Question

How do you find the amplitude and period of `y=3cos(2/3theta)`?

Answer 1

For `y=acos(btheta)=3cos(3/2 theta)`
`Amplitude = a = 3`
`Period = (2π)/|b| = (2pi)/(2/3) = (6pi)/2 = 3pi`

Explanation:

For `amplitude`, we need the factor `A`,
so for `y=3cos(3/2 theta)`, that would be 3.

For `period`, we must plug-in `b` to `(2π)/|b|` and simplify.
`b=3/2`, so `(2π)/|b| = (2pi)/(2/3)`. To simplify this, we must first multiply the lowest denominator by the numerator `((2pi)*3)/((2/3)*3)` getting `(6pi)/2`, which we can then simplify dividing by 2 `((6pi)/2)/(2/2)` getting `3pi`.