sin(x+y)=sin(x)cos(y)-cos(x)sin(y)
also, remember pythagorean rule, [tex]sin^2(x)+cos^2(x)=1[/tex]
given that sin(Θ)=4/5 and cos(x)=-5/13
find sin(x) and cos(Θ)
sin(x)
cos(x)=-5/13
using pythagorean identity
(sin(x))^2+(-5/13)^2=1
sin(x)=+/- 12/13
in the 2nd quadrant, sin is positve so sin(x)=12/13
cos(Θ)
sin(Θ)=4/5
using pythagrean identity
(4/5)^2+(cos(Θ))^2=1
cos(Θ)=+/-3/5
in 1st quadrant, cos is positive
cos(Θ)=3/5
so sin(Θ+x)=sin(Θ)cos(x)+cos(Θ)sin(x)
sin(Θ+x)=(4/5)(-5/13)+(3/5)(12/13)
sin(Θ+x)=16/65
answer is 1st option
