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

# Unit 9: The cosine function

Dylan Busa

### Unit 9 outcomes

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

• Sketch functions of the form $\scriptsize y=a\cos k\theta$.
• Determine the effects of $\scriptsize a$ and $\scriptsize k$ on the cosine graph of the form $\scriptsize y=a\cos k\theta$.
• Find the values of $\scriptsize a$ and $\scriptsize k$ from a given cosine graph of the form $\scriptsize y=a\cos k\theta$.

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

We know that the period of the cosine function is $\scriptsize {{360}^\circ}$ (same as the sine function) (see Figure 1).

We also know that the amplitude is the maximum distance of the function from its midpoint and that the amplitude of $\scriptsize y=\cos x$ is $\scriptsize 1$ (see Figure 2).

In $\scriptsize y=\cos x$, because the amplitude is $\scriptsize 1$ and the graph’s midpoint is the x-axis, the maximum value of $\scriptsize y=\cos \theta$ is $\scriptsize 1$ and the minimum value is $\scriptsize -1$. In the interval $\scriptsize {{0}^\circ} \leq x \leq {{360}^\circ}$, the graph of $\scriptsize y=\cos x$ has a maximum turning point at $\scriptsize ({{0}^\circ},1)$ and $\scriptsize ({{360}^\circ},1)$, and a minimum turning point at $\scriptsize ({{180}^\circ},-1)$ (see Figure 3).

In level 2 we were introduced to the cosine function of the form $\scriptsize y=a\sin \theta +q$. We discovered that the value of $\scriptsize q$ shifted the graph vertically up and down while the value of $\scriptsize a$ changed the amplitude of the graph.

### Take note!

The period of $\scriptsize y=a\cos \theta +q$ is $\scriptsize {{360}^\circ}$.

The effect of $\scriptsize q$:

• if $\scriptsize q \gt 0$, the function is shifted up by $\scriptsize q$ units
• if $\scriptsize q \lt 0$, the function is shifted down by $\scriptsize q$ units.

The effect of $\scriptsize a$:

• if $\scriptsize a \gt 1$, the amplitude increases to $\scriptsize a$
• if $\scriptsize 0 \lt a \lt 1$, the amplitude decreases to $\scriptsize a$
• if $\scriptsize a \lt 0$, the function is reflected about the x-axis
• if $\scriptsize a \lt -1$, the amplitude increases to $\scriptsize a$
• if $\scriptsize -1 \lt a \lt 0$, the amplitude decreases to $\scriptsize a$.

### Note

Here you will find a cosine function of the form $\scriptsize y=a\cos x+q$ with sliders to change the values of $\scriptsize a$ and $\scriptsize q$. Spend some time playing with it to make sure that you understand how changing the values of $\scriptsize a$ and $\scriptsize q$ affect the shape and location of the cosine function of the form $\scriptsize y=a\cos x+q$.

## The cosine function of the form $\scriptsize y=\cos kx$

In Activity 7.1 in unit 7, we saw that the value of $\scriptsize k$ in $\scriptsize y=\sin kx$ affects the period of the function and that we can calculate the period as:
$\scriptsize \text{Period}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| k \right|}}$ where $\scriptsize \left| k \right|$ is the absolute value of $\scriptsize k$, in other words the positive value of $\scriptsize k$.
This means that if $\scriptsize k \gt 1$, the period decreases and if $\scriptsize 0 \lt k \lt 1$, the period increases.

The same is true for the cosine function $\scriptsize y=\cos kx$. The period can be calculated as follows:
$\scriptsize \text{Period}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| k \right|}}$ where $\scriptsize \left| k \right|$ is the absolute value of $\scriptsize k$, in other words the positive value of $\scriptsize k$.
This means that if $\scriptsize k \gt 1$, the period decreases and if $\scriptsize 0 \lt k \lt 1$, the period increases.

If $\scriptsize k \lt 0$, it changes the period, but the graph is not reflected about the x-axis. This is because the graph of $\scriptsize y=\cos x$ is symmetrical about the y-axis. Therefore $\scriptsize \cos(-x)=\cos x$ (see Figure 5).

