##-60##

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 ##(10)## ##(1/2sqrt(3)/2)## ##(-1/2 sqrt(3)/2)## ##(-10)## ##(-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.