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