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

Question

asked Apr 27, 2022 108k views

1 Answer

Akshay Shrivastava · Answer 1 · 2022-05-04T13:33:50+0000

Answer:

$Period = 2\pi$

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
$2\pi$