Question 1179610


 Find the exact value of the trigonometric function given that 

{{{sin( u) = -3/5}}} and 
{{{cos (v )= -7/25}}}

 (Both u and v are in Quadrant III.)

first find {{{ cos(u)}}} and {{{sin( v) }}}

if {{{sin( u) = -3/5}}} then {{{opp=3}}} and {{{hyp=5}}}


use Pythagorean theorem to find {{{adj}}} side

{{{adj=sqrt(5^2-3^2)}}}

{{{adj=sqrt(25-9)}}}

{{{adj=sqrt(16)}}}

{{{adj=4}}} or{{{adj= -4}}}

since {{{u }}} in Quadrant III, sine and cosine are negative, so use {{{adj=-4}}}

and{{{ cos(u)=-4/5}}}

if {{{cos (v )= -7/25}}} then {{{adj=7}}} and {{{hyp=25}}}

{{{opp=sqrt(25^2-7^2)}}}

{{{opp=sqrt(576)}}}

{{{opp=24}}} or {{{opp=-24}}}...in Quadrant III sine and cosine are negative

{{{sin(v)=-24/25}}}


since {{{sin( u) = -3/5}}},{{{ cos(u)=-4/5}}}, {{{sin(v)=-24/25}}}, {{{cos (v )= -7/25}}},  then,

{{{sin(u+v) =sin(u) cos(v) +cos(u) sin(v) }}}

{{{sin(u+v) =(-3/5) (-7/25) +  (-4/5)(-24/25)}}}

{{{sin(u+v) =21/125+  96/125}}}

{{{sin(u+v) =117/125}}}