sin x=1/9sinx=19. We need the ratio for cos x so we can calculate the (x/2)(x2). 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 = 4sqrt5a=√92−12=√81−1=√80=4√5. Hence,
cos x = (4sqrt5)/9, 0 < x< pi/2cosx=4√59,0<x<π2. We also need the domain for x/2 x2 and we can get that by halving the given domain 0 < x/2 < pi/40<x2<π4 so x/2x2 is in Quadrant I. Since it is in quadrant I sin (x/2), cos(x/2) and tan(x/2)sin(x2),cos(x2)andtan(x2) will all have positive answers.
Now lets calculate sin (x/2), cos(x/2) and tan(x/2)sin(x2),cos(x2)andtan(x2),
sin (x/2)=sqrt(1/2 (1-cosx)sin(x2)=√12(1−cosx)
=sqrt(1/2 (1-(4sqrt5)/9)=
⎷12(1−4√59)
=sqrt((9-4sqrt5)/18=√9−4√518
=sqrt(162-72sqrt5)/18=√162−72√518
cos (x/2)=sqrt(1/2 (1+cosx))cos(x2)=√12(1+cosx)
=sqrt(1/2(1+(4sqrt5)/9)=
⎷12(1+4√59)
=sqrt((9+4sqrt5)/18=√9+4√518
=sqrt(162+72sqrt5)/18=√162+72√518
tan (x/2)=sinx/(1+cosx)tan(x2)=sinx1+cosx
=(1/9)/(1+(4sqrt5)/9=191+4√59
=(1/9)/((9+4sqrt5)/9)=199+4√59
=1/cancel9 * cancel9/(9+4sqrt5)
=1/(9+4sqrt5) *(9-4sqrt5)/(9-4sqrt5)
=(9-4sqrt5)/(81-80
=9-4sqrt5
