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

# Unit 1: Trigonometric ratios

Dylan Busa

### Unit 1: Trigonometric ratios

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

• Define and use the trigonometric ratios of $\scriptsize \cos \theta$, $\scriptsize \sin \theta$ and $\scriptsize \tan \theta$.
• Calculate the trigonometric ratios in each quadrant of the Cartesian plane.
• Calculate the value of expressions containing trigonometric ratios.

## What you should know

Before you start this unit, make sure you can:

• Use the Theorem of Pythagoras to find missing lengths in right-angled triangles. Go over Subject outcome 3.5 Unit 1 if you need help with this.

Here is a short self-assessment to make sure you have the skills you need to proceed with this unit.

In $\scriptsize \Delta \text{ABC}$, $\scriptsize \hat{\text{A}}=90{}^\circ$, $\scriptsize \text{AB}=4.5\,\text{cm}$ and $\scriptsize \text{BC}=7\,\text{cm}$. Calculate the length of $\scriptsize \text{AC}$.

Solution

$\scriptsize \Delta \text{ABC}$ is a right-angled triangle. Therefore, we can use the Theorem of Pythagoras to calculate the length of the $\scriptsize \text{AC}$.
\scriptsize \begin{align*}\text{B}{{\text{C}}^{2}} & =\text{A}{{\text{B}}^{2}}+\text{A}{{\text{C}}^{2}}\\\therefore \text{A}{{\text{C}}^{2}} & =\text{B}{{\text{C}}^{2}}-\text{A}{{\text{B}}^{2}}\\\therefore \text{AC} & =\sqrt{{\text{B}{{\text{C}}^{2}}-\text{A}{{\text{B}}^{2}}}}\\ & =\sqrt{{{{7}^{2}}-{{{4.5}}^{2}}}}\\ & =\sqrt{{49-20.25}}\\ & =\sqrt{{28.75}}\\ & =5.36\end{align*}

## Introduction

What is trigonometry (or ‘trig’ as it is sometimes called)? A clue lies in the first part of the word ‘trigonometry’. ‘Tri’ means three; as in tricycle (three wheels) and triangle (three angles and three sides). Trigonometry deals with the relationships between the angles and sides of triangles.

### Did you know?

The word trigonometry comes from the Greek words for triangle (trigōnon) and measure (metron).

## Trigonometry

From about the year 150 A.D., Egyptians and Greeks used trigonometry to measure the distances between objects and places they could not measure directly. They even used it to measure the distances between stars and the circumference of the Earth. Trigonometry uses a technique called triangulation to measure distances.

Trigonometry has, therefore, always played a key part in navigation. Many of the modern applications of trigonometry still have to do with navigation. Modern satellite navigation systems still use trigonometry and triangulation to find the distances between landmarks.

Other fields that use trigonometry include acoustics, optics, financial market analysis, electronics, probability, statistics, biology, medical imaging, chemistry, cryptology, meteorology, oceanography, land surveying, architecture, phonetics, engineering, computer graphics, and game development, amongst many others.

To lay the foundation for understanding what trigonometry is and how it works, try this next activity.

### Activity 1.1: The sides in similar triangles

Time required: 20 minutes

What you need:

• a pen or pencil
• paper
• a protractor
• a ruler

What to do:

1. Draw three triangles of different sizes using a protractor and a ruler so that each triangle has interior angles equal to $\scriptsize 30{}^\circ$, $\scriptsize 60{}^\circ$ and $\scriptsize 90{}^\circ$ as shown below. We call these ‘similar triangles’ because they are all the same shape, but not the same size. If you do not have a protractor or ruler, you can measure the lengths of the triangles below instead of your own.
2. Now measure the angles and lengths accurately and complete the following table. Leave all your ratios as fractions for now.
3. What do you notice about each of the ratios? You can convert each ratio to a decimal if this helps you to compare. Does it matter what the lengths of the sides of the triangles are if the angles inside the three triangles stay the same size?
4. Draw one more right-angled triangle but this time make the other two angles $\scriptsize 40{}^\circ$ and $\scriptsize 50{}^\circ$. Measure the lengths of the sides as before. Are the sides in the same ratio as the first three triangles that you drew?

