Functions and algebra: Use a variety of techniques to sketch and interpret information from graphs of functions

# Unit 8: Horizontal transformation of the sine graph

Dylan Busa

### Unit outcomes

By the end of this unit you will be able to:

• Sketch functions of the form $\scriptsize y=\sin (\theta +p)$.
• Determine the effects of positive and negative values of $\scriptsize p$ on the sine graph $\scriptsize y=\sin (\theta +p)$.
• Find the value of $\scriptsize p$ from a given sine graph of the form $\scriptsize y=\sin (\theta +p)$.

Remember that the domain of trigonometric functions can be represented as $\scriptsize x$ or $\scriptsize \theta$. Therefore, $\scriptsize y=\sin x$ and $\scriptsize y=\sin \theta$ are the same function.

## What you should know

Before you start this unit, make sure you can:

## Introduction

So far in this subject outcome, we have been investigating the effects of the values of $\scriptsize a$, $\scriptsize p$ and $\scriptsize q$ on various functions. For example, for the functions $\scriptsize y=a{{(x+p)}^{2}}+q$, $\scriptsize y=\displaystyle \frac{a}{{(x+p)}}+q$ and $\scriptsize y=a.{{b}^{{x+p}}}+q$ we have discovered that $\scriptsize q$ is responsible for a vertical shift of the graph of the function while $\scriptsize p$ is responsible for a horizontal shift such that:

• if $\scriptsize q \gt 0$, the graph shifts $\scriptsize q$ units up
• if $\scriptsize q \lt 0$, the graph shifts $\scriptsize q$ units down
• if $\scriptsize p \gt 0$, the graph shifts $\scriptsize q$ units to the left
• if $\scriptsize p \lt 0$, the graph shifts $\scriptsize q$ units to the right.

### Why $\scriptsize q$ results in a vertical shift

If we consider the basic quadratic function, we can see why this is the case. Consider $\scriptsize f(x)={{x}^{2}}$. If we add a constant of $\scriptsize q$ to the equation such that it becomes $\scriptsize f(x)={{x}^{2}}+q$, we can see that we will be adding this constant to the value of $\scriptsize f(x)$ (the y-value) for every value of $\scriptsize x$ we feed into the function. In other words, if $\scriptsize f(x)={{x}^{2}}$, then $\scriptsize {{x}^{2}}+q=f(x)+q$. If $\scriptsize q=2$, for example, we will add this same positive constant to each function and shift the whole graph $\scriptsize 2$ units up (see Figure 1).

### Why $\scriptsize p$ results in a horizontal shift

Now let’s consider the role of $\scriptsize p$. Again, let $\scriptsize f(x)={{x}^{2}}$. When $\scriptsize x=2$ the function value is $\scriptsize f(2)=2^2=4$. When $\scriptsize x=4$, the function value is $\scriptsize f(4)=4^2=16$. But if we replaced $\scriptsize x$ with $\scriptsize (x-2)$ in the function, then the function value when $\scriptsize x=4$ would be $\scriptsize f(4-2)=f(2)=2^2=4$. In other words, the function value that used to occur when $\scriptsize x=2$ now occurs when $\scriptsize x=4$. The graph has been shifted two units to the right.

Therefore, replacing $\scriptsize x$ with $\scriptsize (x+p)$ shifts the graph horizontally by $\scriptsize -p$ units. If $\scriptsize p\lt0$, the graph shifts $\scriptsize p$ units to the right. If $\scriptsize p\gt0$, the graph shifts $\scriptsize p$ units to the left.

## The sine function of the form $\scriptsize y=\sin (x+p)$

By looking at the effects of $\scriptsize p$ on the quadratic function above, you should be able to predict the effect that $\scriptsize p$ will have on the graph of $\scriptsize y=\sin (x+p)$. But let’s investigate.

### Activity 8.1: Shifts?

Time required: 40 minutes

What you need:

• a pen or pencil
• a calculator
• several pieces of paper

What to do:

1. Start by making a sketch of $\scriptsize y=\sin x$ for the interval $\scriptsize {{0}^\circ}\le x\le {{360}^\circ}$ on your piece of paper. If you need to, you can complete the following table of values first.
 $\scriptsize x$ $\scriptsize {{0}^\circ}$ $\scriptsize {{90}^\circ}$ $\scriptsize {{180}^\circ}$ $\scriptsize {{270}^\circ}$ $\scriptsize {{360}^\circ}$ $\scriptsize \sin x$
2. Now complete the following tables of values and sketch the additional functions on the same set of axes for the interval $\scriptsize {{0}^\circ}\le x\le {{360}^\circ}$.
 $\scriptsize x$ $\scriptsize {{0}^\circ}$ $\scriptsize {{90}^\circ}$ $\scriptsize {{180}^\circ}$ $\scriptsize {{270}^\circ}$ $\scriptsize {{360}^\circ}$ $\scriptsize \sin \left( {x+{{{30}}^\circ}} \right)$
 $\scriptsize x$ $\scriptsize {{0}^\circ}$ $\scriptsize {{90}^\circ}$ $\scriptsize {{180}^\circ}$ $\scriptsize {{270}^\circ}$ $\scriptsize {{360}^\circ}$ $\scriptsize \sin \left( {x-{{{60}}^\circ}} \right)$
3. Using your sketches, describe the effect of $\scriptsize p$ on the graph of the function $\scriptsize y=\sin (x+p)$.

What did you find?

1. Below is the completed table of values and the sketch of $\scriptsize y=\sin x,\text{ }{{0}^\circ}\le x\le {{360}^\circ}$ (see Figure 3).
 $\scriptsize x$ $\scriptsize {{0}^\circ}$ $\scriptsize {{90}^\circ}$ $\scriptsize {{180}^\circ}$ $\scriptsize {{270}^\circ}$ $\scriptsize {{360}^\circ}$ $\scriptsize \sin x$ $\scriptsize 0$ $\scriptsize 1$ $\scriptsize 0$ $\scriptsize -1$ $\scriptsize 0$
2. Here are the completed tables of values for the other functions.
 $\scriptsize x$ $\scriptsize {{0}^\circ}$ $\scriptsize {{90}^\circ}$ $\scriptsize {{180}^\circ}$ $\scriptsize {{270}^\circ}$ $\scriptsize {{360}^\circ}$ $\scriptsize \sin \left( {x+{{{30}}^\circ}} \right)$ $\scriptsize \displaystyle \frac{1}{2}$ $\scriptsize 0.866$ $\scriptsize -\displaystyle \frac{1}{2}$ $\scriptsize -0.866$ $\scriptsize \displaystyle \frac{1}{2}$
 $\scriptsize x$ $\scriptsize {{0}^\circ}$ $\scriptsize {{90}^\circ}$ $\scriptsize {{180}^\circ}$ $\scriptsize {{270}^\circ}$ $\scriptsize {{360}^\circ}$ $\scriptsize \sin \left( {x-{{{60}}^\circ}} \right)$ $\scriptsize -0.866$ $\scriptsize \displaystyle \frac{1}{2}$ $\scriptsize 0.866$ $\scriptsize -\displaystyle \frac{1}{2}$ $\scriptsize -0.866$

Below are the graphs of $\scriptsize y=\sin x$, $\scriptsize y=\sin \left( {x+{{{30}}^\circ}} \right)$ and $\scriptsize y=\sin \left( {x-{{{60}}^\circ}} \right)$ (see Figure 4).

3. We can see from the sketches that the value of $\scriptsize p$ in $\scriptsize y=\sin (x+p)$ results in a horizontal shift of the graph left or right by $\scriptsize p$ units. In the case of $\scriptsize y=\sin (x+{{30}^\circ})$ the graph is shifted $\scriptsize {{30}^\circ}$ to the left. In the case of $\scriptsize y=\sin (x-{{60}^\circ})$, the graph is shifted $\scriptsize {{60}^\circ}$ to the right. We can see this by looking at the maximum and minimum turning points of the graphs as well as the x-intercepts

### Take note!

In $\scriptsize y=\sin (x+p)$:

• if $\scriptsize p \gt 0$, the graph is shifted $\scriptsize p$ units to the left
• if $\scriptsize p \lt 0$, the graph is shifted $\scriptsize p$ units to the right.

### Note

If you have an internet connection, spend some time playing with this interactive simulation.

Here you will find a graph of the function $\scriptsize \displaystyle y=\sin (x+p)$ with a slider to change the value of $\scriptsize p$. Pay particular attention to how changing the value of $\scriptsize p$ affects the location of the turning points and the intercepts with the x-axis.

## Sketch functions of the form $\scriptsize y=\sin (x+p)$

The best way to sketch functions of the form $\scriptsize y=\sin (x+p)$ is to transform the basic function of $\scriptsize y=\sin x$ depending on the value of $\scriptsize p$. To do this, you need to know the set of ‘anchor points’ of $\scriptsize y=\sin x$, as transformation of these points will help you to sketch functions of the form$\scriptsize y=\sin (x+p)$.

