#arcsin(sqrt(3)/2) = pi/3# radians #= 60^o#
So #cos(2 arcsin(sqrt(3)/2)) = cos((2 pi)/3) = -1/2#
Picture an equilateral triangle with sides of length 1. Cut the triangle in two to get two right-angled triangles.
Each of these right-angled triangles has internal angles of #30^o#, #60^o# and #90^o# (or #pi/6#, #pi/3# and #pi/2# radians if you prefer).
The shortest side (which is opposite the #30^o# (pi/6) angle) of one of these triangles has length #1/2# and the hypotenuse is of length #1#.
So #sin 30^o = sin (pi/6) = cos 60^o = cos (pi/3) = 1/2#.
The remaining side of the right angled triangle can be found using Pythagoras theorem to be
#sqrt(1^2 - (1/2)^2) = sqrt(3/4) = sqrt(3)/2#.
Hence #sin 60^o = sin (pi/3) = cos 30^o = cos (pi/6) = sqrt(3)/2#.
So when you see values of #sin theta# or #cos theta# of the form #+-1/2# or #+-sqrt(3)/2# then you know that #theta# is a multiple of #30^o# but not a multiple of #90^o#. Then look at a rough graph of #sin theta# or #cos theta# to spot which multiples you need.