What did you find?

You should have found that the value of each of the ratios of corresponding sides in each triangle was always the same. In other words, $\scriptsize \displaystyle \frac{{\text{AB}}}{{\text{BC}}}=\displaystyle \frac{{\text{DE}}}{{\text{EF}}}=\displaystyle \frac{{\text{GH}}}{{\text{HK}}}$, $\scriptsize \displaystyle \frac{{\text{AB}}}{{\text{AC}}}=\displaystyle \frac{{\text{DE}}}{{\text{DF}}}=\displaystyle \frac{{\text{GH}}}{{\text{GK}}}$ and $\scriptsize \displaystyle \frac{{\text{BC}}}{{\text{AC}}}=\displaystyle \frac{{\text{EF}}}{{\text{DF}}}=\displaystyle \frac{{\text{HK}}}{{\text{GK}}}$.

This means that it is the size of the angles in a triangle that determines the ratios between the lengths of the sides. It does not matter how big or small the triangle is. If the sizes of the angles stay the same, the sides will always be in the same ratio. If we change the sizes of the angles, we change the ratio of the sides.

## The basic trigonometric ratios

We use the fact that it is the size of the angles in a triangle that determine the ratio of the sides as the basis of trigonometry and to define the three basic trigonometric ratios.

Have a look at the right-angled triangle in Figure 2. We can name the sides of the triangle with reference to the angle called $\scriptsize \theta$ (theta) at $\scriptsize \text{A}$ as follows. The hypotenuse ($\scriptsize \text{AB}$) is always the side opposite the right-angle. The side next to $\scriptsize \theta$ ($\scriptsize \text{AC}$) is called the adjacent side (adjacent means next to) and the side opposite $\scriptsize \theta$ ($\scriptsize \text{BC}$) is called the opposite side.

If we defined the sides with respect to $\scriptsize \beta$ (beta) at $\scriptsize \text{B}$, then the adjacent side would be $\scriptsize \text{BC}$ and the opposite side would be $\scriptsize \text{AC}$. The hypotenuse would remain unchanged.

### Note

The definitions of opposite, adjacent and hypotenuse are only applicable when working with right-angled triangles. Always check to make sure your triangle has a right-angle before you use them.

In Activity 1.1, we saw that we can define the three ratios of the lengths of the sides within any right-angled triangle. We can give each of these three ratios a special name – sine, cosine and tangent – all with respect to the angle $\scriptsize \theta$:

$\scriptsize \sin \theta =\displaystyle \frac{{\text{length of the opposite side}}}{{\text{length of the hypotenuse}}}$
$\scriptsize \cos \theta =\displaystyle \frac{{\text{length of the adjacent side}}}{{\text{length of the hypotenuse}}}$
$\scriptsize \tan \theta =\displaystyle \frac{{\text{length of the adjacent side}}}{{\text{length of the opposite side}}}$

In $\scriptsize \Delta \text{ABC}$ in Figure 2, these three ratios would be defined as:

$\scriptsize \sin \theta =\displaystyle \frac{{\text{opposite}}}{{\text{hypotenuse}}}=\displaystyle \frac{{\text{BC}}}{{\text{AB}}}$
$\scriptsize \cos \theta =\displaystyle \frac{{\text{adjacent}}}{{\text{hypotenuse}}}=\displaystyle \frac{{\text{AC}}}{{\text{AB}}}$
$\scriptsize \tan \theta =\displaystyle \frac{{\text{opposite}}}{{\text{adjacent}}}=\displaystyle \frac{{\text{BC}}}{{\text{AC}}}$

In $\scriptsize \Delta \text{ABC}$ in Figure 2, we could have also defined the three ratios with respect to $\scriptsize \beta$. Remembering that the sides BC and AC would then be ‘adjacent’ and ‘opposite’, reflecting their position relative to angle $\scriptsize \beta$, the ratios of $\scriptsize \beta$ are:

$\scriptsize \sin \beta =\displaystyle \frac{{\text{opposite}}}{{\text{hypotenuse}}}=\displaystyle \frac{{\text{AC}}}{{\text{AB}}}$
$\scriptsize \cos \beta =\displaystyle \frac{{\text{adjacent}}}{{\text{hypotenuse}}}=\displaystyle \frac{{\text{BC}}}{{\text{AB}}}$
$\scriptsize \tan \beta =\displaystyle \frac{{\text{opposite}}}{{\text{adjacent}}}=\displaystyle \frac{{\text{AC}}}{{\text{BC}}}$

These three ratios of sine, cosine and tangent form the basis of all of trigonometry.

### Note

The three basic trigonometric ratios are:

$\scriptsize \sin \theta =\displaystyle \frac{{\text{opposite}}}{{\text{hypotenuse}}}$ $\scriptsize \cos \theta =\displaystyle \frac{{\text{adjacent}}}{{\text{hypotenuse}}}$ $\scriptsize \tan \theta =\displaystyle \frac{{\text{opposite}}}{{\text{adjacent}}}$

We never define the trigonometric ratios with respect to the right-angle.

### Example 1.1

1. State the three basic trigonometric ratios for the following triangles with respect to the angle $\scriptsize \theta$.
1. .
2. .
3. .
2. Complete each of the following for the given triangle:
1. $\scriptsize \sin \hat{\text{A}}\,$
2. $\scriptsize \cos \hat{\text{A}}\,$
3. $\scriptsize \tan \hat{\text{A}}\,$
4. $\scriptsize \sin \hat{\text{C}}\,$
5. $\scriptsize \cos \hat{\text{C}}\,$
6. $\scriptsize \tan \hat{\text{C}}\,$

Solutions

1. .
1. $\scriptsize \sin \theta =\displaystyle \frac{{\text{opp}}}{{\text{hyp}}}=\displaystyle \frac{{\text{EF}}}{{\text{DE}}}$
$\scriptsize \cos \theta =\displaystyle \frac{{\text{adj}}}{{\text{hyp}}}=\displaystyle \frac{{\text{DF}}}{{\text{DE}}}$
$\scriptsize \tan \theta =\displaystyle \frac{{\text{opp}}}{{\text{adj}}}=\displaystyle \frac{{\text{EF}}}{{\text{DF}}}$
2. $\scriptsize \sin \theta =\displaystyle \frac{{\text{opp}}}{{\text{hyp}}}=\displaystyle \frac{{\text{XZ}}}{{\text{XY}}}$
$\scriptsize \cos \theta =\displaystyle \frac{{\text{adj}}}{{\text{hyp}}}=\displaystyle \frac{{\text{YZ}}}{{\text{XY}}}$
$\scriptsize \tan \theta =\displaystyle \frac{{\text{opp}}}{{\text{adj}}}=\displaystyle \frac{{\text{XZ}}}{{\text{YZ}}}$
3. $\scriptsize \sin \theta =\displaystyle \frac{{\text{opp}}}{{\text{hyp}}}=\displaystyle \frac{{\text{AB}}}{{\text{AC}}}$
$\scriptsize \cos \theta =\displaystyle \frac{{\text{adj}}}{{\text{hyp}}}=\displaystyle \frac{{\text{BC}}}{{\text{AC}}}$
$\scriptsize \tan \theta =\displaystyle \frac{{\text{opp}}}{{\text{adj}}}=\displaystyle \frac{{\text{AB}}}{{\text{BC}}}$
2. .
1. $\scriptsize \sin \text{A}=\displaystyle \frac{{\text{opp}}}{{\text{hyp}}}=\displaystyle \frac{{\text{BC}}}{{\text{AC}}}$
2. $\scriptsize \cos \text{A}=\displaystyle \frac{{\text{adj}}}{{\text{hyp}}}=\displaystyle \frac{{\text{AB}}}{{\text{AC}}}$
3. $\scriptsize \tan \text{A}=\displaystyle \frac{{\text{opp}}}{{\text{adj}}}=\displaystyle \frac{{\text{BC}}}{{\text{AB}}}$
4. $\scriptsize \sin \text{C}=\displaystyle \frac{{\text{opp}}}{{\text{hyp}}}=\displaystyle \frac{{\text{AB}}}{{\text{AC}}}$
5. $\scriptsize \cos \text{C}=\displaystyle \frac{{\text{adj}}}{{\text{hyp}}}=\displaystyle \frac{{\text{BC}}}{{\text{AC}}}$
6. $\scriptsize \tan \text{C}=\displaystyle \frac{{\text{opp}}}{{\text{adj}}}=\displaystyle \frac{{\text{AB}}}{{\text{BC}}}$

