Space, shape and measurement: Solve problems by constructing and interpreting trigonometric models

# Unit 4: An introduction to radians

Dylan Busa

### Unit 4 outcomes

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

• Convert from degrees to radians.
• Convert from radians to degrees.

## What you should know

Before you start this unit, make sure you can:

## Introduction

Throughout levels 2, 3 and 4, whenever we have measured an angle or given its size, we have done so in degrees. We are very familiar with degrees. There are $\scriptsize {{360}^\circ}$ in one full revolution. A half revolution is $\scriptsize {{180}^\circ}$. A quarter revolution is $\scriptsize {{90}^\circ}$, and so on.

So, in rotating through a revolution or circle, you have turned $\scriptsize {{360}^\circ}$. But have you ever asked yourself why there are $\scriptsize {{360}^\circ}$ in one revolution? Why not $\scriptsize {{400}^\circ}$ or $\scriptsize {{25}^\circ}$. The number $\scriptsize 360$ was chosen on purpose, although we are not quite sure why. Some theories are that it is based on old lunar calendars or the fact that the earth takes about $\scriptsize 360$days to travel around the sun ($\scriptsize {{1}^\circ}$ per day).

Now degrees work fine for some applications but it is not the best method because it is somewhat arbitrary. Mathematicians hate things being arbitrary, so they came up with a better way to measure angles. It is called radian measure. A radian is based on the radius of a circle.

An angle of $\scriptsize 1$ radian (or rad) is made when we wrap a line the same length as the radius of a circle around the circumference of the circle. In other words, an angle of $\scriptsize 1$ radian is created by an arc length equal to the radius of a circle (see figure 1).

### Note

If you have an internet connection, watch these excellent simple explanations of radians:

For a fuller explanation, you can also watch the video called “What are Radians?”.

Unlike degrees, radians have no unit. Radians are pure numbers. Radians are also the official method of measuring angles.

Now, we know that the circumference of a circle is $\scriptsize 2\pi r$. If to move through an angle of one radian, we have to move a distance around the circumference of $\scriptsize r$ units, this means that there are $\scriptsize 2\pi$ radians in one full revolution, or $\scriptsize \pi$ radians in half a revolution (see figure 2). Click on figure 2 to play the animation.

We can use this to calculate that there are $\scriptsize \displaystyle \frac{{{{{180}}^\circ}}}{\pi }\approx {{57.2958}^\circ}$ in one radian.

### Take note!

To convert from degrees to radians, multiply by $\scriptsize \pi$ and divide by $\scriptsize {{180}^\circ}$.
To convert from radians to degrees, multiply by $\scriptsize {{180}^\circ}$ and divide by $\scriptsize \pi$.

 Degrees Radians $\scriptsize {{30}^\circ}$ $\scriptsize \displaystyle \frac{\pi }{6}$ $\scriptsize {{45}^\circ}$ $\scriptsize \displaystyle \frac{\pi }{4}$ $\scriptsize {{60}^\circ}$ $\scriptsize \displaystyle \frac{\pi }{3}$ $\scriptsize {{90}^\circ}$ $\scriptsize \displaystyle \frac{\pi }{2}$ $\scriptsize {{180}^\circ}$ $\scriptsize \pi$ $\scriptsize {{270}^\circ}$ $\scriptsize \displaystyle \frac{{3\pi }}{2}$ $\scriptsize {{360}^\circ}$ $\scriptsize 2\pi$

### Example 4.1

1. Convert the following degrees to radians:
1. $\scriptsize {{60}^\circ}$
2. $\scriptsize {{50}^\circ}$
3. $\scriptsize {{300}^\circ}$
4. $\scriptsize {{272}^\circ}$
2. Convert the following radians to degrees:
1. $\scriptsize 2\pi$
2. $\scriptsize \displaystyle \frac{{8\pi }}{3}$
3. $\scriptsize 16\pi$
4. $\scriptsize 0.6512975$
5. $\scriptsize 3$

