We have the sine of a difference, so step one will be the difference angle formula,
#sin(a-b) = sin a cos b - cos a sin b#
#sin ( arccos(sqrt{2}/2) - arcsin(2x)) #
#= sin arccos(sqrt{2}/2) cos arcsin(2x) + cos arccos(sqrt{2}/2) sin arcsin(2x)
#
Well the sine of arcsine and the cosine of arccosine are easy, but what about the others? Well we recognize #arccos(\sqrt{2}/2)# as #\pm 45^circ#, so
#sin arccos(\sqrt{2}/2)= \pm \sqrt{2}/2#
I'll leave the #pm# there; I try to follow the convention that arccos is all inverse cosines, versus Arccos, the principal value.
If we know the sine of an angle is #2x#, that's a side of #2x# and a hypotenuse of #1# so the other side is #\sqrt{1-4x^2}#.
# cos arcsin(2x) = \pm sqrt{1-4x^2}#
Now,
#sin ( arccos(sqrt{2}/2) - arcsin(2x)) #
#=\pm \sqrt{2}/2 sqrt{1-4x^2} + (sqrt{2}/2)(2x)#
#= { 2x \pm sqrt{1 - 4x^2}}/{sqrt{2}} #