### Note

You can use Soh Cah Toa to help you remember how each of the trig ratios are defined.

### Exercise 1.1

1. Define $\scriptsize x$ in this triangle in three different ways.
2. Which of the following statements about $\scriptsize \beta$ in the triangle is correct?
1. $\scriptsize \cos \beta =\displaystyle \frac{4}{5}$
2. $\scriptsize \sin \beta =\displaystyle \frac{4}{5}$
3. $\scriptsize \cos \left( {\displaystyle \frac{3}{5}} \right)=\beta$
4. $\scriptsize \tan \beta =0.75$

The full solutions are at the end of the unit.

## Evaluate basic trigonometric ratios

Have a look at this triangle from exercise 1.1 question 2 again.

We can see that $\scriptsize \sin \beta =\displaystyle \frac{4}{5}$. This means that the expression $\scriptsize \sin \beta$ has a definite value and it depends on the size of $\scriptsize \beta$. In the triangle above, what do you think $\scriptsize \sin \beta +\sin \beta$ is equal to?

Hopefully you said that $\scriptsize \sin \beta +\sin \beta =\displaystyle \frac{4}{5}+\displaystyle \frac{4}{5}=\displaystyle \frac{8}{5}$.
We could also say that $\scriptsize \sin \beta +\sin \beta =2\sin \beta =2\times \displaystyle \frac{4}{5}=\displaystyle \frac{8}{5}$.

When we do calculations involving the three basic trig ratios, we can treat them the same way we treat numbers and variables. So long as we do not split the trig ratio from its angle, they obey all the normal arithmetic rules.

Therefore, $\scriptsize \sin \theta +\sin \theta =2\sin \theta$ the same way that $\scriptsize x+x=2x$. We treat $\scriptsize \sin \theta$ as a single entity. We do not split it up. This means that $\scriptsize \sin \theta +\sin \theta \ne \sin 2\theta$. $\scriptsize \sin 2\theta$ means that we are finding the sine of a bigger angle the size of two times the angle $\scriptsize \theta$.

$\scriptsize \cos \alpha \times \cos \alpha \times \cos \alpha ={{\cos }^{3}}\alpha$ the same way that $\scriptsize x\times x\times x={{x}^{3}}$. We position the exponent above the trig ratio.

This means that $\scriptsize \cos \alpha \times \cos \alpha \times \cos \alpha \ne \cos {{\alpha }^{3}}$. $\scriptsize \cos {{\alpha }^{3}}$ means that we are finding the cosine of the cube of angle $\scriptsize \alpha$. We could say that $\scriptsize \cos \alpha \times \cos \alpha \times \cos \alpha ={{\left( {\cos \alpha } \right)}^{3}}$ but we tend not to write it this way.

### Note

• $\scriptsize \sin \theta +\sin \theta =2\sin \theta$
• $\scriptsize 3\sin \theta -\sin \theta =2\sin \theta$
• $\scriptsize \sin \theta \times \sin \theta ={{\sin }^{2}}\theta$
• $\scriptsize \displaystyle \frac{{{{{\sin }}^{3}}\theta }}{{\sin \theta }}={{\sin }^{2}}\theta$
• $\scriptsize \displaystyle \frac{{2\sin \theta }}{{\sin \theta }}=2$

If you ever get confused, you can always replace the trig ratio and its operative angle with $\scriptsize x$.

