`arccos(cos((7pi)/2))` Use the properties of inverse trigonometric functions to evaluate the expression.
this could be thought of as 3 whole rounds `(6pi/2) ` plus` pi/2`
as `pi/2` is between [-1,1] we don't need to do anything. the arccos is the inverse of cos, so
`7pi/2` is the same as `pi/2` as shown above, and cos `(pi/2) = 0 ` (because on the unit circle pi/2 has the coordinates (0,1), and the x is the cos)
`arccos(0) = 90^@` or `pi/2`