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

# Unit 10: Horizontal transformation of the cosine graph

Dylan Busa

### Unit outcomes

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

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

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

## What you should know

Before you start this unit, make sure you can:

## Introduction

It should not surprise you that the effect of $\scriptsize p$ in $\scriptsize y=\cos (x+p)$ is to shift the graph horizontally by $\scriptsize p$ units. 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.

### Take note!

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

• 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 $\scriptsize \displaystyle y=\cos (x+p)$ 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 $\scriptsize y=\cos (x+p)$

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

### Example 10.1

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

1. Sketch the function 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 period.
5. State the amplitude.

Solutions

1. The function is of the form $\scriptsize y=\cos (x+p)$ with $\scriptsize p=-{{30}^\circ}$.
.
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.
.
 $\scriptsize \cos x$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{180}^\circ},-1)$ $\scriptsize ({{270}^\circ},0)$ $\scriptsize ({{360}^\circ},1)$ $\scriptsize \cos (x-{{30}^\circ})$ $\scriptsize ({{30}^\circ},1)$ $\scriptsize ({{120}^\circ},0)$ $\scriptsize ({{210}^\circ},-1)$ $\scriptsize ({{300}^\circ},0)$ $\scriptsize ({{390}^\circ},1)$

.
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 & =\cos ({{0}^\circ}-{{30}^\circ})\\\therefore y & =\displaystyle \frac{{\sqrt{3}}}{2}=0.866\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 $\scriptsize 0.866$.
.
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: $\scriptsize ({{0}^\circ},0.866)$
x-intercepts: $\scriptsize ({{120}^\circ},0)$ and $\scriptsize ({{300}^\circ},0)$
3. Domain: $\scriptsize \{x|x\in \mathbb{R},\text{ }{{0}^\circ}\le x\le {{360}^\circ}\}$
Range: $\scriptsize \{y|y\in \mathbb{R},\text{ }-1\le x\le 1\}$
4. The period is $\scriptsize {{360}^\circ}$.
5. The amplitude is $\scriptsize 1$.

### Take note!

In general, the domain of the function $\scriptsize y=\cos (x+p)$ is $\scriptsize x\in \mathbb{R}\text{ }$. We had to restrict the domain in Example 10.1, because the interval in which we were working was restricted.

### Exercise 10.1

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

1. $\scriptsize f(x)=\cos (x+{{30}^\circ})$ for $\scriptsize {{0}^\circ}\le x\le {{360}^\circ}$
2. $\scriptsize g(x)=\cos (x-{{45}^\circ})$ for $\scriptsize -{{360}^\circ}\le x\le {{360}^\circ}$
3. $\scriptsize h(x)=\cos (x-{{40}^\circ})$ for $\scriptsize -{{360}^\circ}\le x\le {{360}^\circ}$

The full solutions are at the end of the unit.

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

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

### Example 10.2

Sketch the graph of $\scriptsize f(x)=2.5\cos (x+{{60}^\circ})$ for the interval $\scriptsize -{{360}^\circ}\le x\le {{360}^\circ}$.

Solution

The function is of the form $\scriptsize y=a\cos (x+p)$ where $\scriptsize a=2.5$ and $\scriptsize p={{60}^\circ}$.

We need to transform one period of ‘anchor points’ noting that each of these points is going to be shifted $\scriptsize {{60}^\circ}$ to the left and each of the y-values is going to be multiplied by $\scriptsize 2.5$.

 $\scriptsize \cos x$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{180}^\circ},-1)$ $\scriptsize ({{270}^\circ},0)$ $\scriptsize ({{360}^\circ},1)$ $\scriptsize 2.5\cos (x+{{60}^\circ})$ $\scriptsize (-{{60}^\circ},2.5)$ $\scriptsize ({{30}^\circ},0)$ $\scriptsize ({{120}^\circ},-2.5)$ $\scriptsize ({{210}^\circ},0)$ $\scriptsize ({{300}^\circ},2.5)$

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 & =2.5\cos (0+{{60}^\circ})\\\therefore y & =2.5\times \displaystyle \frac{1}{2}\\=1.25\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 $\scriptsize 1.25$.

