Question

if sin(theta)=-3/5 and theta is in quadrant III, find tan(theta/2)

Answer 1

use t = tan (theta/2) and sin(theta) = 2t/(1 + t^2)

-3/5 = 2t/(1+t^2)

or -3(1+t^2) = 5(2t)

simplify

3t^2 +10t + 3= 0

(3t +1)(t + 9) = 0

t = -1/3 and -9

Answer 2

-1/3 = tan(theta/2) and -9 = tan(theta/2)

ummm something is not quite right

Answer 3

figured it out...need to use \[\tan \theta/2=(\sin \theta)/(1+\cos \theta)\]