### Example 8.1

Given the function $\scriptsize y=\sin (x-{{45}^\circ})$:

1. Sketch the function for the interval $\scriptsize {{0}^\circ}\le x\le {{360}^\circ}$.
2. State the domain and range.
3. State the period.
4. State the amplitude.

Solution

1. The function is of the form $\scriptsize y=\sin (x+p)$ with, $\scriptsize p=-{{45}^\circ}$.
We need to transform one period of ‘anchor points’ noting that each of these points is going to be shifted $\scriptsize {{45}^\circ}$ to the right.
.
 $\scriptsize \sin x$ $\scriptsize ({{0}^\circ},0)$ $\scriptsize ({{90}^\circ},1)$ $\scriptsize ({{180}^\circ},0)$ $\scriptsize ({{270}^\circ},-1)$ $\scriptsize ({{360}^\circ},0)$ $\scriptsize \sin (x-{{45}^\circ})$ $\scriptsize ({{45}^\circ},0)$ $\scriptsize ({{135}^\circ},1)$ $\scriptsize ({{225}^\circ},0)$ $\scriptsize ({{315}^\circ},-1)$ $\scriptsize ({{405}^\circ},0)$

.
To make sure that the shape of the graph is as accurate as possible, we must also find the y-intercept.

y-intercept (let $\scriptsize x=0$):
\scriptsize \begin{align*}y & =\sin ({{0}^\circ}-{{45}^\circ})\\\therefore y & =-0.707\end{align*}
.
Because the period of the function is $\scriptsize {{360}^\circ}$ we know that the function value at $\scriptsize {{360}^\circ}$ will also be $\scriptsize -0.707$.
.
We can now plot our transformed ‘anchor points’ and draw the graph for the interval $\scriptsize {{0}^\circ}\le x\le {{360}^\circ}$.
.

2. Domain: $\scriptsize \{x|x\in \mathbb{R},{{0}^\circ}\le x\le {{360}^\circ}\}\text{ }$
Range: $\scriptsize \{y|y\in \mathbb{R},-1\le x\le 1\}\text{ }$
3. The period is $\scriptsize {{360}^\circ}$.
4. The amplitude is $\scriptsize 1$.

### Take note!

In general, the domain of the function $\scriptsize y=\sin (x+p)$ is $\scriptsize x\in \mathbb{R}\text{ }$. We had to restrict the domain in Example 8.1, because the interval in which we were working was restricted.

### Exercise 8.1

Sketch the following functions for the indicated intervals on separate sets of axes:

1. $\scriptsize f(x)=\sin (x+{{30}^\circ})$ for $\scriptsize {{0}^\circ}\le x\le {{360}^\circ}$
2. $\scriptsize g(x)=\sin (x-{{45}^\circ})$ for $\scriptsize -{{360}^\circ}\le x\le {{360}^\circ}$
3. $\scriptsize h(x)=\sin (x+{{20}^\circ})$ for $\scriptsize -{{360}^\circ}\le x\le {{360}^\circ}$

The full solutions are at the end of the unit.

## Sketch functions of the form $\scriptsize y=a\sin (x+p)$

Now that we know that the value of $\scriptsize p$ shifts the function of the form $\scriptsize y=\sin (x+p)$ horizontally by $\scriptsize p$ degrees, we can combine our knowledge of the effects of $\scriptsize a$ with this and examine functions of the form $\scriptsize y=a\sin (x+p)$.

### Example 8.2

Sketch the graph of $\scriptsize f(x)=2\sin (x+{{25}^\circ})$ for the interval $\scriptsize -{{360}^\circ}\le x\le {{360}^\circ}$.

Solution

The function is of the form $\scriptsize y=a\sin (x+p)$ where $\scriptsize a=2$ and $\scriptsize p={{25}^\circ}$.

We need to transform one period of ‘anchor points’ noting that each of these points is going to be shifted $\scriptsize {{25}^\circ}$ to the left and each of the y-values is going to be multiplied by $\scriptsize 2$.

 $\scriptsize \sin x$ $\scriptsize ({{0}^\circ},0)$ $\scriptsize ({{90}^\circ},1)$ $\scriptsize ({{180}^\circ},0)$ $\scriptsize ({{270}^\circ},-1)$ $\scriptsize ({{360}^\circ},0)$ $\scriptsize 2\sin (x+{{25}^\circ})$ $\scriptsize (-{{25}^\circ},0)$ $\scriptsize ({{65}^\circ},2)$ $\scriptsize ({{155}^\circ},0)$ $\scriptsize ({{245}^\circ},-2)$ $\scriptsize ({{335}^\circ},0)$

To make sure that the shape of the graph is as accurate as possible, we must also find the y-intercept.

y-intercept (let $\scriptsize x=0$):
\scriptsize \begin{align*}y & =2\sin (0+{{25}^\circ})\\\therefore y & =0.845\end{align*}

Because the period of the function is $\scriptsize {{360}^\circ}$ we know that the function values at $\scriptsize -{{360}^\circ}$ and $\scriptsize {{360}^\circ}$ will also be $\scriptsize 0.845$.

