Given `sin(u)=-7/25 , cos(v)=-4/5`
using pythagorean identity,
`sin^2(u)+cos^2(u)=1`
plug in the value of sin(u),
`(-7/25)^2+cos^2(u)=1`
`49/625+cos^2(u)=1`
`cos^2(u)=1-49/625`
`cos^2(u)=(625-49)/625`
`cos^2(u)=576/625`
`cos(u)=sqrt(576/625)`
`cos(u)=+-24/25`
Since u is in Quadrant III ,
`:.cos(u)=-24/25`
Now `sin^2(v)+cos^2(v)=1`
plug in the value of cos(v)=-4/5,
`sin^2(v)+(-4/5)^2=1`
`sin^2(v)+16/25=1`
`sin^2(v)=1-16/25=9/25`
`sin(v)=sqrt(9/25)`
`sin(v)=+-3/5`
since v is in Quadrant III ,
`:.sin(v)=-3/5`
`sin(u+v)=sin(u)cos(v)+cos(u)sin(v)`
plug in the values ,
`sin(u+v)=(-7/25)(-4/5)+(-24/25)(-3/5)`
`sin(u+v)=28/125+72/125=100/125`
`sin(u+v)=4/5`
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