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

### Example 10.3

Given the function $\scriptsize f(x)=2\cos ({{30}^\circ}-x)$:

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 $\scriptsize y=a\cos (x+p)$. We first need to get it into this form.
\scriptsize \begin{align*}f(x)&=2\cos ({{30}^\circ}-x)\\&=2\cos (-x+{{30}^\circ})\\&=2\cos \left( {-\left( {x-{{{30}}^\circ}} \right)} \right)\\&=2\cos (x-{{30}^\circ})\end{align*}
.
Remember we learnt in unit 9 that $\scriptsize \cos (-x)=\cos x$ because the function is symmetrical about the y-axis and this meant that we get the same function value for $\scriptsize -x$ as we do for $\scriptsize x$. This is why $\scriptsize 2\cos \left( {-\left( {x-{{{30}}^\circ}} \right)} \right)=2\cos (x-{{30}^\circ})$.
.
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 need to be multiplied by $\scriptsize 2$.
.
 $\scriptsize \cos x$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{180}^\circ},-1)$ $\scriptsize ({{270}^\circ},0)$ $\scriptsize ({{360}^\circ},1)$ $\scriptsize 2\cos (x-{{30}^\circ})$ $\scriptsize ({{30}^\circ},2)$ $\scriptsize ({{120}^\circ},0)$ $\scriptsize ({{210}^\circ},-2)$ $\scriptsize ({{300}^\circ},0)$ $\scriptsize ({{390}^\circ},2)$

.

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 \displaystyle \begin{align*}f(0) & =2\cos ({{0}^\circ}-{{30}^\circ})\\\therefore f(0) & =2\times \displaystyle \frac{{\sqrt{3}}}{2}=\sqrt{3}=1.732\text{ }\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 $\scriptsize 1.732$.
.
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: $\scriptsize ({{0}^\circ},\sqrt{3})$ or $\scriptsize ({{0}^\circ},1.732)$
x-intercepts: $\scriptsize ({{120}^\circ},0)$ and $\scriptsize ({{300}^\circ},0)$
3. Domain: $\scriptsize \{x|x\in \mathbb{R},\text{ }{{0}^\circ}\le x\le {{360}^\circ}\}$
Range: $\scriptsize \{f(x)|f(x)\in \mathbb{R},\text{ }-2\le f(x)\le 2\}$
4. The period is $\scriptsize {{360}^\circ}$.
The amplitude is $\scriptsize 2$.

### Exercise 10.2

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

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

The full solutions are at the end of the unit.

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

By examining the amplitude and the horizontal shift of $\scriptsize y=a\cos (x+p)$, we can determine the values of $\scriptsize a$ and $\scriptsize p$.

### Example 10.4

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

Solution

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

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 $\scriptsize y=\cos x$ has a maximum turning point at $\scriptsize ({{0}^\circ},1)$. The function above has a maximum turning point at $\scriptsize (-{{15}^\circ},2)$. Therefore, the function has been shifted $\scriptsize {{15}^\circ}$ to the left and $\scriptsize p={{15}^\circ}$.

The amplitude of the graph is $\scriptsize 2$ and has not been reflected about the x-axis. Therefore, $\scriptsize a=2$.

The function is $\scriptsize y=2\cos (x+{{15}^\circ})$.

### Example 10.5

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

Solution

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

Firstly, we can see that the amplitude of the graph is $\scriptsize 1.5$ units. We do not yet know whether $\scriptsize a=1.5$ or $\scriptsize a=-1.5$.

We know that the function $\scriptsize y=\cos x$ has a maximum turning point at $\scriptsize ({{0}^\circ},1)$ and then cuts the x-axis at $\scriptsize ({{90}^\circ},0)$. This function instead has a maximum turning point at $\scriptsize ({{130}^\circ},1.5)$. Therefore, we can say that the point $\scriptsize ({{0}^\circ},1)$ has been transformed into the point $\scriptsize ({{130}^\circ},1.5)$. This would indicate that that graph has been shifted $\scriptsize {{130}^\circ}$ to the right and therefore that$\scriptsize p=-{{130}^\circ}$.

