A little more detail.  It's good to draw triangles
on x and y axes for such problems:

sin(arccos(3/5) - arctan(5/12))

Let A = arccos(3/5) and B = arctan(5/12)

The we use the identity:

sin(A-B)= sin(A)cos(B)-cos(A)sin(B)

First we draw A = arccos(3/5).

arccos(3/5) means 

"the angle in the first quadrant whose cosine is 3/5".

So we draw a triangle in the first quadrant.  Since 
cosine = adj/hyp = x/r we make the adjacent side, x, the
same as the numerator of 3/5, which is x=3 and make 
the hypotenuse, r, the denominator of 3/5 which is r=5.


 

Next we draw arctan(5/12).

arctan(5/12) means 

"the angle in the first quadrant whose tangent is 5/12".

So we draw another right triangle in the first quadrant.  
Since tangent = opp/adj = y/x we make the opposite side, y, the
same as the numerator of 5/12, which is y=5 and make 
the adjacent side, x, the denominator of 5/12 which is x=12.


 

Now we can easily finish:

sin(A-B)= sin(A)cos(B)-cos(A)sin(B)
        = (4/5)(12/13)-(3/5)(5/13)
        = 48/65 - 15/65
        = 33/65


Edwin