If θ is an angle in standard position that terminates in Quadrant III such that tanθ = 5/12, then sinθ/2 = _____.

Answer:sin θ/2=5√26/26=0.196Step-by-step explanation:θ ∈(π,3π/2)such thatθ/2 ∈(π/2,3π/4)As a result,0<sin θ/2<1, and-1<cos θ/2<0tan θ/2=sin  θ/2/cos θ/2such thattan θ/2<0Lett=tan θ/2t<0By the double angle identity for tangents2 tan θ/2/1-(tanθ /2)^2  = tanθ2t/1-t^2=5/1224t=5 - 5t^2Solve this quadratic equation for t :t1=1/5 andt2= -5Discard t1 because t is not smaller than 0Let s= sin θ/20<s<1.By the definition of tangents.tan θ/2= sin θ/2/ cos θ/2Apply the Pythagorean Algorithm to express the cosine of θ/2 in terms of s. Note the cos θ/2 is expected to be smaller than zero.cos θ/2 = -√1-(sin θ/2)^2 = - √1-s^2Solve for s.s/-√1-s^2 = -5s^2=25(1-s^2)s=√25/26 = 5√26/26Thereforesin θ/2=5√26/26=0.196....