Question

Which function has a period equal to half the period of the function in y = -3sin(2/3x - 2π) + 2?

a. y = 3cos(2/3x - π) + 2
b.y = -3/2cos(2/3x - 2π) + 2
c.y = -3cos(2/3x - 2π) + 2
d.y = 3cos(4/3x - 2π) + 2

Answer 1

d.y = 3cos(4/3x - 2π) + 2

Explanation:

Answer 2

D.
`y=3\cos((4)/(3)x-2\pi)+2`

Explanation:

The given function is:

`y=-3\sin((2)/(3)x-2\pi)+2`

This function is of the form:

`y=A\sin(Bx-C)+D`, where
`B=(2)/(3)`

The period is given by:

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

`T=(2\pi)/((2)/(3))=3\pi`

Half of this period is
`(3\pi)/(2)`.

The function that has a period of
`(3\pi)/(2)` is

`y=3\cos((4)/(3)x-2\pi)+2`