However, we generally prefer $\scriptsize -{{90}^\circ}\le p\le {{90}^\circ}$. We also know that the function $\scriptsize y=\cos x$ has a minimum turning point at $\scriptsize ({{180}^\circ},-1)$. Therefore, we can also say that the point $\scriptsize ({{130}^\circ},1.5)$ is the result of a $\scriptsize {{50}^\circ}$ shift to the left and that the function has been reflected about the x-axis. Because this gives us a value for $\scriptsize p$ that is $\scriptsize -{{90}^\circ}\le p\le {{90}^\circ}$, this is the transformation we will use.

This means that $\scriptsize p={{50}^\circ}$, $\scriptsize a=-1.5$ and that the function is $\scriptsize y=-1.5\cos (x+{{50}^\circ})$.

### Exercise 10.3

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

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.

## The relationship between $\scriptsize y=\sin x$ and $\scriptsize y=\cos x$

You may have noticed that the names of the functions sine and cosine are similar. Well, there is a very good reason for this as we will discover in Activity 10.1.

### Activity 10.1: Co-functions

Time required: 20 minutes

What you need:

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

What to do:

1. Consider the function $\scriptsize f(x)=\cos (x-{{90}^\circ})$. Make a neat sketch of the graph of this function for the interval $\scriptsize {{0}^\circ}\le x\le {{360}^\circ}$ on a piece of paper. Use the following table of ‘anchor points’ to help you.
 $\scriptsize \cos x$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{180}^\circ},-1)$ $\scriptsize ({{270}^\circ},0)$ $\scriptsize ({{360}^\circ},1)$ $\scriptsize \cos (x-{{90}^\circ})$
2. What do you notice about the graph of the function $\scriptsize f(x)=\cos (x-{{90}^\circ})$ and the function $\scriptsize y=\sin x$?
3. Write a mathematical expression that captures this relationship.
4. Now consider the function $\scriptsize g(x)=\sin (x-{{90}^\circ})$. Make a neat sketch of the graph of this function $\scriptsize {{0}^\circ}\le x\le {{360}^\circ}$ on a new piece of paper. Use the following table of ‘anchor points’ to help you.
 $\scriptsize \sin x$ $\scriptsize ({{0}^\circ},0)$ $\scriptsize ({{90}^\circ},1)$ $\scriptsize ({{180}^\circ},0)$ $\scriptsize ({{270}^\circ},-1)$ $\scriptsize ({{360}^\circ},0)$ $\scriptsize \sin (x-{{90}^\circ})$
5. What do you notice about the graph of the function $\scriptsize g(x)=\sin (x-{{90}^\circ})$ and the function $\scriptsize y=\cos x$?
6. Write a mathematical expression that captures this relationship.
7. Do these relationships between the sine and cosine functions hold for $\scriptsize y=\cos (x+{{90}^\circ})$ and $\scriptsize y=\sin (x+{{90}^\circ})$? Draw the graphs of these functions to help you.

What did you find?

1. Here is the completed table of values. The graph of $\scriptsize f(x)=\cos (x-{{90}^\circ})$ for the interval $\scriptsize {{0}^\circ}\le x\le {{360}^\circ}$ is shown in Figure 7.
 $\scriptsize \cos x$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{180}^\circ},-1)$ $\scriptsize ({{270}^\circ},0)$ $\scriptsize ({{360}^\circ},1)$ $\scriptsize \cos (x-{{90}^\circ})$ $\scriptsize ({{90}^\circ},1)$ $\scriptsize ({{180}^\circ},0)$ $\scriptsize ({{270}^\circ},-1)$ $\scriptsize ({{360}^\circ},0)$ $\scriptsize ({{450}^\circ},1)$
2. The graph of $\scriptsize f(x)=\cos (x-{{90}^\circ})$ is the same as the graph of $\scriptsize y=\sin x$.
3. $\scriptsize \cos (x-{{90}^\circ})=\sin x$
4. Here is the completed table of values. The graph of $\scriptsize g(x)=\sin (x-{{90}^\circ})$ for the interval $\scriptsize {{0}^\circ}\le x\le {{360}^\circ}$ is shown in Figure 8.
 $\scriptsize \sin x$ $\scriptsize ({{0}^\circ},0)$ $\scriptsize ({{90}^\circ},1)$ $\scriptsize ({{180}^\circ},0)$ $\scriptsize ({{270}^\circ},-1)$ $\scriptsize ({{360}^\circ},0)$ $\scriptsize \sin (x-{{90}^\circ})$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{180}^\circ},1)$ $\scriptsize ({{270}^\circ},0)$ $\scriptsize ({{360}^\circ},-1)$ $\scriptsize ({{450}^\circ},0)$
5. The graph of $\scriptsize g(x)=\sin (x-{{90}^\circ})$ is the same as the graph of $\scriptsize y=-\cos x$.
6. $\scriptsize \sin (x-{{90}^\circ})=-\cos x$
7. If we sketch the function $\scriptsize y=\cos (x+{{90}^\circ})$ we get the graph in Figure 9. We can see that this is the same as the graph of $\scriptsize y=-\sin x$. Therefore, $\scriptsize \cos (x+{{90}^\circ})=-\sin x$.

