Chapter 5, Problem 31RE

(a)

To determine

To find: TheAmplitude, period and phase shift of the function.

Answer to Problem 31RE

TheAmplitude= $1$ , $\text{Period=4}$ and Phase shift = $0$ .

Explanation of Solution

Given:

  $\text{y=-Sin}\frac{1}{2}\text{x}$ .

Concept used:

  $\text{y=ASin}\left(\text{B}\left(\text{x+C}\right)\right)\text{+D}$ .

Where $\text{A}$ is amplitude.

Period = $\frac{\text{2π}}{\text{B}}$ .

Phase shift= $\text{C}$

  $\left(\text{Positive}\text{is}\text{to}\text{the}\text{left}\right)$ .

Vertical shift= $\text{D}$ .

Calculation:

  $\text{y=-Sin}\frac{1}{2}\text{x}$ .

It can be written as

  $\text{y=-Sin}\frac{1}{2}\left(\text{x-0}\right)\text{+0}$

Comparing with general equation $\text{y=Asin}\left(\text{B}\left(\text{x+C}\right)\right)\text{+D}$ .

Here Amplitude $\text{A=}\left|\text{A}\right|\text{=}\left|\text{-1}\right|\text{=1}\text{.}$

Amplitude= $1$ .

Period= $\frac{\text{2π}}{\frac{\text{1}}{\text{2}}}\text{=4π}\text{.}$

  $\text{Period=4}\text{.}$

Phase shift by comparing with the general equation

Phase shift is $\text{C=0}$ .

Phase shift = $0$ .

Hence, Amplitude= $1$ , $\text{Period=4}$ and Phase shift = $0$ .

(b)

To determine

To Sketch:The graph of the trigonometric function.

Answer to Problem 31RE

From the graph of $\text{y=-Sin}\frac{1}{2}\text{x}$ amplitude and period can be verified which is

Amplitude= $\text{1}$ , $\text{Period=4π}$ .

Explanation of Solution

Given: $\text{y=-Sin}\frac{1}{2}\text{x}$ .

Concept used:

The graph will be Sine function.

Where $\text{A}$ is Amplitude.

Amplitude here is the maximum point.

Period is showed in the graph $\left[\text{0,4π}\right]$ after than its repetitions starts.

Calculation:

The trigonometric period and amplitude graph of the function $\text{y=-Sin}\frac{1}{2}\text{x}$ .

  

As from the graph of $\text{y=-Sin}\frac{1}{2}\text{x}$ amplitude and period can be verified which is

Amplitude= $\text{1}$ , $\text{Period=4π}$ .