Question

Determine the amplitude or period as requested. Period of a y = - 1/3 * sin x - 1/3 pi/3 1/3 d 2pi

Answer 1

`Period = 2\pi`

Explanation:

Given

`y = -(1)/(3) * \sin(x - (1)/(3))`

Required

Determine the period

A sine function is represented as:

`y =A \sin(Bx + C) + D`

Where

`Period = (2\pi)/(B)`

By comparing:

`y =A \sin(Bx + C) + D` and
`y = -(1)/(3) * \sin(x - (1)/(3))`

`Bx = x`

So:

`B = 1`

So, we have:

`Period = (2\pi)/(B)`

`Period = (2\pi)/(1)`

`Period = 2\pi`

Hence, the period of the function is
`2\pi`