.
If we sketch the function $\scriptsize y=\sin (x+{{90}^\circ})$ we get the graph in Figure 10. We can see that this is the graph of $\scriptsize y=\cos x$. Therefore, $\scriptsize \sin (x+{{90}^\circ})=\cos x$.

In Activity 10.1 we learnt of the following relationships between the sine and cosine functions:

• $\scriptsize \cos (x-{{90}^\circ})=\sin x$
• $\scriptsize \sin (x-{{90}^\circ})=-\cos x$
• $\scriptsize \cos (x+{{90}^\circ})=-\sin x$
• $\scriptsize \sin (x+{{90}^\circ})=\cos x$.

These relationships between the sine and cosine functions are precisely the reason for their names. We say that cosine is the co-function of sine because we can change one function into the other by adding or subtracting $\scriptsize {{90}^\circ}$ from the function input values. The reason for the change in signs has to do with whether the original function value is positive or negative.

Take $\scriptsize \cos (x-{{90}^\circ})$ for example. If we assume that $\scriptsize x$ is an acute angle ($\scriptsize {{0}^\circ} \lt x \lt {{90}^\circ}$), then $\scriptsize (x-{{90}^\circ})$ will lie in the fourth quadrant where cosine is positive (see Figure 11). Therefore, because $\scriptsize \cos (x-{{90}^\circ})$is positive, it is equal to $\scriptsize \sin x$ $\scriptsize \cos (x-{{90}^\circ})=\sin x$.

However, if we take $\scriptsize \sin (x-{{90}^\circ})$ and still assume that $\scriptsize {{0}^\circ} \lt x \lt {{90}^\circ}$, then $\scriptsize (x-{{90}^\circ})$ will still lie in the fourth quadrant where sine is negative. Therefore, we need the negative value of $\scriptsize \cos x$ to ensure that the ratios are still the same $\scriptsize \sin (x-{{90}^\circ})=-\cos x$.

### Did you know?

The co-function of tangent is called cotangent or COT for short. Where $\scriptsize \tan \theta =\displaystyle \frac{y}{x}$ on the Cartesian plane, $\scriptsize \cot \theta =\displaystyle \frac{x}{y}$.

## Summary

In this unit you have learnt the following:

• The effects of $\scriptsize a$ and $\scriptsize p$ on the cosine graph of the form $\scriptsize y=a\cos (x+p)$.
• How to sketch functions of the form $\scriptsize y=a\cos (x+p)$.
• How to find the values of $\scriptsize a$ and $\scriptsize p$ from a given cosine graph of the form $\scriptsize y=a\cos (x+p)$.
• That sine and cosine are co-functions, and that:
• $\scriptsize \cos (x-{{90}^\circ})=\sin x$
• $\scriptsize \sin (x-{{90}^\circ})=-\cos x$
• $\scriptsize \cos (x+{{90}^\circ})=-\sin x$
• $\scriptsize \sin (x+{{90}^\circ})=\cos x$.

# Unit 10: Assessment

#### Suggested time to complete: 55 minutes

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

The full solutions are at the end of the unit.