### Example 1.2

Given $\scriptsize \Delta \text{ABC}$ with $\scriptsize \overset{\wedge }{\mathop{\text{C}}}\,=90{}^\circ$, $\scriptsize \overset{\wedge }{\mathop{\text{A}}}\,=\theta$ and $\scriptsize \overset{\wedge }{\mathop{\text{B}}}\,=\alpha$.

Determine the value of the following, showing all calculations.

1. $\scriptsize \cos \alpha$
2. $\scriptsize \tan \theta$
3. $\scriptsize 1+{{\tan }^{2}}\theta$
4. $\scriptsize \displaystyle \frac{{\sin \alpha }}{{\cos \alpha }}$

Solutions

1. $\scriptsize \cos \alpha =\displaystyle \frac{{\text{adj}}}{{\text{hyp}}}=\displaystyle \frac{{\text{BC}}}{{\text{AB}}}=\displaystyle \frac{4}{5}$
2. $\scriptsize \tan \theta =\displaystyle \frac{\text{opp}}{\text{adj}}=\displaystyle \frac{{\text{BC}}}{{\text{AC}}}$. But we do not have the length of $\scriptsize \text{AC}$, so we can use Pythagoras’ Theorem to find $\scriptsize \text{AC}$.\scriptsize \begin{align*}\text{A}{{\text{B}}^{2}} & =\text{A}{{\text{C}}^{2}}+\text{B}{{\text{C}}^{2}}\\\therefore \text{A}{{\text{C}}^{2}} & =\text{A}{{\text{B}}^{2}}-\text{B}{{\text{C}}^{2}}\\\therefore \text{AC} & =\sqrt{{\text{A}{{\text{B}}^{2}}-\text{B}{{\text{C}}^{2}}}}\\ & =\sqrt{{{{5}^{2}}-{{4}^{2}}}}\\ & =\sqrt{{25-16}}\\ & =\sqrt{9}\\ & =3\end{align*}
Therefore, $\scriptsize \tan \theta =\displaystyle \frac{{opp}}{{adj}}=\displaystyle \frac{{\text{BC}}}{{\text{AC}}}=\displaystyle \frac{4}{3}$
3. .
\scriptsize \begin{align*}1+{{\tan }^{2}}\theta & =1+{{\left( {\displaystyle \frac{4}{3}} \right)}^{2}}\\=1+\displaystyle \frac{{16}}{9}\\=\displaystyle \frac{{9+16}}{9}\\=\displaystyle \frac{{25}}{9}\end{align*}
4. .
$\scriptsize \displaystyle \frac{{\sin \alpha }}{{\cos \alpha }}=\displaystyle \frac{{\displaystyle \frac{3}{5}}}{{\displaystyle \frac{4}{5}}}=\displaystyle \frac{3}{5}\times \displaystyle \frac{5}{4}=\displaystyle \frac{3}{4}$

### Exercise 1.2

Given $\scriptsize \Delta \text{ABC}$ with $\scriptsize \hat{\text{C}}\,=90^{\circ}$, $\scriptsize \hat{\text{A}}\,=\theta$ and $\scriptsize \hat{\text{B}}\,=\alpha$.

Determine the value of the following, showing all calculations.

1. $\scriptsize \sin \alpha$
2. $\scriptsize \cos \theta +\tan \theta$
3. $\scriptsize 1+{{\sin }^{2}}\theta$
4. $\scriptsize {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha$

The full solutions are at the end of the unit.

## The trigonometric ratios on the Cartesian plane

One of the many powerful features of trigonometry is that it helps us explore things that repeat or cycle. To see why this is the case, we need to extend the definitions of the basic trig ratios to the Cartesian plane.

