1. ## Given sin(x) and tan(y) Can someone please tell me if I did this correct?

Given sin(x) = 3/5 with x in the second quadrant and tan (y) = 12/5 with y in the 3rd quad.

sin (x+y) = 819/65 <----- Is this correct?

cos(x-y) = 728/65 <------ Is this correct?

2. The answers you propose cannot possibly be correct as they are both greater than 1, but sin and cos of an angle must lie between -1 and +1. It would have helped if you had shown your working, but here are the ingredients.

sin(x+y) = sin(x)cos(y) + cos(x)sin(y)
cos(x-y) = cos(x)cos(y) + sin(x)sin(y)

So you need the four quantities sin(x),cos(x),sin(y),cos(y).

Given sin(x) = 3/5 and x in the second (top left) quadrant (pi/2 < x < pi)
x is the angle in a 3-4-5 triangle and drawing the picture cos(x) = -4/5.
This is a short-cut based on recognising 3-4-5: in general just say cos(x)^2 = 1-sin(x)^2 = 1 - (3/5)^2 = 1-9/25 = 16/25 so cos(x) = +- 4/5: second and thrid quadrant have negative cos.
Similarly, y is the angle in a 5-12-13 triangle so sin(x) = -12/13, cos(x) = -5/13.
Again, draw the picture with tan = opposite/adjacent = 5/12.

Now substitute into the two formulae above. Finally, tan(x) = sin(x)/cos(x) = -3/4.

3. ## Don't understand

Hi I am sorry but I don't understand the way you explained how to get the solution of this problem.