# Unit 10: Solutions

### Exercise 10.1

1. $\scriptsize f(x)=\cos (x+{{30}^\circ})$ 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.
 $\scriptsize \cos x$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{180}^\circ},-1)$ $\scriptsize ({{270}^\circ},0)$ $\scriptsize ({{360}^\circ},1)$ $\scriptsize \cos (x+{{30}^\circ})$ $\scriptsize (-{{30}^\circ},1)$ $\scriptsize ({{60}^\circ},0)$ $\scriptsize ({{150}^\circ},-1)$ $\scriptsize ({{240}^\circ},0)$ $\scriptsize ({{330}^\circ},1)$

.
y-intercept (let $\scriptsize x=0$):
\scriptsize \begin{align*}f(0) & =\cos ({{0}^\circ}+{{30}^\circ})\\\therefore f(0) & =\displaystyle \frac{{\sqrt{3}}}{2}=0.866\text{ }\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 $\scriptsize 0.866$.
.

2. $\scriptsize g(x)=\cos (x-{{45}^\circ})$ 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.
 $\scriptsize \cos x$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{180}^\circ},-1)$ $\scriptsize ({{270}^\circ},0)$ $\scriptsize ({{360}^\circ},1)$ $\scriptsize \cos (x-{{45}^\circ})$ $\scriptsize ({{45}^\circ},1)$ $\scriptsize ({{135}^\circ},0)$ $\scriptsize ({{225}^\circ},-1)$ $\scriptsize ({{315}^\circ},0)$ $\scriptsize ({{405}^\circ},1)$

.
y-intercept (let $\scriptsize x=0$):
\scriptsize \begin{align*}g(0) & =\cos ({{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. $\scriptsize h(x)=\cos (x-{{40}^\circ})$ for $\scriptsize -{{360}^\circ}\le x\le {{360}^\circ}$
$\scriptsize p=-{{40}^\circ}$. Therefore, the graph will be shifted $\scriptsize {{40}^\circ}$ to the right.
 $\scriptsize \cos x$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{180}^\circ},-1)$ $\scriptsize ({{270}^\circ},0)$ $\scriptsize ({{360}^\circ},1)$ $\scriptsize \cos (x-{{40}^\circ})$ $\scriptsize ({{40}^\circ},1)$ $\scriptsize ({{130}^\circ},0)$ $\scriptsize ({{220}^\circ},-1)$ $\scriptsize ({{310}^\circ},0)$ $\scriptsize ({{400}^\circ},1)$

.
y-intercept (let $\scriptsize x=0$):
\scriptsize \begin{align*}h(0) & =\cos ({{0}^\circ}-{{40}^\circ})\\\therefore h(0) & =0.766\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 $\scriptsize 0.766$.
.

Back to Exercise 10.1

### Exercise 10.2

1. $\scriptsize y=-2\cos (x+{{30}^\circ})$ 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.
 $\scriptsize \cos x$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{180}^\circ},-1)$ $\scriptsize ({{270}^\circ},0)$ $\scriptsize ({{360}^\circ},1)$ $\scriptsize -2\cos (x+{{30}^\circ})$ $\scriptsize (-{{30}^\circ},-2)$ $\scriptsize ({{60}^\circ},0)$ $\scriptsize ({{150}^\circ},2)$ $\scriptsize ({{240}^\circ},0)$ $\scriptsize ({{330}^\circ},-2)$

.
y-intercept (let $\scriptsize x=0$):
\scriptsize \begin{align*}y & =-2\cos ({{0}^\circ}+{{30}^\circ})\\\therefore y & =-2\times \displaystyle \frac{{\sqrt{3}}}{2}\\=-\sqrt{3}=-1.732\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 $\scriptsize -1.732$.
.