Another way of looking at this is that angles of $\scriptsize -\theta$ lie in the fourth quadrant where the cosine function is positive, as it is in the first quadrant. Therefore, there is no change of sign for the function values.

### Note

If you have an internet connection, visit this online simulation to compare the effect values of $\scriptsize k \lt 0$ have on the sine and cosine functions.

### Take note!

In $\scriptsize y=\cos kx$ the $\scriptsize \text{period}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| k \right|}}$.
$\scriptsize \cos (-x)=\cos (x)$, so the graph is not reflected about the x-axis for $\scriptsize k \lt 0$.

### Example 9.1

Given the function $\scriptsize y=\cos \left( {\displaystyle \frac{{2x}}{3}} \right)$ for the interval $\scriptsize {{0}^\circ}\le x\le {{360}^\circ}$, determine the:

1. domain and range
2. period.

Solutions

1. Domain: $\scriptsize \{x|x\in \mathbb{R},\text{ }{{0}^\circ}\le x\le {{360}^\circ}\}$
Range: $\scriptsize \{y|y\in \mathbb{R},\text{ }-1\le y\le 1\}$
2. $\scriptsize k=\displaystyle \frac{2}{3}$. Therefore, the period of the graph is $\scriptsize \displaystyle \frac{{{{{360}}^\circ}}}{{\left| {\displaystyle \frac{2}{3}} \right|}}=\displaystyle \frac{{3\times {{{360}}^\circ}}}{2}={{540}^\circ}$.

### Exercise 9.1

1. Determine the periods of the following functions:
1. $\scriptsize y=\cos (2x)$
2. $\scriptsize y=\cos \left( {-\displaystyle \frac{3}{5}x} \right)$
3. $\scriptsize y=\cos \left( {-\displaystyle \frac{3}{2}x} \right)$
2. For each function of the form $\scriptsize y=\cos kx$, determine the value of $\scriptsize k$:
1. .
2. .
3. .

The full solutions are at the end of the unit.

## Sketch functions of the form $\scriptsize y=a\cos kx$

Now that we know that the value of $\scriptsize k$ affects the period of the function $\scriptsize y=\cos kx$, we can combine this with our knowledge of the effects of $\scriptsize a$ and examine functions of the form $\scriptsize y=a\cos kx$.

The effects of $\scriptsize a$ on $\scriptsize y=a\cos kx$ are exactly the same as they were on $\scriptsize y=a\sin x+q$. Before we learn how to sketch functions of the form $\scriptsize y=a\cos kx$, spend some time playing with this online simulation.

In this simulation you will find a graph of the function $\scriptsize y=a\cos kx+q$ for the interval $\scriptsize -{{360}^\circ}\le x\le {{360}^\circ}$. Pay particular attention to how the amplitude, period and the values of the turning points and intercepts with the axes change as you change the values of $\scriptsize a$, $\scriptsize k$ and $\scriptsize q$.

The best way to sketch functions of the form $\scriptsize y=a\cos kx$ is to transform the basic function of $\scriptsize y=\cos x$ depending on the values of $\scriptsize a$ and $\scriptsize k$. 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 $\scriptsize y=a\cos kx$.

### Take note!

The ‘anchor points’ for $\scriptsize y=\cos x$ are illustrated in Figure 6 for the interval $\scriptsize -{{360}^\circ}\le x\le {{360}^\circ}$.

### Example 9.2

Sketch the function $\scriptsize y=2\cos 2x$ for the interval $\scriptsize -{{180}^\circ}\le x\le {{180}^\circ}$.

Solution

The function is of the form $\scriptsize y=a\cos kx$, with$\scriptsize a=2$ and $\scriptsize k=2$.

