need help on trig please explain to me question in link http

need help on trig. please explain to me. question in link https://gyazo.com/e9a556669bab5788d980fb548219c4c0

cosA= 0.83

by sin^2A +cos^2A =1 we will find sinA value

sin^2A = 1 - cos^2A= 1 -(0.83)^2

sin^2A= 1 -0.69

sinA = -0.56

so A must be in QIV since cos has to be positive

cos2A = cos^A -sin^2A

cos2A = cos^2A - (1 -cos^2A)

cos2A = 2cos^2A-1

plug A=A/2

cosA = 2cos^(A/2) -1

0.83 +1 = 2cos^2(A/2)

1.83 = 2cos^2(A/2)

cos^(A/2) =0.915

cos(A/2) =-0.96 ( since A/2 is in QII)

now sin2A =2sinA.cosA

plug A=A/2

sinA = 2sin(A/2) .cos(A/2)

-0.56 = 2sin(A/2). (-0.96)

sin(A/2) = -0.56/-1.92

sin(A/2) =0.29

tan(A/2) = 0.29/-0.96

=-0.30