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

Answer:

$Period = 2\pi$

Step-by-step explanation:

Given

$y = -\frac{1}{3} * \sin(x - \frac{1}{3})$

Required

Determine the period

A sine function is represented as:

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

Where

$Period = \frac{2\pi}{B}$

By comparing:

$y =A \sin(Bx + C) + D$ and $y = -\frac{1}{3} * \sin(x - \frac{1}{3})$

$Bx = x$

So:

$B = 1$

So, we have:

$Period = \frac{2\pi}{B}$

$Period = \frac{2\pi}{1}$

$Period = 2\pi$

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