Using the half-angle formula for cosine:
#cos(theta/2) = +-sqrt((1+costheta)/2)#
Therefore...with
#theta = arcsin(3/5)#
#cos((1/2)arcsin(3/5)) = +-sqrt(((1+cos(arcsin(3/5)))/2)#
Now, set up a right triangle having #theta# as one of the angles that is not the right angle. Since #theta = arcsin(3/5)#, we realize that #theta# is an angle whose sine is #3/5#. Since "sine equals opposite over hypotenuse," mark the hypotenuse of the triangle with a 5, and the side opposite angle #theta# with a 3.
This is a 3-4-5 triangle. The length of the remaining side is 4.
The cosine of the angle is adjacent over hypotenuse:
#cos(arcsin(3/5)) = 4/5#.
Now #1 + 4/5 = 9/5#. Dividing #9/5# by 2 gives #9/10#.
This is the radicand. Back to the half-angle formula,
#cos((1/2)arcsin(3/5)) = +-sqrt(((1+cos(arcsin(3/5)))/2)) = +-sqrt(9/10)#.
If we wish to rationalize the radical, #sqrt(9/10)# = #3/sqrt10# = #(3sqrt10)/10#.
So...
#cos((1/2)arcsin(3/5)) = (3sqrt10)/10#.
Since #arcsinx# is in Quadrant 1 when #x > 0#, we need not worry about the #=-# symbol. We know that the angle is in the first quadrant where cosine is positive. Observe that I removed the #=-# symbol for that reason.