Let’s start by plotting a point on the Cartesian plane. Let’s plot the point $\scriptsize (3,4)$ and then join this point to the origin with a straight line (see Figure 3). Let’s also call the angle made between this line and the x-axis $\scriptsize \alpha$.

Now, if we drop a perpendicular line from $\scriptsize \text{A}$ to the x-axis, we will form a right-angled triangle (see Figure 4). With respect to $\scriptsize \alpha$, we can see that the length of the opposite side is given by the y-coordinate of $\scriptsize \text{A}$ which is $\scriptsize 4$ units and the length of the adjacent side is given by the x-coordinate of $\scriptsize \text{A}$ which is $\scriptsize 3$ units. So, in this triangle we can say that $\scriptsize \tan \alpha =\displaystyle \frac{{\text{opp}}}{{\text{adj}}}=\displaystyle \frac{y}{x}=\displaystyle \frac{4}{3}$.

We can now use Pythagoras’ Theorem to find that the length of the hypotenuse is $\scriptsize 5$ units, which we will call $\scriptsize r$. Hence, we see that $\scriptsize \sin \alpha =\displaystyle \frac{{\text{opp}}}{{\text{hyp}}}=\displaystyle \frac{y}{r}=\displaystyle \frac{4}{5}$ and $\scriptsize \cos \alpha =\displaystyle \frac{{\text{adj}}}{{\text{hyp}}}=\displaystyle \frac{x}{r}=\displaystyle \frac{3}{5}$.

Now, what if we draw a point $\scriptsize \text{A }\!\!'\!\!\text{ }$ symmetrical to $\scriptsize \text{A}$ but in the second quadrant? The coordinates of $\scriptsize \text{A }\!\!'\!\!\text{ }$ would be $\scriptsize (4,-3)$. We could drop a perpendicular from $\scriptsize \text{A }\!\!'\!\!\text{ }$ to the x-axis. Obviously, the angle created inside this right-angled triangle would be the same size as $\scriptsize \alpha$. Let’s call it $\scriptsize a$ (see Figure 5). But what would the values of the trig ratios be for this second quadrant triangle?

The length of $\scriptsize r$ (the hypotenuse) would still be $\scriptsize 5$ because it is a length. Therefore, $\scriptsize \sin a=\displaystyle \frac{{\text{opp}}}{{\text{hyp}}}=\displaystyle \frac{y}{r}=\displaystyle \frac{4}{5}$.

However, $\scriptsize \cos a=\displaystyle \frac{{\text{adj}}}{{\text{hyp}}}=\displaystyle \frac{x}{r}=\displaystyle \frac{{-3}}{5}=-\displaystyle \frac{3}{5}$ and $\scriptsize \tan a=\displaystyle \frac{{\text{opp}}}{{\text{adj}}}=\displaystyle \frac{y}{x}=\displaystyle \frac{4}{{-3}}=-\displaystyle \frac{4}{3}$.

The values of the ratios for cosine and tangent are numerically the same but are negative because the value of $\scriptsize x$ in the second quadrant is negative. Only sine is positive.

What do you think will happen to the signs of the three trig ratios if we reflect point $\scriptsize \text{A}$ into the third and fourth quadrants? Make your own sketches to see if you can figure it out before you read on.

In the third quadrant, the values of $\scriptsize x$ and $\scriptsize y$ are both negative. Remember $\scriptsize r$ is a length so is always positive.

$\scriptsize \sin a=\displaystyle \frac{{\text{opp}}}{{\text{hyp}}}=\displaystyle \frac{y}{r}=\displaystyle \frac{{-4}}{5}=-\displaystyle \frac{4}{5}$, $\scriptsize \cos a=\displaystyle \frac{{\text{adj}}}{{\text{hyp}}}=\displaystyle \frac{x}{r}=\displaystyle \frac{{-3}}{5}=-\displaystyle \frac{3}{5}$ and $\scriptsize \tan a=\displaystyle \frac{{\text{opp}}}{{\text{adj}}}=\displaystyle \frac{y}{x}=\displaystyle \frac{{-4}}{{-3}}=\displaystyle \frac{4}{3}$.

