Question

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

Answer 1

Amplitude `= 3/2`, Period `= 4`

Explanation:

Standard form of equation `y = a cos (b theta - c) + d`

Given `y = (3/2) cos ((pix)/2)`

`a = 3/2, b = pi / 2`

Amplitude `= a = (3/2)`

Period = (2pi) / |b| = (2pi) / (pi/2) = 4#

graph{(3/2) cos ((pix) / 2) [-10, 10, -5, 5]}