16: For t from the given interval, the values of x cover the
interval
[0,4π2],
so taking into account the root of it, the argument of sine ranges from 0
through
2π;
that is, the curve runs through the first period of the sine.
Since y attains for t in
[−π,π],
both positive and negative values, the curve does use
both the plus and the minus value of sine in the deduced equation.
Answer