Question 1160204
Evaluate the expression under the given conditions.

{{{sin(theta/2)}}}; 

{{{tan(theta) = -5/12}}}, {{{theta}}} in Quadrant IV (assume {{{0 < theta < 2pi)}}}

{{{tan(theta) =opposite/adjacent }}}=>{{{opposite=-5}}}, {{{adjacent=12}}}

{{{hypothenuse=sqrt((-5)^2+12^2)}}}

{{{hypothenuse=sqrt(25+144)}}}

{{{hypothenuse=sqrt(169)}}}

{{{hypothenuse=13}}}


{{{cos(theta)=adjacent/hypotenuse}}}, then

{{{cos(theta)=12/13}}}

and

{{{sin(theta/2)}}}= ±{{{sqrt(1-cos(theta))/2)}}}...since {{{theta}}} in Quadrant IV, use negative sign

{{{sin(theta/2)= -sqrt(1-12/13)/2)}}}

{{{sin(theta/2)= -sqrt((13-12)/13)/2)}}}

{{{sin(theta/2)= -sqrt((1/13)/2)}}}

{{{sin(theta/2)= -sqrt(2/13))}}}

{{{sin(theta/2)= -sqrt(26)/13}}}



Evaluate the expression under the given conditions.

{{{tan(2theta)}}}; 

since{{{ tan(2theta)= (2tan(theta))/(1-tan^2(theta))}}},  {{{tan(theta)=sin(theta)/cos(theta)}}}, and {{{cos(theta)}}} is given,  we have to find {{{sin(theta)}}}
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{{{cos(theta) = 3/5}}}, {{{theta}}}  in Quadrant I


{{{cos(theta)=adjacent/hypotenuse}}}->{{{adjacent=3}}} and {{{hypotenuse=5}}}

then 
{{{opposite=sqrt(5^2-3^2)}}}
{{{opposite=sqrt(25-9)}}}
{{{opposite=sqrt(16)}}}
{{{opposite=4}}}

so, {{{sin(theta)=4/5}}}

then

{{{tan(theta)=opposite/adjacent}}}

{{{tan(theta)=4/3}}}

now we can find

{{{ tan(2theta)= (2tan(theta))/(1-tan^2(theta))}}}......substitute {{{tan(theta)}}}

{{{ tan(2theta)= (2(4/3))/(1-(4/3)^2)}}}

{{{ tan(2theta)= (8/3)/(1-16/9)}}}

{{{ tan(2theta)= (8/3)/(9/9-16/9)}}}

{{{ tan(2theta)= (8/cross(3)1)/(-7/cross(9)3)}}}

{{{ tan(2theta)= (8/1)/(-7/3)}}}

{{{ tan(2theta)=-24/7}}}