Solutions

1. To convert from degrees to radians, we have to multiply the angle by $\scriptsize \pi$ and then divide by $\scriptsize {{180}^\circ}$.
1. $\scriptsize {{60}^\circ}=\displaystyle \frac{{{{{60}}^\circ}\times \pi }}{{{{{180}}^\circ}}}=\displaystyle \frac{\pi }{3}$
2. $\scriptsize {{50}^\circ}=\displaystyle \frac{{{{{50}}^\circ}\times \pi }}{{{{{180}}^\circ}}}=0.278\pi \approx 0.873$
3. $\scriptsize {{300}^\circ}=\displaystyle \frac{{{{{300}}^\circ}\times \pi }}{{{{{180}}^\circ}}}=\displaystyle \frac{{5\pi }}{3}\approx 5.236$
4. $\scriptsize {{272}^\circ}=\displaystyle \frac{{{{{272}}^\circ}\times \pi }}{{{{{180}}^\circ}}}=\displaystyle \frac{{68\pi }}{{45}}\approx 4.747$
2. To convert from radians to degrees, we have to multiply by $\scriptsize {{180}^\circ}$ and then divide by $\scriptsize \pi$.
1. $\scriptsize 2\pi =\displaystyle \frac{{2\pi \times {{{180}}^\circ}}}{\pi }={{360}^\circ}$
2. $\scriptsize \displaystyle \frac{{8\pi }}{3}=\displaystyle \frac{{8\pi \times {{{180}}^\circ}}}{{3\pi }}={{480}^\circ}$
3. $\scriptsize 16\pi =\displaystyle \frac{{16\pi \times {{{180}}^\circ}}}{\pi }=2\ {{880}^\circ}$
4. $\scriptsize 0.6512975=\displaystyle \frac{{0.6512975\times {{{180}}^\circ}}}{\pi }\approx {{37.317}^\circ}$
5. $\scriptsize 3=\displaystyle \frac{{3\times {{{180}}^\circ}}}{\pi }\approx {{171.887}^\circ}$

### Exercise 4.1

1. Convert the following angles from degrees to radians:
1. $\scriptsize {{30}^\circ}$
2. $\scriptsize {{75}^\circ}$
3. $\scriptsize {{135}^\circ}$
4. $\scriptsize {{215}^\circ}$
2. Convert the following angles from radians to degrees:
1. $\scriptsize 5\pi$
2. $\scriptsize \displaystyle \frac{\pi }{6}$
3. $\scriptsize \displaystyle \frac{{7\pi }}{2}$
4. $\scriptsize {{\pi }^{2}}$

The full solutions are at the end of the unit.

Everything that we have learnt to do in degrees, we can do in radians. Some of it is even easier to do in radians. However, switching from degrees to radians can sometimes be a little confusing. Therefore, if you are ever asked to give your final answer in radians, you can work in degrees and then simply convert your final answer to radians.

### Example 4.2

Solve for $\scriptsize \theta$ in radians if $\scriptsize 0\le \theta \le 6.282$ and $\scriptsize 2\cos \theta -1=0$.
(Note: This interval restriction is very often given as $\scriptsize 0\le \theta \le 2\pi$.)

Solution

We have been asked to solve for $\scriptsize \theta$ in radians. If we like, we can work in degrees and convert the final answer. However, let’s try work in radians.

Firstly, we need to recognise that the interval for $\scriptsize \theta$ is one revolution. $\scriptsize 6.282\approx 2\times \pi$ or one revolution. Therefore, this is the same interval as $\scriptsize {{0}^\circ}\le \theta \le {{360}^\circ}$.

\scriptsize \begin{align*}2\cos \theta -1 & =0\\\therefore \cos \theta & =\displaystyle \frac{1}{2}\end{align*}
Ref angle: $\scriptsize \theta =\displaystyle \frac{\pi }{3}$

Cosine is positive in the first and fourth quadrants.
$\scriptsize \theta =2\pi -\displaystyle \frac{\pi }{3}=\displaystyle \frac{{5\pi }}{3}$
General solution: $\scriptsize \theta =\displaystyle \frac{\pi }{3}+k.2\pi \text{ or }\theta =\displaystyle \frac{{5\pi }}{3}+k.2\pi ,k\in \mathbb{Z}$
In the interval $\scriptsize 0\le \theta \le 6.282$: $\scriptsize \theta =\displaystyle \frac{\pi }{3}\text{ or }\theta =\displaystyle \frac{{5\pi }}{3}$

Alternatively, you could have done all your working in degrees and converted your final answers of $\scriptsize \theta ={{60}^\circ}\text{ or }\theta ={{300}^\circ}$ to radians.

### Take note!

If you would like to work in radians, you need to change your calculator to work in radian mode. To change a Casio calculator, follow these steps:

1. If you see a little D or DEG at the top of your screen, it means that your calculator is working in degree mode.
2. Press the SHIFT key and then the MODE/SETUP key. You will see a full list of the available modes. Radians is normally option 4, so press 4 (or whatever option your calculator says is radians).
3. After this, you will see a little R or RAD at the top of your screen.
4. Go through the same process to change your calculator back to degree mode.