.
y-intercept: $\scriptsize ({{0}^\circ},-1.732)$
x-intercepts: $\scriptsize (-{{300}^\circ},0)$, $\scriptsize (-{{120}^\circ},0)$, $\scriptsize ({{60}^\circ},0)$, $\scriptsize ({{240}^\circ},0)$
Domain: $\scriptsize \{x|x\in \mathbb{R},-{{360}^\circ}\le x\le {{360}^\circ}\}\text{ }$
Range: $\scriptsize \{y|y\in \mathbb{R},-2\le y\le 2\}\text{ }$

2. $\scriptsize 2y=\cos (x+{{60}^\circ})$ for $\scriptsize {{0}^\circ}\le x\le {{360}^\circ}$
\scriptsize \begin{align*}2y & =\cos (x+{{60}^\circ})\\\therefore y & =\displaystyle \frac{1}{2}\cos (x+{{60}^\circ})\end{align*}
$\scriptsize a=\displaystyle \frac{1}{2}$. Therefore, the amplitude will be $\scriptsize \displaystyle \frac{1}{2}$.
$\scriptsize p={{60}^\circ}$. Therefore, the graph will be shifted $\scriptsize {{60}^\circ}$ to the left.
 $\scriptsize \cos x$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{180}^\circ},-1)$ $\scriptsize ({{270}^\circ},0)$ $\scriptsize ({{360}^\circ},1)$ $\scriptsize \displaystyle \frac{1}{2}\cos (x+{{60}^\circ})$ $\scriptsize (-{{60}^\circ},\displaystyle \frac{1}{2})$ $\scriptsize ({{30}^\circ},0)$ $\scriptsize ({{120}^\circ},-\displaystyle \frac{1}{2})$ $\scriptsize ({{210}^\circ},0)$ $\scriptsize ({{300}^\circ},\displaystyle \frac{1}{2})$

.
y-intercept (let $\scriptsize x=0$):
\scriptsize \begin{align*}y & =\displaystyle \frac{1}{2}\cos ({{0}^\circ}+{{60}^\circ})\\\therefore y & =\displaystyle \frac{1}{2}\times \displaystyle \frac{1}{2}=\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 $\scriptsize \displaystyle \frac{1}{4}$.
.

.
y-intercept: $\scriptsize ({{0}^\circ},0.25)$ or $\scriptsize ({{0}^\circ},\displaystyle \frac{1}{4})$
x-intercepts: $\scriptsize ({{30}^\circ},0)$, $\scriptsize ({{210}^\circ},0)$
Domain: $\scriptsize \{x|x\in \mathbb{R},{{0}^\circ}\le x\le {{360}^\circ}\}\text{ }$
Range: $\scriptsize \{y|y\in \mathbb{R},-\displaystyle \frac{1}{2}\le y\le \displaystyle \frac{1}{2}\}\text{ }$

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

.
y-intercept (let $\scriptsize x=0$):
\scriptsize \begin{align*}y & =3\cos ({{0}^\circ}-{{60}^\circ})\\\therefore y & =-3\times \displaystyle \frac{1}{2}\\=\displaystyle \frac{3}{2}=1.5\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 $\scriptsize 1.5$.
.

.
y-intercept: $\scriptsize ({{0}^\circ},1.5)$ or $\scriptsize ({{0}^\circ},\displaystyle \frac{3}{2})$
x-intercepts: $\scriptsize (-{{210}^\circ},0)$, $\scriptsize (-{{30}^\circ},0)$, $\scriptsize ({{150}^\circ},0)$, $\scriptsize ({{330}^\circ},0)$
Domain: $\scriptsize \{x|x\in \mathbb{R},-{{360}^\circ}\le x\le {{360}^\circ}\}$
Range: $\scriptsize \{y|y\in \mathbb{R},-3\le y\le 3\}$

Back to Exercise 10.2

### Exercise 10.3

1. Amplitude is $\scriptsize 0.5=\displaystyle \frac{1}{2}$
The maximum turning point has been transformed to $\scriptsize ({{45}^\circ},-\displaystyle \frac{1}{2})$. The graph has been shifted $\scriptsize {{45}^\circ}$ to the right and reflected about the x-axis. Therefore, $\scriptsize a=-\displaystyle \frac{1}{2}$ and $\scriptsize p=-{{45}^\circ}$.
$\scriptsize y=-\displaystyle \frac{1}{2}\cos (x-{{45}^\circ})$
2. Domain: $\scriptsize \{x|x\in \mathbb{R},\text{ }{{0}^\circ}\le x\le {{360}^\circ}\}$
Range: $\scriptsize \{y|y\in \mathbb{R},\text{ }-\displaystyle \frac{1}{2}\le y\le \displaystyle \frac{1}{2}\}$

Back to Exercise 10.3

### Unit 10: Assessment

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

.
y-intercept (let $\scriptsize x=0$):
\scriptsize \begin{align*}y & =\displaystyle \frac{1}{4}\cos (0-{{60}^\circ})\\\therefore y & =\displaystyle \frac{1}{4}\times \displaystyle \frac{1}{2}=\displaystyle \frac{1}{8}=0.125\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 $\scriptsize 0.125$.
.

