Question 1127013
{{{csc(theta) = 9 }}}

=>{{{sin(theta) = 1/9 = y/z}}}

{{{x^2 + y^2 = z^2}}}
{{{x^2 + 1^2 = 9^2}}}
{{{x^2 = 81 - 1}}}
{{{x^2 = 80}}}
{{{x^2 = 16*5}}}
{{{x = sqrt(16*5)}}}
{{{x = 4sqrt(5)}}}


since, we have {{{x}}} as in the second quadrant, we will have
{{{cos(theta) = -x/z = - 4sqrt(5)/ 9 }}}<----- it is negative


since we are in the second quadrant as {{{90}}}° < {{{theta}}} < {{{180}}}° , and we have {{{theta/2}}} which changes the interval to :

 {{{90/2 }}}°< {{{theta/2}}} < {{{180/2}}}°

{{{45 }}}°< {{{theta/2}}} < {{{90}}}° <----- first quadrant


{{{sin(x/2)= sqrt((1-cos(x))/2)}}} ---> we are going to take only the positive root since the interval now in the first quadrant.

{{{sin(x/2)= sqrt((1+4sqrt(5)/ 9)/2)}}}

{{{sin(x/2)= sqrt((9+4sqrt(5))/ 9)/2)}}}

{{{sin(x/2)= sqrt((9+4sqrt(5))/ 9)/2)}}}

{{{sin(x/2)=(1/6 )sqrt(9 + 4 sqrt(5))}}}



{{{cos(theta/2) = sqrt(( 1 + cos(theta )) / 2 ) }}}---> we are going to take only the positive root since the interval now in the first quadrant.

{{{cos(theta/2) = sqrt( (1 + 4sqrt(5)/ 9 ) / 2 ) }}}

{{{cos(theta/2) = sqrt( ((9 + 4sqrt(5))/ 9 ) / 2 ) }}}

{{{cos(theta/2) =sqrt((1/2)(1/9) (9 + (4 sqrt(5))))}}}

{{{cos(theta/2) =(1/3)sqrt((1/2)(9 + (4 sqrt(5))))}}}



{{{tan(theta/2) = sin(theta/2)/cos(theta/2)}}} ---> we are going to take only the positive root since the interval now in the first quadrant


{{{tan(theta/2) = ((1/6 )sqrt(9 + 4 sqrt(5)))/((1/3)sqrt((1/2)(9 + (4 sqrt(5)))))}}}


{{{tan(theta/2) = (3/6 )((sqrt(9 + 4 sqrt(5)))/(sqrt((1/2)(9 + (4 sqrt(5))))))}}}

{{{tan(theta/2) = (1/2 )((sqrt(9 + 4 sqrt(5)))/((sqrt(1/2))sqrt(9 + 4 sqrt(5)) )))}}}


{{{tan(theta/2) = (1/2 )(cross((sqrt(9 + 4 sqrt(5)))1)/((sqrt(1/2))cross(sqrt(9 + 4 sqrt(5))) )))}}}


{{{tan(theta/2) = (1/2 )(1/sqrt(1/2))}}}


{{{tan(theta/2) = (1/2 )(1/(sqrt(1)/sqrt(2))))}}}


{{{tan(theta/2) = (1/2 )(sqrt(2))}}}


{{{tan(theta/2) = sqrt(2)/2}}}