### Exercise 4.2

In each of the following equations, determine the value(s) of $\scriptsize \theta$ in radians if $\scriptsize 0\le \theta \le 2\pi$:

1. $\scriptsize \displaystyle \frac{1}{2}\cos \theta =0.435$
2. $\scriptsize \tan \left( {\theta -\displaystyle \frac{\pi }{6}} \right)=1.57$
3. $\scriptsize 2{{\sin }^{2}}\theta +\sin \theta =1$

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• That one radian is the angle created by moving one radius arc length along the circumference of a circle.
• To convert from degrees to radians, multiply by $\scriptsize \pi$ and divide by $\scriptsize {{180}^\circ}$.
• To convert from radians to degrees, multiply by $\scriptsize {{180}^\circ}$ and divide by $\scriptsize \pi$.

# Unit 4: Assessment

#### Suggested time to complete: 15 minutes

Question 1 and 2 adapted from NC(V) Level 4 Paper 2 November 2011 question 3.2 and November 2012 question 3.2 respectively

In each of the following, determine the value(s) of $\scriptsize \theta$ in radians if $\scriptsize 0\le \theta \le 6.282$.

1. $\scriptsize 6\cos \theta -5=\displaystyle \frac{4}{{\cos \theta }}$
2. $\scriptsize 4\sin \theta +3\tan \theta =\displaystyle \frac{3}{{\cos \theta }}+4$
3. $\scriptsize \cos 2\theta -\cos \theta +1=0$
4. $\scriptsize \cos 2\theta =1-3\cos \theta$

The full solutions are at the end of the unit.

# Unit 4: Solutions

### Exercise 4.1

1. .
1. $\scriptsize {{30}^\circ}=\displaystyle \frac{{{{{30}}^\circ}\times \pi }}{{{{{180}}^\circ}}}=\displaystyle \frac{\pi }{6}$
2. $\scriptsize {{75}^\circ}=\displaystyle \frac{{{{{75}}^\circ}\times \pi }}{{{{{180}}^\circ}}}=\displaystyle \frac{{5\pi }}{{12}}$
3. $\scriptsize {{135}^\circ}=\displaystyle \frac{{{{{135}}^\circ}\times \pi }}{{{{{180}}^\circ}}}=\displaystyle \frac{{3\pi }}{4}$
4. $\scriptsize {{215}^\circ}=\displaystyle \frac{{{{{215}}^\circ}\times \pi }}{{{{{180}}^\circ}}}=\displaystyle \frac{{43\pi }}{{36}}=1.194\pi$
2. .
1. $\scriptsize 5\pi =\displaystyle \frac{{5\pi \times {{{180}}^\circ}}}{\pi }={{900}^\circ}$
2. $\scriptsize \displaystyle \frac{\pi }{6}=\displaystyle \frac{{\pi \times {{{180}}^\circ}}}{{6\pi }}={{30}^\circ}$
3. $\scriptsize \displaystyle \frac{{7\pi }}{2}=\displaystyle \frac{{7\pi \times {{{180}}^\circ}}}{{2\pi }}={{630}^\circ}$
4. $\scriptsize {{\pi }^{2}}=\displaystyle \frac{{{{\pi }^{2}}\times {{{180}}^\circ}}}{\pi }=\pi \times {{180}^\circ}={{565.49}^\circ}$