We can now plot our transformed ‘anchor points’ and draw the graph for the interval $\scriptsize -{{360}^\circ}\le x\le {{360}^\circ}$.

### Example 8.3

Given the function $\scriptsize f(x)=\sin ({{30}^\circ}-x)$:

1. Sketch the graph for the interval $\scriptsize {{0}^\circ}\le x\le {{360}^\circ}$.
2. State the intercepts with the axes.
3. State the domain and range.
4. State the amplitude and period.

Solutions

1. The function is not in the form $\scriptsize y=\sin (x+p)$. We first need to get it into this form.
\scriptsize \begin{align*}f(x)&=\sin ({{30}^\circ}-x)\\&=\sin (-x+{{30}^\circ})\\&=\sin \left( {-\left( {x-{{{30}}^\circ}} \right)} \right)\\&=-\sin (x-{{30}^\circ})\end{align*}We know from the previous unit that $\scriptsize \sin (-x)=-\sin (x)$. In other words, the graph of $\scriptsize f(x)=\sin ({{30}^\circ}-x)$ is going to be a reflection of the standard sine graph about the x-axis because $\scriptsize a=-1$.
.
We need to transform one period of ‘anchor points’ noting that each of these points is going to be shifted $\scriptsize {{30}^\circ}$ to the right and each of the y-values of these points needs to be multiplied by $\scriptsize -1$ .
.
 $\scriptsize \sin x$ $\scriptsize ({{0}^\circ},0)$ $\scriptsize ({{90}^\circ},1)$ $\scriptsize ({{180}^\circ},0)$ $\scriptsize ({{270}^\circ},-1)$ $\scriptsize ({{360}^\circ},0)$ $\scriptsize \sin (x-{{30}^\circ})$ $\scriptsize ({{30}^\circ},0)$ $\scriptsize ({{120}^\circ},-1)$ $\scriptsize ({{210}^\circ},0)$ $\scriptsize ({{300}^\circ},1)$ $\scriptsize ({{390}^\circ},0)$

.
To make sure that the shape of the graph is as accurate as possible, we must also find the y-intercept.
y-intercept (let $\scriptsize x=0$):
\scriptsize \begin{align*}f(0) & =\sin ({{30}^\circ}-0)\\\therefore f(0) & =\displaystyle \frac{1}{2}\end{align*}
.
Because the period of the function is $\scriptsize {{360}^\circ}$ we know that the function value at $\scriptsize {{360}^\circ}$ will also be $\scriptsize \displaystyle \frac{1}{2}$.
.
We can now plot our transformed ‘anchor points’ and draw the graph for the interval $\scriptsize {{0}^\circ}\le x\le {{360}^\circ}$.
.

2. y-intercept: $\scriptsize ({{0}^\circ},\displaystyle \frac{1}{2})$
x-intercepts: $\scriptsize ({{30}^\circ},0)$ and $\scriptsize ({{210}^\circ},0)$
3. Domain: $\scriptsize \{x|x\in \mathbb{R},\text{ }{{0}^\circ}\le x\le {{360}^\circ}\}\text{ }$
Range: $\scriptsize \{y|y\in \mathbb{R},\text{ }-1\le x\le 1\}\text{ }$
4. The period is $\scriptsize {{360}^\circ}$.
The amplitude is $\scriptsize 1$.

### Exercise 8.2

For each of the following functions, sketch the graph for the indicated interval and state the domain and range:

1. $\scriptsize y=-2\sin (x+{{30}^\circ})$ for $\scriptsize -{{360}^\circ}\le x\le {{360}^\circ}$
2. $\scriptsize 2y=\sin (x+{{30}^\circ})$ for $\scriptsize {{0}^\circ}\le x\le {{360}^\circ}$
3. $\scriptsize \displaystyle \frac{1}{3}y=\sin ({{60}^\circ}-x)$ for $\scriptsize -{{360}^\circ}\le x\le {{360}^\circ}$

The full solutions are at the end of the unit.

### Find the equation of a sine function of the form $\scriptsize y=a\sin (x+p)$

If we examine the amplitude and the horizontal shift of $\scriptsize y=a\sin (x+p)$, we can determine the values of $\scriptsize a$ and $\scriptsize p$.