Because $\scriptsize k=2$, we know that the period of the function will be $\scriptsize \displaystyle \frac{{{{{360}}^\circ}}}{{\left| 2 \right|}}={{180}^\circ}$. We need to transform one period of ‘anchor points’ as follows:

 $\scriptsize \cos x$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{180}^\circ},-1)$ $\scriptsize ({{270}^\circ},0)$ $\scriptsize ({{360}^\circ},1)$ $\scriptsize \cos 2x$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{45}^\circ},0)$ $\scriptsize ({{90}^\circ},-1)$ $\scriptsize ({{135}^\circ},0)$ $\scriptsize ({{180}^\circ},1)$

Because the period has been halved, the value of each of the x-coordinates is also halved. We can see that the function will start at$\scriptsize 1$, which is its maximum turning point $\scriptsize ({{0}^\circ},1)$, fall to zero , fall to its minimum turning point $\scriptsize ({{90}^\circ},-1)$, rise back to zero and rise back to its maximum turning point $\scriptsize ({{180}^\circ},1)$ in $\scriptsize {{180}^\circ}$, the new period.

Because $\scriptsize a=2$, we know that each function value (or y-value) is going to be multiplied by $\scriptsize 2$ thereby making the amplitude of the function $\scriptsize 2$. We need to further transform one period of ‘anchor points’ as follows:

 $\scriptsize \cos x$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{180}^\circ},-1)$ $\scriptsize ({{270}^\circ},0)$ $\scriptsize ({{360}^\circ},1)$ $\scriptsize \cos 2x$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{45}^\circ},0)$ $\scriptsize ({{90}^\circ},-1)$ $\scriptsize ({{135}^\circ},0)$ $\scriptsize ({{180}^\circ},1)$ $\scriptsize 2\cos 2x$ $\scriptsize ({{0}^\circ},2)$ $\scriptsize ({{45}^\circ},0)$ $\scriptsize ({{90}^\circ},-2)$ $\scriptsize ({{135}^\circ},0)$ $\scriptsize ({{180}^\circ},2)$

We can see that the function will start at $\scriptsize 2$, which is its new maximum turning point $\scriptsize ({{0}^\circ},2)$, fall to zero, fall to its new minimum turning point $\scriptsize ({{90}^\circ},-2)$, rise back to zero and rise back to its new maximum turning point $\scriptsize ({{180}^\circ},2)$ in $\scriptsize {{180}^\circ}$, the new period.

We can now plot our transformed ‘anchor points’ and draw the graph. We first draw it for the interval $\scriptsize {{0}^\circ}\le x\le {{180}^\circ}$ and then continue the pattern for the full interval of $\scriptsize -{{180}^\circ}\le x\le {{180}^\circ}$.

### Example 9.3

Given the function $\scriptsize f(x)=3\cos \left( {-\displaystyle \frac{1}{2}x} \right)$:

1. Sketch the graph of $\scriptsize f(x)$ for the interval $\scriptsize -{{360}^\circ}\le x\le {{360}^\circ}$.
2. State the domain and range of $\scriptsize f(x)$.
3. List the turning points of $\scriptsize f(x)$.
4. What is the amplitude of $\scriptsize f(x)$?
5. What is the period of $\scriptsize f(x)$?

Solutions

1. The function is of the form $\scriptsize y=a\cos kx$ with $\scriptsize a=3$ and $\scriptsize k=-\displaystyle \frac{1}{2}$.
.
Because $\scriptsize k=-\displaystyle \frac{1}{2}$, we know that the period of the function will be $\scriptsize \displaystyle \frac{{{{{360}}^\circ}}}{{\left| {-\displaystyle \frac{1}{2}} \right|}}=\displaystyle \frac{{{{{360}}^\circ}}}{{\displaystyle \frac{1}{2}}}={{720}^\circ}\text{ }$.
.
We need to transform one period of ‘anchor points’ as follows:
 $\scriptsize \cos x$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{180}^\circ},-1)$ $\scriptsize ({{270}^\circ},0)$ $\scriptsize ({{360}^\circ},1)$ $\scriptsize \cos \left( {-\displaystyle \frac{1}{2}x} \right)$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{180}^\circ},0)$ $\scriptsize ({{360}^\circ},-1)$ $\scriptsize ({{540}^\circ},0)$ $\scriptsize ({{720}^\circ},1)$