This time only tangent is positive (see Figure 6).

In the fourth quadrant, the $\scriptsize y$ value is negative. Therefore,
$\scriptsize \sin a=\displaystyle \frac{{\text{opp}}}{{\text{hyp}}}=\displaystyle \frac{y}{r}=\displaystyle \frac{{-4}}{5}=-\displaystyle \frac{4}{5}$, $\scriptsize \cos a=\displaystyle \frac{{\text{adj}}}{{\text{hyp}}}=\displaystyle \frac{x}{r}=\displaystyle \frac{3}{5}=\displaystyle \frac{3}{5}$ and $\scriptsize \tan a=\displaystyle \frac{{\text{opp}}}{{\text{adj}}}=\displaystyle \frac{y}{x}=\displaystyle \frac{{-4}}{3}=-\displaystyle \frac{4}{3}$.

This time only cosine is positive (see Figure 6).

What we have actually done is to draw a circle, radius $\scriptsize r=5$, with its centre at the origin (see Figure 7). As we move point $\scriptsize \text{A}$ around the circumference of the circle, the radius makes an angle $\scriptsize \alpha$ with the positive x-axis. At any time, we can drop a perpendicular from $\scriptsize \text{A}$ to the x-axis to create a right-angled triangle and calculate the trig ratios of this angle $\scriptsize \alpha$.

As we have discovered, if $\scriptsize \text{A}$ is in the first quadrant ($\scriptsize 0{}^\circ \lt \alpha \lt 90{}^\circ$), then all the trig ratios are positive. If $\scriptsize \text{A}$ is in the second quadrant ($\scriptsize 90{}^\circ \lt \alpha \lt 180{}^\circ$) then only sine is positive. If $\scriptsize \text{A}$ is in the third quadrant ($\scriptsize 180{}^\circ \lt \alpha \lt 270{}^\circ$) then only tangent is positive and if $\scriptsize \text{A}$ is in the fourth quadrant ($\scriptsize 270{}^\circ \lt \alpha \lt 360{}^\circ$), then only cosine is positive.

There is a useful tool called the CAST diagram (see Figure 8) that helps us remember which trig ratios are positive in which quadrants.

### Take note!

On the Cartesian plane, we define the trig ratios based on the coordinates of a point $\scriptsize \text{P}(x,y)$ and the angle ($\scriptsize \theta$) made between the positive x-axis and the line ($\scriptsize r$) drawn from the origin to $\scriptsize \text{P}$.

$\scriptsize \sin \theta =\displaystyle \frac{y}{r}$ $\scriptsize \cos \theta =\displaystyle \frac{x}{r}$ $\scriptsize \tan \theta =\displaystyle \frac{y}{x}$

If you have an internet connection, spend some time playing with the interactive simulation called CAST diagram.

Click and drag point $\scriptsize \text{P}$ to move it around the circle and notice how the values of the trig ratios change. Pay special attention to how their signs change as $\scriptsize \text{P}$ moves from one quadrant to another.

### Example 1.3

1. If $\scriptsize \sin \theta =\displaystyle \frac{3}{5}$ and $\scriptsize 90{}^\circ \le \theta \le 180{}^\circ$ determine:
1. $\scriptsize \cos \theta$
2. $\scriptsize {{\tan }^{2}}\theta +1$
2. If $\scriptsize \tan \beta =-\sqrt{3}$ and $\scriptsize 270{}^\circ \le \beta \le 360{}^\circ$ determine:
1. $\scriptsize \sin \beta +\cos \beta$
2. $\scriptsize {{\sin }^{2}}\beta +{{\cos }^{2}}\beta$
3. $\scriptsize \displaystyle \frac{1}{{2\cos \beta }}$

Solutions

1. It is always best to make a quick sketch of the situation. We are told that $\scriptsize 90{}^\circ \le \theta \le 180{}^\circ$. This means that we are working in the second quadrant. We are also told that $\scriptsize \sin \theta =\displaystyle \frac{3}{5}$. This means that $\scriptsize r=5$ and the y-coordinate of the point on the circumference of the circle is $\scriptsize 3$.

