Question

What is the amplitude and period of `y=3cos (1/2 x)`?

Answer 1

Consider the basic form:
`y = cos(theta)`
which has an amplitude of `1`
(`y` has a range of `[-1,+1]`)
and
a period of `2pi`
(`cos(0) " to " cos(2pi)` forms one recurring cycle)

In `3 cos(1/2 x)`
the coefficient `3` stretches the range of `y` to `[-3,+3]`
so its amplitude is `3`

The coefficient `1/2 " of "x` forces `x` to have to extend from
`0 " to " 4pi` to cover the set of argument values from `0 " to " 2pi`
so the period of `y=3cos(1/2x)` is `4pi`