#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#