 1/10# --> #sin (x/2) = +- 1/sqrt10 = +- sqrt10/10#.
x is in Quadrant 4, #x/2# is also in quadrant 4, #sin (x/2)# is negative.
#sin (x /2) = - sqrt10/10#
Check by calculator.
#sin x = - 3/5# --> #x = - 36^@87# and #x = 216^@87#
a. #x = - 36^@87# --> #x/2 = - 18^@46# --> #sin (x/2) = - 0.316# = #= - sqrt10/10#. OK
b. #x = 216.87# --> #x/2 = 108.46# --> #sin (x/2) = 0.948 = (3sqrt10/10)#. OK
