#arcsin(3/5)# is some #theta# between #-pi/2# and #pi/2# with #sintheta = 3/5#.

Furthermore, with #-pi/2 <= theta <= pi/2# and #sin theta# a positive number, we conclude that #theta# is between #0# and #pi/2#.

We want to find #sin2 theta# and we already know #sin theta#, so if we find #cos theta#, then we can ue the double angle formula for sine.

You've probably done this kind of problem many times by now. #theta# is in the first quadrant and #sin theta = 3/5#, find #cos theta#.

Use your favorite method -- draw a triangle, or a unit circle, or an angle in standard position, or skip the picture and use #cos theta = +-sqrt(1-sin^2 theta)# (recall that our #theta# is in Quadrant 1, so its cosine is positive.)

All of the above is really explanation of our thought process.

**All we really need to write** is something like:

Let #theta = arcsin(3/5)#, then # sin theta = 3/5# and

#cos theta = 4/5#

And #sin(2 theta) = 2 sin theta cos theta#.

So, putting it all together we get:

#sin(2arcsin(3/5)) = 2(3/5)(4/5) = 24/25#