A right-angled triangle with hypotenuse #1# and one side #sqrt(3)/2# has the third side of length #1/4# by Pythagoras. Draw a unit circle centred on the origin #O# and sketch a regular hexagon in it, with vertices #(1,0)#, #(1/2,sqrt(3)/2)#, #(-1/2, sqrt(3)/2)#, #(-1,0)#, #(-1/2,-sqrt(3)/2)#, #(1/2,-sqrt(3)/2)#. Let #P# the last vertex in this list and let #N# be the foot of the perpendicular from #P# to the #x#-axis. Then, disregarding signs, angle #hat{PON}=sin^-1(sqrt(3)/2)#. But clearly #hat{PON}#=360°/6# by symmetry.
Applying the definition of "principal value" of the inverse sin function (which requires the angle to be within the inclusive range #+-90°#) you get #-60°#.
There are many other equivalent ways of visualizing this result.