### Example 8.4

The graph below is a function of the form $\scriptsize y=a\sin (x+p)$. Determine the values of $\scriptsize a$ and $\scriptsize p$.

Solution

We are told that the function is of the form $\scriptsize y=a\sin (x+p)$.

Firstly, we can see that the amplitude of the graph is $\scriptsize 2$ units. We do not yet know whether $\scriptsize a=2$ or $\scriptsize a=-2$.

We know that the function $\scriptsize y=\sin x$ passes through the origin and then reaches a maximum turning point at $\scriptsize ({{90}^\circ},1)$. This function instead reaches a minimum turning point at $\scriptsize ({{30}^\circ},-2)$. Therefore, we can say that the point $\scriptsize ({{90}^\circ},1)$ has been transformed into the point $\scriptsize ({{30}^\circ},-2)$.

This means that the point has been shifted horizontally $\scriptsize {{60}^\circ}$ to the left and the y-value has been multiplied by $\scriptsize -2$. Therefore, $\scriptsize a=-2$ and $\scriptsize p={{60}^\circ}$, and the function is $\scriptsize y=-2\sin (x+{{60}^\circ})$.

### Exercise 8.3

Given the graph of the form $\scriptsize y=a\sin (x+p)$:

1. Determine the values of $\scriptsize a$ and $\scriptsize p$.
2. State the domain and range of the function.

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• The effects of $\scriptsize a$ and $\scriptsize p$ on the sine graph of the form $\scriptsize y=a\sin (x+p)$.
• How to sketch functions of the form$\scriptsize y=a\sin (x+p)$.
• How to find the values of $\scriptsize a$ and $\scriptsize p$ from a given sine graph of the form$\scriptsize y=a\sin (x+p)$.

# Unit 8: Assessment

#### Suggested time to complete: 45 minutes

1. Sketch the following functions for the given intervals:
1. $\scriptsize 3y=\sin \left( {x-{{{45}}^\circ}} \right)$ for $\scriptsize -{{360}^\circ}\le x\le {{360}^\circ}$
2. $\scriptsize g(x)=-4\sin ({{60}^\circ}-x)$ for $\scriptsize {{0}^\circ}\le x\le {{360}^\circ}$
2. From the graph below of the form $\scriptsize y=a\sin (x+p)$, determine the values of $\scriptsize a$ and $\scriptsize p$.

The full solutions are at the end of the unit.

# Unit 8: Solutions

### Exercise 8.1

1. $\scriptsize f(x)=\sin (x+{{30}^\circ})$ for $\scriptsize {{0}^\circ}\le x\le {{360}^\circ}$
$\scriptsize p={{30}^\circ}$. Therefore, the graph will be shifted $\scriptsize {{30}^\circ}$ to the left.
 $\scriptsize \sin x$ $\scriptsize ({{0}^\circ},0)$ $\scriptsize ({{90}^\circ},1)$ $\scriptsize ({{180}^\circ},0)$ $\scriptsize ({{270}^\circ},-1)$ $\scriptsize ({{360}^\circ},0)$ $\scriptsize \sin (x+{{30}^\circ})$ $\scriptsize (-{{30}^\circ},0)$ $\scriptsize ({{60}^\circ},1)$ $\scriptsize ({{150}^\circ},0)$ $\scriptsize ({{240}^\circ},-1)$ $\scriptsize ({{330}^\circ},0)$

.
y-intercept (let $\scriptsize x=0$):
\scriptsize \begin{align*}f(0) & =\sin ({{0}^\circ}+{{30}^\circ})\\\therefore f(0) & =\displaystyle \frac{1}{2}\end{align*}
.
Because the period of the function is $\scriptsize {{360}^\circ}$ we know that the function value at $\scriptsize {{360}^\circ}$ will also be $\scriptsize \displaystyle \frac{1}{2}$.
.