Back to Exercise 4.1

### Exercise 4.2

1. .
\scriptsize \begin{align*}\displaystyle \frac{1}{2}\cos \theta &=0.435\\\therefore \cos \theta &=0.87\end{align*}
Ref angle: $\scriptsize 0.516$ (or $\scriptsize \theta ={{29.54}^\circ}$)
Cosine is positive in the first and fourth quadrants.
$\scriptsize \theta =2\pi -0.516=5.767$ radians (or $\scriptsize \theta ={{360}^\circ}-{{29.54}^\circ}={{330.46}^\circ}$)
Therefore, in the interval $\scriptsize 0\le \theta \le 6.282$, $\scriptsize \theta =0.516\text{ or }\theta =5.767$ radians.
2. .
$\scriptsize \tan \left( {\theta -\displaystyle \frac{\pi }{6}} \right)=1.57$
Ref angle: $\scriptsize \theta -\displaystyle \frac{\pi }{6}=1.004$ (or $\scriptsize \theta ={{57.51}^\circ}$)
Tangent is positive in the first and third quadrants.
$\scriptsize \theta -\displaystyle \frac{\pi }{6}=\pi +1.004=4.146$ (or $\scriptsize \theta ={{237.51}^\circ}$)
\scriptsize \begin{align*}\theta -\displaystyle \frac{\pi }{6} & =4.146+k.\pi ,k\in \mathbb{Z}\\\theta & =4.146+\displaystyle \frac{\pi }{6}+k.\pi ,k\in \mathbb{Z}\\\therefore \theta & =4.669+k.\pi ,k\in \mathbb{Z}\end{align*}
In the interval $\scriptsize 0\le \theta \le 6.282$: $\scriptsize \theta =1.528\text{ or }\theta =4.670$
3. .
\scriptsize \begin{align*}2{{\sin }^{2}}\theta +\sin \theta & =1\\\therefore 2{{\sin }^{2}}\theta +\sin \theta -1 & =0\\\therefore (2\sin \theta -1)(\sin \theta +1) & =0\\\therefore \sin \theta =\displaystyle \frac{1}{2}\quad & \text{or}\quad \text{sin}\theta =-1\end{align*}
$\scriptsize \sin \theta =\displaystyle \frac{1}{2}$:
Ref angle: $\scriptsize \theta =\displaystyle \frac{\pi }{6}$
Sine is positive in the first and second quadrants.
$\scriptsize \theta =\pi -\displaystyle \frac{\pi }{6}=\displaystyle \frac{{5\pi }}{6}$
General solution: $\scriptsize \theta =\displaystyle \frac{\pi }{6}+k.2\pi \text{ or }\theta =\displaystyle \frac{{5\pi }}{6}+k.2\pi ,k\in \mathbb{Z}$
In the interval $\scriptsize 0\le \theta \le 2\pi$: $\scriptsize \theta =\displaystyle \frac{\pi }{6}\text{ or }\theta =\displaystyle \frac{{5\pi }}{6}$
$\scriptsize \sin \theta =-1$:
Ref angle: $\scriptsize \theta =-\displaystyle \frac{\pi }{2}$
General solution: $\scriptsize \theta =-\displaystyle \frac{\pi }{2}+k.2\pi ,k\in \mathbb{Z}$
In the interval $\scriptsize 0\le \theta \le 2\pi$: $\scriptsize \theta =\displaystyle \frac{{3\pi }}{2}$

