#sin x=1/9#. We need the ratio for cos x so we can calculate the #(x/2)#. So since we know that the opposite is 1 and the hypotenuse is 9 and we are in quadrant 1 from the given domain we can calculate the adjacent side using pythagorean theorem.
That is,
#a=sqrt(9^2 -1^2)=sqrt(81-1)=sqrt80 = 4sqrt5#. Hence,
#cos x = (4sqrt5)/9, 0 < x< pi/2#. We also need the domain for# x/2 # and we can get that by halving the given domain #0 < x/2 < pi/4# so #x/2# is in Quadrant I. Since it is in quadrant I #sin (x/2), cos(x/2) and tan(x/2)# will all have positive answers.
Now lets calculate #sin (x/2), cos(x/2) and tan(x/2)#,
#sin (x/2)=sqrt(1/2 (1-cosx)#
#=sqrt(1/2 (1-(4sqrt5)/9)#
#=sqrt((9-4sqrt5)/18#
#=sqrt(162-72sqrt5)/18#
#cos (x/2)=sqrt(1/2 (1+cosx))#
#=sqrt(1/2(1+(4sqrt5)/9)#
#=sqrt((9+4sqrt5)/18#
#=sqrt(162+72sqrt5)/18#
#tan (x/2)=sinx/(1+cosx)#
#=(1/9)/(1+(4sqrt5)/9#
#=(1/9)/((9+4sqrt5)/9)#
#=1/cancel9 * cancel9/(9+4sqrt5)#
#=1/(9+4sqrt5) *(9-4sqrt5)/(9-4sqrt5)#
#=(9-4sqrt5)/(81-80#
#=9-4sqrt5#