.
Because the period has been doubled, the value of each of the x-coordinates is also doubled. We can see that the function will start at $\scriptsize 1$, which is its maximum turning point $\scriptsize ({{0}^\circ},1)$, fall to zero , fall to its minimum turning point $\scriptsize ({{360}^\circ},-1)$, rise back to zero and rise back to its maximum turning point $\scriptsize ({{720}^\circ},1)$ in $\scriptsize {{720}^\circ}$, the new period.
.
Because $\scriptsize a=3$, we know that each function value (or y-value) is going to be multiplied by $\scriptsize 3$ thereby making the amplitude of the function $\scriptsize 3$. We need to further transform one period of ‘anchor points’ as follows:

 $\scriptsize \cos x$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{180}^\circ},-1)$ $\scriptsize ({{270}^\circ},0)$ $\scriptsize ({{360}^\circ},1)$ $\scriptsize \cos \left( {-\displaystyle \frac{1}{2}x} \right)$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{180}^\circ},0)$ $\scriptsize ({{360}^\circ},-1)$ $\scriptsize ({{540}^\circ},0)$ $\scriptsize ({{720}^\circ},1)$ $\scriptsize 3\cos \left( {-\displaystyle \frac{1}{2}x} \right)$ $\scriptsize ({{0}^\circ},3)$ $\scriptsize ({{180}^\circ},0)$ $\scriptsize ({{360}^\circ},-3)$ $\scriptsize ({{540}^\circ},0)$ $\scriptsize ({{720}^\circ},3)$

.
We can see that the function will start at , which is its new maximum turning point $\scriptsize ({{0}^\circ},3)$, fall to zero , fall to its new minimum turning point $\scriptsize ({{360}^\circ},-3)$, rise back to zero and rise back to its new maximum turning point $\scriptsize ({{720}^\circ},3)$ in $\scriptsize {{720}^\circ}$, the new period.
.
We can now plot our transformed ‘anchor points’ and draw the graph for the interval $\scriptsize -{{360}^\circ}\le x\le {{360}^\circ}$.
.

2. Domain: $\scriptsize \{x|x\in \mathbb{R},\text{ }-{{360}^\circ}\le x\le {{360}^\circ}\}$
Range$\scriptsize \{f(x)|f(x)\in \mathbb{R},\text{ }-3\le f(x)\le 3\}$
3. Turning points: $\scriptsize (-{{360}^\circ},3)$, $\scriptsize ({{0}^\circ},3)$ and $\scriptsize ({{360}^\circ},-3)$
4. Amplitude is $\scriptsize 3$
5. Period is $\scriptsize {{720}^\circ}$

### Exercise 9.2

Sketch the following functions for the indicated intervals:

1. $\scriptsize y=-2\cos \displaystyle \frac{\theta }{2}$ for $\scriptsize -{{360}^\circ}\le \theta \le {{360}^\circ}$
2. $\scriptsize \displaystyle y=\cos \left( {-\displaystyle \frac{{3x}}{2}} \right)$ for $\scriptsize -{{180}^\circ}\le x\le {{180}^\circ}$
3. $\scriptsize y=\displaystyle \frac{1}{2}\cos 3\theta$ for $\scriptsize -{{180}^\circ}\le \theta \le {{180}^\circ}$

The full solutions are at the end of the unit.

## Find the equation of a cosine function of the form $\scriptsize y=a\cos kx$

By examining the amplitude and period of a graph of the form $\scriptsize y=a\cos kx$, we can determine the values of $\scriptsize a$ and $\scriptsize k$.

### Example 9.4

Given the following graph of the form $\scriptsize y=a\cos kx$, determine the values of $\scriptsize a$ and $\scriptsize k$:

Solution

From the graph we can see that the amplitude is $\scriptsize 4$. We can also see that the graph is not reflected about the x-axis. At zero, it is at its maximum turning point in the same way as the graph of $\scriptsize y=\cos x$. Therefore, $\scriptsize a=4$ and $\scriptsize y=4\cos kx$.