2. $\scriptsize g(x)=-4\cos ({{45}^\circ}-x)$ for $\scriptsize -{{360}^\circ}\le x\le {{360}^\circ}$
\scriptsize \begin{align*}g(x) & =-4\cos ({{45}^\circ}-x)\\&=-4\cos (-x+{{45}^\circ})\\&=-4\cos \left( {-\left( {x-{{{45}}^\circ}} \right)} \right)\\&=-4\cos (x-{{45}^\circ})\end{align*}
$\scriptsize a=-4$. Therefore, the amplitude will be $\scriptsize 4$ and the graph will be reflected about the x-axis.
$\scriptsize \displaystyle p=-{{45}^\circ}$. Therefore, the graph will be shifted $\scriptsize {{45}^\circ}$ to the right.
 $\scriptsize \cos x$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{180}^\circ},-1)$ $\scriptsize ({{270}^\circ},0)$ $\scriptsize ({{360}^\circ},1)$ $\scriptsize -4\cos (x-{{45}^\circ})$ $\scriptsize ({{45}^\circ},-4)$ $\scriptsize ({{135}^\circ},0)$ $\scriptsize ({{225}^\circ},4)$ $\scriptsize ({{315}^\circ},0)$ $\scriptsize ({{405}^\circ},-4)$

.
y-intercept (let $\scriptsize x=0$):
\scriptsize \begin{align*}g(0)&=-4\cos (0-{{45}^\circ})\\\therefore y & =-4\times \displaystyle \frac{1}{{\sqrt{2}}}=-2.828\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 $\scriptsize -2.828$.
.

2. The function is of the form $\scriptsize y=a\cos (x+p)$. The amplitude is $\scriptsize 2$. The minimum turning point $\scriptsize ({{180}^\circ},-1)$ has been transformed to $\scriptsize ({{170}^\circ},-2)$. The graph has been shifted $\scriptsize {{10}^\circ}$ to the left. Therefore, $\scriptsize a=2$ and $\scriptsize p={{10}^\circ}$. The function is $\scriptsize y=2\cos (x+{{10}^\circ})$.
3. .
1. $\scriptsize y=a\cos x$ shows no horizontal shift. The maximum turning point that was at $\scriptsize ({{0}^\circ},1)$ is now at $\scriptsize ({{0}^\circ},-1)$. Therefore, the amplitude is $\scriptsize 1$ but the graph has been reflected about the x-axis. Therefore, $\scriptsize y=-\cos x$.
2. $\scriptsize y=\cos (x+p)$ means that the function has undergone a horizontal shift. The maximum turning point that was at $\scriptsize ({{0}^\circ},1)$ is now at $\scriptsize ({{180}^\circ},1)$. In other words, the graph has been shifted $\scriptsize {{180}^\circ}$ to the right. Therefore $\scriptsize y=\cos (x-{{180}^\circ})$.
Alternatively, we can say that the maximum turning point that was at $\scriptsize ({{360}^\circ},1)$ is now at $\scriptsize ({{180}^\circ},1)$ and that the graph has been shifted $\scriptsize {{180}^\circ}$ to the left and $\scriptsize y=\cos (x+{{180}^\circ})$
3. $\scriptsize y=\sin (x+p)$ means that the function has undergone a horizontal shift. The maximum turning point that was at $\scriptsize ({{90}^\circ},1)$ is now at $\scriptsize ({{180}^\circ},1)$. Therefore, the graph has been shifted $\scriptsize {{90}^\circ}$ to the right and $\scriptsize y=\sin (x-{{90}^\circ})$

Back to Unit 10: Assessment