2. $\scriptsize g(x)=\sin (x-{{45}^\circ})$ for $\scriptsize -{{360}^\circ}\le x\le {{360}^\circ}$
$\scriptsize p=-{{45}^\circ}$. Therefore, the graph will be shifted $\scriptsize {{45}^\circ}$ to the right.
 $\scriptsize \sin x$ $\scriptsize ({{0}^\circ},0)$ $\scriptsize ({{90}^\circ},1)$ $\scriptsize ({{180}^\circ},0)$ $\scriptsize ({{270}^\circ},-1)$ $\scriptsize ({{360}^\circ},0)$ $\scriptsize \sin (x-{{45}^\circ})$ $\scriptsize ({{45}^\circ},0)$ $\scriptsize ({{135}^\circ},1)$ $\scriptsize ({{225}^\circ},0)$ $\scriptsize ({{315}^\circ},-1)$ $\scriptsize ({{405}^\circ},0)$

.
y-intercept (let $\scriptsize x=0$):
\scriptsize \begin{align*}g(0) & =\sin ({{0}^\circ}-{{45}^\circ})\\\therefore g(0) & =-\displaystyle \frac{1}{{\sqrt{2}}}=-0.707\end{align*}
.
Because the period of the function is $\scriptsize {{360}^\circ}$ we know that the function values at $\scriptsize -{{360}^\circ}$ and $\scriptsize {{360}^\circ}$ will also be $\scriptsize -0.707$.
.

3. $\scriptsize h(x)=\sin (x+{{20}^\circ})$ for $\scriptsize -{{360}^\circ}\le x\le {{360}^\circ}$
$\scriptsize p={{20}^\circ}$. Therefore, the graph will be shifted $\scriptsize {{20}^\circ}$ to the left.
 $\scriptsize \sin x$ $\scriptsize ({{0}^\circ},0)$ $\scriptsize ({{90}^\circ},1)$ $\scriptsize ({{180}^\circ},0)$ $\scriptsize ({{270}^\circ},-1)$ $\scriptsize ({{360}^\circ},0)$ $\scriptsize \sin (x+{{20}^\circ})$ $\scriptsize (-{{20}^\circ},0)$ $\scriptsize ({{70}^\circ},1)$ $\scriptsize ({{160}^\circ},0)$ $\scriptsize ({{250}^\circ},-1)$ $\scriptsize ({{340}^\circ},0)$

.
y-intercept (let $\scriptsize x=0$):
\scriptsize \begin{align*}h(0) & =\sin ({{0}^\circ}+{{20}^\circ})\\\therefore h(0) & =0.342\end{align*}
.
Because the period of the function is $\scriptsize {{360}^\circ}$ we know that the function values at $\scriptsize -{{360}^\circ}$ and $\scriptsize {{360}^\circ}$ will also be $\scriptsize 0.342$.
.

Back to Exercise 8.1

### Exercise 8.2

1. $\scriptsize y=-2\sin (x+{{30}^\circ})$ for $\scriptsize -{{360}^\circ}\le x\le {{360}^\circ}$
$\scriptsize a=-2$. Therefore, the amplitude will be $\scriptsize 2$ and the graph will be reflected about the x-axis.
$\scriptsize p={{30}^\circ}$. Therefore, the graph will be shifted $\scriptsize {{30}^\circ}$ to the left.
 $\scriptsize \sin x$ $\scriptsize ({{0}^\circ},0)$ $\scriptsize ({{90}^\circ},1)$ $\scriptsize ({{180}^\circ},0)$ $\scriptsize ({{270}^\circ},-1)$ $\scriptsize ({{360}^\circ},0)$ $\scriptsize -2\sin (x+{{30}^\circ})$ $\scriptsize (-{{30}^\circ},0)$ $\scriptsize ({{60}^\circ},-2)$ $\scriptsize ({{150}^\circ},0)$ $\scriptsize ({{240}^\circ},2)$ $\scriptsize ({{330}^\circ},0)$

.
y-intercept (let $\scriptsize x=0$):
\scriptsize \begin{align*}y & =-2\sin ({{0}^\circ}+{{30}^\circ})\\\therefore y & =-1\end{align*}
.
Because the period of the function is $\scriptsize {{360}^\circ}$ we know that the function values at $\scriptsize -{{360}^\circ}$ and $\scriptsize {{360}^\circ}$ will also be $\scriptsize -1$.
.

.
Domain: $\scriptsize \{x|x\in \mathbb{R},\text{ }-{{360}^\circ}\le x\le {{360}^\circ}\}$
Range: $\scriptsize \{y|y\in \mathbb{R},\text{ }-2\le y\le 2\}$