The period is $\scriptsize {{120}^\circ}$.
\scriptsize \begin{align*}\text{Period}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| k \right|}}\\{{120}^\circ}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| k \right|}}\\\therefore \left| k \right|=3\text{ }\end{align*}

$\scriptsize y=4\cos 3x$

### Example 9.5

Given the following graph of the form $\scriptsize y=a\cos kx$, determine the values of $\scriptsize a$ and $\scriptsize k$:

Solution

From the graph we can see that the amplitude is $\scriptsize 0.8=\displaystyle \frac{4}{5}$. We can also see that the graph is reflected about the x-axis. At zero, it is at its minimum turning point unlike the graph of $\scriptsize \displaystyle y=\cos x$ which is at its maximum turning point. Therefore, $\scriptsize a=-\displaystyle \frac{4}{5}$ and $\scriptsize y=-\displaystyle \frac{4}{5}\cos kx$.
\scriptsize \begin{align*}\text{Period}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| k \right|}}\\{{540}^\circ}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| k \right|}}\\\therefore \left| k \right|=\displaystyle \frac{{{{{360}}^\circ}}}{{{{{540}}^\circ}}}=\displaystyle \frac{2}{3}\text{ }\end{align*}

$\scriptsize y=-\displaystyle \frac{4}{5}\cos \left( {\displaystyle \frac{2}{3}x} \right)$

### Exercise 9.3

Given the graph of the form $\scriptsize y=a\cos kx$:

1. Determine the values of $\scriptsize a$ and $\scriptsize k$.
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 $\scriptsize k$ on the cosine graph of the form $\scriptsize y=a\cos k\theta$.
• How to sketch functions of the form $\scriptsize y=a\cos k\theta$.
• How to find the values of $\scriptsize a$ and $\scriptsize k$ from a given cosine graph of the form $\scriptsize y=a\cos k\theta$.

# Unit 9: Assessment

#### Suggested time to complete: 45 minutes

1. Sketch the following functions for the given intervals:
1. $\scriptsize 3y=\cos \left( {\displaystyle \frac{1}{2}x} \right)$ for $\scriptsize -{{720}^\circ}\le x\le {{360}^\circ}$
2. $\scriptsize g(x)=-3\cos 2x$ for $\scriptsize {{0}^\circ}\le x\le {{270}^\circ}$
2. From the graph below of the form $\scriptsize y=a\cos kx$, determine the values of $\scriptsize a$ and $\scriptsize k$:

The full solutions are at the end of the unit.

# Unit 9: Solutions

### Exercise 9.1

1. .
1. $\scriptsize y=\cos (2x)$
$\scriptsize k=2$
$\scriptsize \text{Period}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| 2 \right|}}=\displaystyle \frac{{{{{360}}^\circ}}}{2}={{180}^\circ}\text{ }$
2. $\scriptsize y=\cos \left( {-\displaystyle \frac{3}{5}x} \right)$
$\scriptsize k=-\displaystyle \frac{3}{5}$
$\scriptsize \text{Period}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| {-\displaystyle \frac{3}{5}} \right|}}=\displaystyle \frac{{{{{360}}^\circ}}}{{\displaystyle \frac{3}{5}}}=\displaystyle \frac{{5\times {{{360}}^\circ}}}{3}={{600}^\circ}$
3. $\scriptsize y=\cos \left( {-\displaystyle \frac{3}{2}x} \right)$
$\scriptsize k=-\displaystyle \frac{3}{2}$
$\scriptsize \text{Period}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| {-\displaystyle \frac{3}{2}} \right|}}=\displaystyle \frac{{2\times {{{360}}^\circ}}}{3}={{240}^\circ}$
2. .
1. The period of the graph is $\scriptsize {{720}^\circ}$.
\scriptsize \begin{align*}\text{Period} & =\displaystyle \frac{{{{{360}}^\circ}}}{{\left| k \right|}}\\{{720}^\circ} & =\displaystyle \frac{{{{{360}}^\circ}}}{{\left| k \right|}}\\\therefore \left| k \right| & =\displaystyle \frac{{{{{360}}^\circ}}}{{{{{720}}^\circ}}}\\\therefore \left| k \right| & =\displaystyle \frac{1}{2}\end{align*}
2. The period of the graph is $\scriptsize {{288}^\circ}$.
\scriptsize \displaystyle \begin{align*}\text{Period}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| k \right|}}\\{{288}^\circ}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| k \right|}}\\\therefore \left| k \right|=\displaystyle \frac{{{{{360}}^\circ}}}{{{{{288}}^\circ}}}=\displaystyle \frac{5}{4}\text{ }\end{align*}
3. The period of the graph is .
\scriptsize \begin{align*}\text{Period}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| k \right|}}\\{{120}^\circ}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| k \right|}}\\\therefore \left| k \right|=\displaystyle \frac{{{{{360}}^\circ}}}{{{{{120}}^\circ}}}=3\text{ }\end{align*}

