How do you evaluate #arccos(1/2)# without a calculator?

#arccos(1/2)# means give me the angle that gives a #cos(1/2)#.

If you look at a trig circle in the first quadrant, the angle with a #cos(1/2)# is #60^@#. This is found in the first quadrant because of the restricted domain and range of the function.

Il grafico di #f(x) = arccos(x)# has domain:#[-1,1]#, & range: #[0, pi]#
graph{arccos(x) [-5.687, 5.413, -1.287, 4.263]}