2. $\scriptsize 2y=\sin (x+{{30}^\circ})$ for $\scriptsize {{0}^\circ}\le x\le {{360}^\circ}$
\scriptsize \begin{align*}2y & =\sin (x+{{30}^\circ})\\\therefore y & =\displaystyle \frac{1}{2}\sin (x+{{30}^\circ})\end{align*}
$\scriptsize a=\displaystyle \frac{1}{2}$. Therefore, the amplitude will be $\scriptsize \displaystyle \frac{1}{2}$.
$\scriptsize p={{30}^\circ}$. Therefore, the graph will be shifted $\scriptsize {{30}^\circ}$ to the left.
 $\scriptsize \sin x$ $\scriptsize ({{0}^\circ},0)$ $\scriptsize ({{90}^\circ},1)$ $\scriptsize ({{180}^\circ},0)$ $\scriptsize ({{270}^\circ},-1)$ $\scriptsize ({{360}^\circ},0)$ $\scriptsize \displaystyle \frac{1}{2}\sin (x+{{30}^\circ})$ $\scriptsize (-{{30}^\circ},0)$ $\scriptsize ({{60}^\circ},\displaystyle \frac{1}{2})$ $\scriptsize ({{150}^\circ},0)$ $\scriptsize ({{240}^\circ},-\displaystyle \frac{1}{2})$ $\scriptsize ({{330}^\circ},0)$

.
y-intercept (let $\scriptsize x=0$):
\scriptsize \begin{align*}y & =\displaystyle \frac{1}{2}\sin ({{0}^\circ}+{{30}^\circ})\\\therefore y & =\displaystyle \frac{1}{4}\end{align*}
.
Because the period of the function is $\scriptsize {{360}^\circ}$ we know that the function value at $\scriptsize {{360}^\circ}$ will also be $\scriptsize \displaystyle \frac{1}{4}$.
.

.
Domain: $\scriptsize \{x|x\in \mathbb{R},\text{ }{{0}^\circ}\le x\le {{360}^\circ}\}$
Range: $\scriptsize \{y|y\in \mathbb{R},\text{ }-\displaystyle \frac{1}{2}\le y\le \displaystyle \frac{1}{2}\}$

3. $\scriptsize \displaystyle \frac{1}{3}y=\sin ({{60}^\circ}-x)$ for $\scriptsize -{{360}^\circ}\le x\le {{360}^\circ}$
\scriptsize \begin{align*}\displaystyle \frac{1}{3}y & =\sin ({{60}^\circ}-x)\\\therefore y & =3\sin ({{60}^\circ}-x)\\&=3\sin (-x+{{60}^\circ})\\&=3\sin \left( {-\left( {x-{{{60}}^\circ}} \right)} \right)\\&=-3\sin (x-{{60}^\circ})\end{align*}
$\scriptsize a=-3$. Therefore, the amplitude will be $\scriptsize 3$ and the graph will be reflected about the x-axis.
$\scriptsize p=-{{60}^\circ}$. Therefore, the graph will be shifted $\scriptsize {{60}^\circ}$ to the right.
 $\scriptsize \sin x$ $\scriptsize ({{0}^\circ},0)$ $\scriptsize ({{90}^\circ},1)$ $\scriptsize ({{180}^\circ},0)$ $\scriptsize ({{270}^\circ},-1)$ $\scriptsize ({{360}^\circ},0)$ $\scriptsize -3\sin (x-{{60}^\circ})$ $\scriptsize ({{60}^\circ},0)$ $\scriptsize ({{150}^\circ},-3)$ $\scriptsize ({{240}^\circ},0)$ $\scriptsize ({{330}^\circ},3)$ $\scriptsize ({{420}^\circ},0)$

.
y-intercept (let $\scriptsize x=0$):
\scriptsize \begin{align*}y & =-3\sin ({{0}^\circ}-{{60}^\circ})\\\therefore y & =-3\times -\displaystyle \frac{{\sqrt{3}}}{2}=2,598\end{align*}
.
Because the period of the function is $\scriptsize {{360}^\circ}$ we know that the function values at $\scriptsize -{{360}^\circ}$ and $\scriptsize {{360}^\circ}$ will also be $\scriptsize 2.598$.
.

.
Domain: $\scriptsize \{x|x\in \mathbb{R},\text{ }-{{360}^\circ}\le x\le {{360}^\circ}\}$
Range: $\scriptsize \{y|y\in \mathbb{R},\text{ }-3\le y\le 3\}$

Back to Exercise 8.2

### Exercise 8.3

1. Amplitude is $\scriptsize 0.25=\displaystyle \frac{1}{4}$
$\scriptsize ({{90}^\circ},1)$ has been transformed to $\scriptsize ({{40}^\circ},-\displaystyle \frac{1}{4})$. Therefore, $\scriptsize a=-\displaystyle \frac{1}{4}$ and $\scriptsize p={{50}^\circ}$.
$\scriptsize y=-\displaystyle \frac{1}{4}\sin (x+{{50}^\circ})$
2. Domain: $\scriptsize \{x|x\in \mathbb{R},\text{ }{{0}^\circ}\le x\le {{360}^\circ}\}$
Range: $\scriptsize \{y|y\in \mathbb{R},\text{ }-\displaystyle \frac{1}{4}\le y\le \displaystyle \frac{1}{4}\}$

