Question 1160411


if {{{cos (theta) = -8/17}}} and theta is in quadrant 3,


{{{cos ^2(theta)= (-8/17)^2=64/289 }}}


answer:

C.{{{64/289 }}}



and {{{tan^2 (theta)=sin(theta)/ cos (theta)}}}

{{{sin(theta)=opposite/hypotenuse}}}
 {{{cos(theta)=adjacent/hypotenuse }}}
{{{tan(theta)=opposite/adjacent }}}

if  {{{cos (theta) = -8/17}}} =>{{{adjacent=-8}}} and {{{hypotenuse=17 }}}

using Pythagorean theorem, we have

{{{opposite=sqrt(17^2-(-8)^2)}}}

{{{opposite=sqrt(289-64)}}}

{{{opposite=sqrt(225)}}}

{{{opposite=15}}}

and if {{{theta}}} is in quadrant {{{III}}}, in quadrant {{{III}}}, {{{cos(theta) < 0}}}, {{{sin(theta) < 0}}} and {{{tan(theta) > 0 }}}

{{{sin(theta)=-opposite/hypotenuse}}}....substitute opposite and hypotenuse

{{{sin(theta)=-15/17}}}

then {{{tan(theta) = (-15/17)/ (-8/17) }}}

{{{tan(theta) = (15/cross(17))/ (8/cross(17)) }}}

{{{tan(theta) = 15/ 8 }}}

and
{{{tan^2(theta) = (15/ 8 )^2}}}

{{{tan^2(theta) = 225/ 64}}}



then, 

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

{{{tan(2theta) = 2(15/ 8 ) / (1 - 225/ 64)}}}

{{{tan(2theta) = (15/ 4 ) / ((64 - 225)/ 64)}}}

{{{tan(2theta) = (15/ 1 ) / (-161)/ 16)}}}

{{{tan(2theta) = -240/161}}}


answer:

D.{{{-240/161}}}