We need to make use of the property that #cos(arccos(x))=x#. At the moment we have a sine instead of a cosine. We also know the formula: #cos^2x+sin^2x=1#.
To make use of both of these, we need to square the expression and immediately take the square root (so we don't do anything illegal):
#sqrt(sin^2(2arccos(3/5)))=sqrt(1-cos^2(2arccos(3/5))#
The only problem that we have now is the #2# in front of the #arccos#. We can solve this by ussing the double angle formula:
#cos(2a)=cos^2a-sin^2a=2cos^2a-1=1-2sin^2a#
We need it in therms of the cosine, so let's take the second one:
#sqrt(1-(2cos^2(arccos(3/5))-1)^2#
Simplifying this:
#sqrt(1-4*cos^4(arccos(3/5))+4cos^2(arccos(3/5))-1)#
Now we can replace #cos(arccos(x))# by #x#:
#sqrt(4*(3/5)^2-4*(3/5)^4)=sqrt(4*(3/5)^2*(1-(3/5)^2)#
#=2*3/5*sqrt(1-9/25)=6/5*sqrt(16/25)=6/5*4/5=24/25#
