Question

How do you find the amplitude and period of `y=5sin(2/3theta)`?

Answer 1

`A=5` and `T=3pi`

Explanation:

A function :

`y=Asinωθ` has amplitude equal to `A` and period equal to

`T=(2pi)/ω`

so the amplitude of the function `y=5sin(2/3θ)` is `A=5`

and the period is `T=(2pi)/(2/3)=3pi`