Question 1179362

Find the exact value of the trigonometric function given that 

{{{sin (u) = -12/13}}} and {{{cos (v )= 24/25}}}


using Pythagorean identity to find {{{cos(u)}}}


{{{sin^2 (u) +cos^2(u)=1}}}

{{{( -12/13)^2 +cos^2(u)=1}}}

{{{cos^2(u)=1-( -12/13)^2}}}

{{{cos^2(u)=1-144/169}}}

{{{cos^2(u)=25/169}}}

{{{cos(u)=sqrt(25/169)}}}

{{{cos(u)=5/13}}} or {{{cos(u)= -5/13 }}}


 In the fourth quadrant, the values for cos are positive only, so use {{{cos(u)= 5/13 }}}



now find {{{sin(v)}}}


{{{sin^2 (v) +cos^2(v)=1}}}

{{{sin^2 (v) +(24/25)^2=1}}}

{{{sin^2 (v) =1-(24/25)^2}}}

{{{sin^2 (v) =1-576/625}}}

{{{sin^2 (v) =49/625}}}

{{{sin (v) =sqrt(49/625)}}}

{{{sin (v) =7/25}}} or  {{{sin (v) =-7/25}}}


 since in Quadrant IV , use {{{sin (v) =-7/25}}}



so we have:

{{{sin (u) = -12/13}}} and {{{cos(u)= 5/13 }}}
{{{sin (v) =-7/25}}} and{{{cos (v )= 24/25}}}



 a)

{{{sin (u+v) =sin(u) cos(v) + cos(u) sin(v)}}}......substitute given values

{{{sin (u+v) =(-12/13 ) (24/25) + (5/13 ) (-7/25)}}}

{{{sin (u+v) =-288/325  -7/65}}}

{{{sin (u+v) =-323/325}}}



b)

{{{sin (u) = -12/13}}} and {{{cos(u)= 5/13 }}}
{{{sin (v) =-7/25}}} and{{{cos (v )= 24/25}}}


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

{{{cos(v-u) =(-12/13) (-7/25) + (5/13) (24/25)}}}

{{{cos(v-u) =84/325 + 24/65}}}

{{{cos(v-u) =204/325}}}



c)

{{{sin (u) = -12/13}}} and {{{cos(u)= 5/13 }}}
{{{sin (v) =-7/25}}} and{{{cos (v )= 24/25}}}


{{{tan(u-v)=sin(u-v)/cos(u-v)}}}

{{{tan(u-v)=(sin(u) cos(v) - cos(u) sin(v)) / (sin(u) sin(v) + cos(u) cos(v))}}}

{{{tan(u-v)=((-12/13) (24/25) - (5/13) (-7/25)) / ((-12/13) (-7/25) +(5/13) (24/25))}}}

{{{tan(u-v)=(-253/325) / (204/325)}}}

{{{tan(u-v)=-253/204}}}