Precalculus, Chapter 5, 5.4, Section 5.4, Problem 48

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