Back to Exercise 9.1

### Exercise 9.2

1. $\scriptsize y=-2\cos \displaystyle \frac{\theta }{2}$ for $\scriptsize -{{360}^\circ}\le \theta \le {{360}^\circ}$
$\scriptsize k=\displaystyle \frac{1}{2}$
$\scriptsize \text{Period}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| k \right|}}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| {\displaystyle \frac{1}{2}} \right|}}={{720}^\circ}$
$\scriptsize a=-2$. Amplitude is $\scriptsize 2$ and graph is reflected about the x-axis.
 $\scriptsize \cos \theta$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{180}^\circ},-1)$ $\scriptsize ({{270}^\circ},0)$ $\scriptsize ({{360}^\circ},1)$ $\scriptsize -2\cos \left( {\displaystyle \frac{\theta }{2}} \right)$ $\scriptsize ({{0}^\circ},-2)$ $\scriptsize ({{180}^\circ},0)$ $\scriptsize ({{360}^\circ},2)$ $\scriptsize ({{540}^\circ},0)$ $\scriptsize ({{720}^\circ},-2)$

2. $\scriptsize y=\cos \left( {-\displaystyle \frac{{3x}}{2}} \right)$ for $\scriptsize -{{180}^\circ}\le x\le {{180}^\circ}$
$\scriptsize k=-\displaystyle \frac{3}{2}$
$\scriptsize \text{Period}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| k \right|}}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| {-\displaystyle \frac{3}{2}} \right|}}=\displaystyle \frac{{{{{360}}^\circ}}}{{\displaystyle \frac{3}{2}}}={{240}^\circ}$
$\scriptsize a=1$. Amplitude is $\scriptsize 1$ and the graph is not reflected about the x-axis.
 $\scriptsize \cos x$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{180}^\circ},-1)$ $\scriptsize ({{270}^\circ},0)$ $\scriptsize ({{360}^\circ},1)$ $\scriptsize \cos \left( {-\displaystyle \frac{{3x}}{2}} \right)$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{60}^\circ},0)$ $\scriptsize ({{120}^\circ},-1)$ $\scriptsize ({{180}^\circ},0)$ $\scriptsize ({{240}^\circ},1)$

3. $\scriptsize y=\displaystyle \frac{1}{2}\cos 3\theta$ for $\scriptsize -{{180}^\circ}\le x\le {{180}^\circ}$
$\scriptsize k=3$
$\scriptsize \text{Period}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| k \right|}}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| 3 \right|}}={{120}^\circ}$
$\scriptsize a=\displaystyle \frac{1}{2}$. Amplitude is $\scriptsize \displaystyle \frac{1}{2}$ and graph is not reflected about the x-axis.
 $\scriptsize \cos \theta$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{180}^\circ},-1)$ $\scriptsize ({{270}^\circ},0)$ $\scriptsize ({{360}^\circ},1)$ $\scriptsize \displaystyle \frac{1}{2}\cos \left( {3\theta } \right)$ $\scriptsize ({{0}^\circ},\displaystyle \frac{1}{2})$ $\scriptsize ({{30}^\circ},0)$ $\scriptsize ({{60}^\circ},-\displaystyle \frac{1}{2})$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{120}^\circ},\displaystyle \frac{1}{2})$