Back to Exercise 8.3

### Unit 8: Assessment

1. .
1. $\scriptsize 3y=\sin \left( {x-{{{45}}^\circ}} \right)$ for $\scriptsize -{{360}^\circ}\le x\le {{360}^\circ}$
\scriptsize \begin{align*}3y & =\sin (x-{{45}^\circ})\\\therefore y & =\displaystyle \frac{1}{3}\sin (x-{{45}^\circ})\end{align*}
$\scriptsize a=\displaystyle \frac{1}{3}$. Therefore, the amplitude will be $\scriptsize \displaystyle \frac{1}{3}$.
$\scriptsize p=-{{45}^\circ}$. Therefore, the graph will be shifted $\scriptsize {{45}^\circ}$ to the right.
 $\scriptsize \sin x$ $\scriptsize ({{0}^\circ},0)$ $\scriptsize ({{90}^\circ},1)$ $\scriptsize ({{180}^\circ},0)$ $\scriptsize ({{270}^\circ},-1)$ $\scriptsize ({{360}^\circ},0)$ $\scriptsize \displaystyle \frac{1}{3}\sin (x-{{45}^\circ})$ $\scriptsize ({{45}^\circ},0)$ $\scriptsize ({{135}^\circ},\displaystyle \frac{1}{3})$ $\scriptsize ({{225}^\circ},0)$ $\scriptsize ({{315}^\circ},-\displaystyle \frac{1}{3})$ $\scriptsize ({{405}^\circ},0)$

.
y-intercept (let $\scriptsize x=0$):
\scriptsize \begin{align*}y & =\displaystyle \frac{1}{3}\sin ({{0}^\circ}-{{45}^\circ})\\\therefore y & =\displaystyle \frac{1}{3}\times -\displaystyle \frac{1}{{\sqrt{2}}}=-0.236\end{align*}
.
Because the period of the function is $\scriptsize {{360}^\circ}$ we know that the function values at $\scriptsize -{{360}^\circ}$ and $\scriptsize {{360}^\circ}$ will also be $\scriptsize 0.236$.
.

2. $\scriptsize g(x)=-4\sin ({{60}^\circ}-x)$ for $\scriptsize {{0}^\circ}\le x\le {{360}^\circ}$
\scriptsize \begin{align*}g(x) & =-4\sin ({{60}^\circ}-x)\\&=-4\sin (-x+{{60}^\circ})\\&=-4\sin \left( {-\left( {x-{{{60}}^\circ}} \right)} \right)\\&=4\sin (x-{{60}^\circ})\end{align*}
$\scriptsize a=4$. Therefore, the amplitude will be $\scriptsize 4$.
$\scriptsize p=-{{60}^\circ}$. Therefore, the graph will be shifted $\scriptsize {{60}^\circ}$ to the right.
 $\scriptsize \sin x$ $\scriptsize ({{0}^\circ},0)$ $\scriptsize ({{90}^\circ},1)$ $\scriptsize ({{180}^\circ},0)$ $\scriptsize ({{270}^\circ},-1)$ $\scriptsize ({{360}^\circ},0)$ $\scriptsize 4\sin (x+{{60}^\circ})$ $\scriptsize ({{60}^\circ},0)$ $\scriptsize ({{150}^\circ},4)$ $\scriptsize ({{240}^\circ},0)$ $\scriptsize ({{330}^\circ},-4)$ $\scriptsize ({{420}^\circ},0)$

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y-intercept (let $\scriptsize x=0$):
\scriptsize \begin{align*}g(0)&=4\sin (0-{{60}^\circ})\\\therefore y & =4\times -\displaystyle \frac{{\sqrt{3}}}{2}=-3.464\end{align*}
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Because the period of the function is $\scriptsize {{360}^\circ}$ we know that the function value at $\scriptsize {{360}^\circ}$ will also be $\scriptsize -3.464$.
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2. The function is of the form $\scriptsize y=a\sin (x+p)$. The amplitude is $\scriptsize 3$. The point $\scriptsize ({{90}^\circ},1)$ has been transformed to $\scriptsize ({{0}^\circ},3)$. Therefore, the graph has been shifted $\scriptsize {{90}^\circ}$to the left. Hence $\scriptsize a=3$ and $\scriptsize p={{90}^\circ}$ and the function is $\scriptsize y=3\sin (x+{{90}^\circ})$.

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