Question 1205224


given:

 angle {{{x}}} is in standard position with its terminal arm in quadrant {{{3 }}}

{{{ tan(x)= 7/5}}}


{{{tan(x)= opp/adj}}}


{{{opp/adj=7/5 }}}=> {{{opp=7}}}, {{{adj=5}}}


then {{{hyp=sqrt(7^2+5^2)=sqrt(74)}}}

since {{{sin(x)=opp/hyp}}} and {{{cos(x)=adj/hyp}}}, we have



{{{sin(x)=7/sqrt(74)}}}  and {{{cos(x)=5/sqrt(74)}}}


since {{{sin(x)}}} and {{{cos(x)}}} are {{{negative}}} in Q III, use 


{{{sin(x)=-7/sqrt(74)}}} and {{{cos(x)=-5/sqrt(74)}}}


to find exact value of {{{cos(2x)}}}, use identity:

{{{cos(2x)=cos^2(x) - sin^2(x)}}}

{{{cos(2x)=(-5/sqrt(74))^2 - (-7/sqrt(74))^2}}}

{{{cos(2x)=25/74 - 49/74}}}

{{{cos(2x)= - 24/74}}}

{{{cos(2x)=-12/37}}}