Back to Exercise 4.2

### Unit 4: Assessment

1. .
\scriptsize \displaystyle \begin{align*}6\cos \theta -5 & =\displaystyle \frac{4}{{\cos \theta }}\quad \quad \cos \theta \ne 0\therefore \theta \ne \displaystyle \frac{\pi }{2}+k.\pi ,k\in \mathbb{Z}\\\therefore 6{{\cos }^{2}}\theta -5\cos \theta -4 & =0\\\therefore (3\cos \theta -4)(2\cos \theta +1) & =0\\\therefore \cos \theta =\displaystyle \frac{4}{3}\text{ } & \text{or }\cos \theta =-\displaystyle \frac{1}{2}\end{align*}
$\scriptsize \cos \theta =\displaystyle \frac{4}{3}$: no solution
$\scriptsize \cos \theta =-\displaystyle \frac{1}{2}$:
Ref angle $\scriptsize \theta =\displaystyle \frac{{2\pi }}{3}$
Cosine is negative in the second and third quadrants.
$\scriptsize \theta =2\pi -\displaystyle \frac{{2\pi }}{3}=\displaystyle \frac{{4\pi }}{3}$
For the interval $\scriptsize 0\le \theta \le 6.282$: $\scriptsize \theta =\displaystyle \frac{{2\pi }}{3}\text{ or }\theta =\displaystyle \frac{{4\pi }}{3}$
2. .
\scriptsize \begin{align*}4\sin \theta +3\tan \theta & =\displaystyle \frac{3}{{\cos \theta }}+4\quad \quad \cos \theta \ne 0\therefore \theta \ne \displaystyle \frac{\pi }{2}+k.\pi ,k\in \mathbb{Z}\\\therefore 4\sin \theta +\displaystyle \frac{{3\sin \theta }}{{\cos \theta }} & =\displaystyle \frac{3}{{\cos \theta }}+4\\\therefore 4\sin \theta \cos \theta +3\sin \theta & =3+4\cos \theta \\\therefore 4\sin \theta \cos \theta +3\sin \theta -3-4\cos \theta & =0\\\therefore 4\cos \theta (\sin \theta -1)+3(\sin \theta -1) & =0\\\therefore (\sin \theta -1)(4\cos \theta +3) & =0\\\therefore \sin \theta =1\text{ } & \text{or }\cos \theta =-\displaystyle \frac{3}{4}\end{align*}
$\scriptsize \sin \theta =1$:
Ref angle: $\scriptsize \theta =\displaystyle \frac{\pi }{2}$
General solution: $\scriptsize \theta =\displaystyle \frac{\pi }{2}+k.2\pi ,k\in \mathbb{Z}$
$\scriptsize \cos \theta =-\displaystyle \frac{3}{4}$
Ref angle: $\scriptsize \theta =2.419$
Cosine is negative in the second and third quadrants.
$\scriptsize \theta =2\pi -2.419=3.864$
General solution: $\scriptsize \theta =2.419+k.2\pi \text{ or }\theta =3.864+k.2\pi ,k\in \mathbb{Z}$
For the interval $\scriptsize 0\le \theta \le 6.282$: $\scriptsize \theta =\displaystyle \frac{\pi }{2}\text{ or }\theta =2.419\text{ or }\theta =3.864$
3. .
\scriptsize \begin{align*}\cos 2\theta -\cos \theta +1 & =0\\\therefore 2{{\cos }^{2}}\theta -1-\cos \theta +1 & =0\\\therefore 2{{\cos }^{2}}-\cos \theta & =0\\\therefore \cos \theta (2\cos \theta -1) & =0\\\therefore \cos \theta =0\text{ or }\cos \theta & =\displaystyle \frac{1}{2}\end{align*}
$\scriptsize \cos \theta =0$:
Ref angle: $\scriptsize \theta =\displaystyle \frac{\pi }{2}$
General solution: $\scriptsize \theta =\displaystyle \frac{\pi }{2}+k.\pi ,k\in \mathbb{Z}$
$\scriptsize \cos \theta =\displaystyle \frac{1}{2}$:
Ref angle: $\scriptsize \theta =\displaystyle \frac{\pi }{3}$
Cosine is positive in the first and fourth quadrants.
$\scriptsize \theta =2\pi -\displaystyle \frac{\pi }{3}=\displaystyle \frac{{5\pi }}{3}$
General solution: $\scriptsize \theta =\displaystyle \frac{\pi }{3}+k.2\pi \text{ or }\theta =\displaystyle \frac{{5\pi }}{3}+k.2\pi ,k\in \mathbb{Z}$
For the interval $\scriptsize 0\le \theta \le 6.282$: $\scriptsize \theta =\displaystyle \frac{\pi }{2}\text{ or }\theta =\displaystyle \frac{{3\pi }}{2}\text{ or }\theta =\displaystyle \frac{\pi }{3}\text{ or }\theta =\displaystyle \frac{{5\pi }}{3}$
4. .
\scriptsize \displaystyle \begin{align*}\cos 2\theta & =1-3\cos \theta \\\therefore 2{{\cos }^{2}}\theta -1-1+3\cos \theta & =0\\\therefore 2{{\cos }^{2}}\theta +3\cos \theta -2 & =0\\\therefore (2\cos \theta -1)(\cos \theta +2) & =0\\\therefore \cos \theta =\displaystyle \frac{1}{2} & \text{ or }\cos \theta =-2\end{align*}
$\scriptsize \cos \theta =\displaystyle \frac{1}{2}$:
Ref angle: $\scriptsize \theta =\displaystyle \frac{\pi }{3}$
Cosine is positive in the first and fourth quadrants.
$\scriptsize \theta =2\pi -\displaystyle \frac{\pi }{3}=\displaystyle \frac{{5\pi }}{3}$
General solution: $\scriptsize \theta =\displaystyle \frac{\pi }{3}+k.2\pi \text{ or }\theta =\displaystyle \frac{{5\pi }}{3}+k.2\pi ,k\in \mathbb{Z}$
$\scriptsize \cos \theta =-2$: No solution
For the interval $\scriptsize 0\le \theta \le 6.282$: $\scriptsize \theta =\displaystyle \frac{\pi }{3}\text{or }\theta =\displaystyle \frac{{5\pi }}{3}$

Back to Unit 4: Assessment