Thread: trig

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?

I also need to fine tan2x using 2tanx/1-tan^2x but got a really jumbled up answer! PLEASE HELP!!!

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.

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