What is \sin(2\arcsin(\frac{3}{5})) ?

${\displaystyle \arcsin \left(\frac{3}{5}\right)}$ is some ${\displaystyle \theta }$ between ${\displaystyle -\frac{\pi }{2}}$ and ${\displaystyle \frac{\pi }{2}}$ with ${\displaystyle \sin \theta =\frac{3}{5}}$
Furthermore, with ${\displaystyle -\frac{\pi }{2}\le \theta \le \frac{\pi }{2}}$ and ${\displaystyle \sin \theta }$ a positive number, we conclude that ${\displaystyle \theta }$ is between 0 and ${\displaystyle \frac{\pi }{2}}$.
We want to find ${\displaystyle \sin 2\theta }$ and we already know ${\displaystyle \sin \theta }$. So if we find ${\displaystyle \cos \theta }$, then we can ue the double angle formula for sine.
You've probably done this kind of problem many times by now. ${\displaystyle \theta }$ is in the first quadrant and ${\displaystyle \sin \theta =\frac{3}{5}}$, find ${\displaystyle \cos \theta }$.
Use your favorite method -- draw a triangle, or a unit circle, or an angle in standard position, or skip the picture and use ${\displaystyle \cos \theta =\pm \sqrt{1-{\sin }^2\theta }}$ (recall that our ${\displaystyle \theta }$ 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 ${\displaystyle \theta =\arcsin \left(\frac{3}{5}\right)}$, then ${\displaystyle \sin \theta =\frac{3}{5}}$ and
${\displaystyle \cos \theta =\frac{4}{5}}$
And ${\displaystyle \sin \left(2\theta \right)=2\sin \theta \cos \theta }$
So, putting it all together we get:
${\displaystyle \sin \left(2\arcsin \left(\frac{3}{5}\right)\right)=2\left(\frac{3}{5}\right)\left(\frac{4}{5}\right)=\frac{24}{25}}$