Back to Exercise 9.2

### Exercise 9.3

1. The amplitude of the function is $\scriptsize 2$and graph is reflected about the x-axis. Therefore, $\scriptsize a=-2$ and $\scriptsize y=-2\cos kx$.
The period is $\scriptsize {{270}^\circ}$. Therefore $\scriptsize k=\displaystyle \frac{{{{{360}}^\circ}}}{{{{{270}}^\circ}}}=\displaystyle \frac{4}{3}$.
$\scriptsize y=-2\cos \left( {\displaystyle \frac{4}{3}x} \right)$
2. Domain: $\scriptsize \{x|x\in \mathbb{R},{{0}^\circ}\le x\le {{360}^\circ}\}\text{ }$
Range: $\scriptsize \{y|y\in \mathbb{R},\text{ }-2\le x\le 2\}$

Back to Exercise 9.3

### Unit 9: Assessment

1. .
1. $\scriptsize 3y=\cos \left( {\displaystyle \frac{1}{2}x} \right)$ for $\scriptsize -{{720}^\circ}\le x\le {{360}^\circ}$
\scriptsize \begin{align*}3y & =\cos \left( {\displaystyle \frac{1}{2}x} \right)\\\therefore y & =\displaystyle \frac{1}{3}\cos \left( {\displaystyle \frac{1}{2}x} \right)\text{ }\end{align*}
$\scriptsize k=\displaystyle \frac{1}{2}$. Therefore,$\scriptsize \text{period}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| {\displaystyle \frac{1}{2}} \right|}}={{720}^\circ}$
$\scriptsize a=\displaystyle \frac{1}{3}$. Therefore, amplitude is $\scriptsize \displaystyle \frac{1}{3}$.
 $\scriptsize \cos x$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{180}^\circ},-1)$ $\scriptsize ({{270}^\circ},0)$ $\scriptsize ({{360}^\circ},1)$ $\scriptsize \displaystyle \frac{1}{3}\cos \left( {\displaystyle \frac{1}{2}x} \right)$ $\scriptsize ({{0}^\circ},\displaystyle \frac{1}{3})$ $\scriptsize ({{180}^\circ},0)$ $\scriptsize \displaystyle ({{360}^\circ},-\displaystyle \frac{1}{3})$ $\scriptsize ({{540}^\circ},0)$ $\scriptsize ({{720}^\circ},\displaystyle \frac{1}{3})$

2. $\scriptsize g(x)=-3\cos 2x$ for $\scriptsize {{0}^\circ}\le x\le {{270}^\circ}$
$\scriptsize k=2$. Therefore,$\scriptsize \text{period}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| 2 \right|}}={{180}^\circ}\text{ }$
$\scriptsize a=-3$. Therefore, amplitude is $\scriptsize 3$ and the graph is reflected about the x-axis.
 $\scriptsize \cos x$ $\scriptsize ({{0}^\circ},1)$ $\scriptsize ({{90}^\circ},0)$ $\scriptsize ({{180}^\circ},-1)$ $\scriptsize ({{270}^\circ},0)$ $\scriptsize ({{360}^\circ},1)$ $\scriptsize -3\cos 2x$ $\scriptsize ({{0}^\circ},-3)$ $\scriptsize ({{45}^\circ},0)$ $\scriptsize \displaystyle ({{90}^\circ},3)$ $\scriptsize ({{135}^\circ},0)$ $\scriptsize ({{180}^\circ},-3)$

2. The amplitude is $\scriptsize 1$ and the graph is reflected about the x-axis. Therefore, $\scriptsize a=-1$ and $\scriptsize y=-\cos kx$.
The graph completes half a period between $\scriptsize -{{360}^\circ}$ and $\scriptsize {{360}^\circ}$. Therefore, the period is $\scriptsize 2\times {{720}^\circ}=1 \ {{440}^\circ}$.
\scriptsize \begin{align*}\text{Period}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| k \right|}}\\1\ {{440}^\circ}=\displaystyle \frac{{{{{360}}^\circ}}}{{\left| k \right|}}\\\therefore \left| k \right|=\displaystyle \frac{{{{{360}}^\circ}}}{{1\ {{{440}}^\circ}}}=\displaystyle \frac{1}{4}\end{align*}
$\scriptsize y=-\cos \left( {\displaystyle \frac{1}{4}x} \right)$

Back to Unit 9: Assessment