Let sin u = 3/5 and sin v = -4/5, where u and v have terminal sides in quadrants ll and lll Then Sin (u-v) =?

Question

explain your answer and step on how you got it, thanks

Answer 1

If sin u = 3/5 and the angle is in the 2nd quadrant, then cos u = -4/5.

If sin v = -4/5 and the angle is in the 3rd quadrant, than cos v = -3/5.

Then using the identity:

sin(u - v) = sin u cos v - cos u sin v
= (3/5)(-3/5) - (-4/5)(-4/5)
= -9/25 - 16/25
= -25/25
= -1

Answer 2

sin(u-v)=sin(u)cos(v)-sin(v)cos(u)

sin u = 3/5 so use pythagorean identity 1= sin squared + cos squared to solve for cos u = -4/5 (negative because u is in quad II and cosine is negative there)

Similarly, cos v = -3/5 (again cosine is negative in quad III)

so we get (3/5)(-3/5) - (-4/5)(-4/5) and simplify this