We can use Pythagoras’ Theorem to calculate the x-coordinate of $\scriptsize \text{A}$.
\scriptsize \begin{align*}{{r}^{2}} & ={{x}^{2}}+{{y}^{2}}\\\therefore {{x}^{2}} & ={{r}^{2}}-{{y}^{2}}\\\therefore x & =\sqrt{{{{r}^{2}}-{{y}^{2}}}}\\&=\sqrt{{25-9}}\\&=\sqrt{16}\\&=\pm4\end{align*}
But this is a point in the second quadrant so $\scriptsize x=-4$.
1. $\scriptsize \cos \theta =\displaystyle \frac{x}{r}=\displaystyle \frac{{-4}}{5}=-\displaystyle \frac{4}{5}$
2. $\scriptsize \tan \theta =\displaystyle \frac{y}{x}=\displaystyle \frac{3}{{-4}}=-\displaystyle \frac{3}{4}$
\scriptsize \begin{align*}\therefore {{\tan }^{2}}\theta +1 & ={{\left( {-\displaystyle \frac{3}{4}} \right)}^{2}}+1\\&=\displaystyle \frac{9}{{16}}+1\\&=\displaystyle \frac{{9+16}}{{16}}\\&=\displaystyle \frac{{25}}{{16}}\end{align*}
2. We are told that $\scriptsize 270{}^\circ \le \beta \le 360{}^\circ$. This means that we are working in the fourth quadrant. We are also told that $\scriptsize \tan \beta =-\sqrt{3}=-\displaystyle \frac{{\sqrt{3}}}{1}$. This means that the y-coordinate of the point on the circumference of the circle is $\scriptsize -\sqrt{3}$ and the x-coordinate is $\scriptsize 1$.

We can use Pythagoras’ Theorem to calculate the length of the radius of the circle.
\scriptsize \begin{align*}{{r}^{2}} & ={{1}^{2}}+{{\left( {-\sqrt{3}} \right)}^{2}}\\\therefore r & =\sqrt{{1+3}}=2\end{align*}
1. .
\scriptsize \begin{align*}\sin \beta +\cos \beta & =\displaystyle \frac{{-\sqrt{3}}}{2}+\displaystyle \frac{1}{2}\\ &=\displaystyle \frac{{-\sqrt{3}+1}}{2}\end{align*}
2. .
\scriptsize \begin{align*}{{\sin }^{2}}\beta +{{\cos }^{2}}\beta & ={{\left( {\displaystyle \frac{{-\sqrt{3}}}{2}} \right)}^{2}}+{{\left( {\displaystyle \frac{1}{2}} \right)}^{2}}\\ &=\displaystyle \frac{3}{4}+\displaystyle \frac{1}{4}\\ &=1\end{align*}
3. .
\scriptsize \begin{align*}\displaystyle \frac{1}{{2\cos \beta }} &=\displaystyle \frac{1}{{2\left( {\displaystyle \frac{1}{2}} \right)}}\\ &=\displaystyle \frac{1}{1}\\ &=1\end{align*}

### Exercise 1.3

1. $\scriptsize \cos \alpha =-\displaystyle \frac{{\sqrt{3}}}{2}$ and $\scriptsize 0{}^\circ \le \alpha \le 180{}^\circ$. Determine:
1. $\scriptsize {{\tan }^{2}}\alpha -1$
2. $\scriptsize 3\sin \alpha +2(1+\cos \alpha )$
2. If $\scriptsize \tan \theta =\displaystyle \frac{6}{8}$ and $\scriptsize 90{}^\circ \le \theta \le 360{}^\circ$, show that $\scriptsize \sin \theta (1+\cos \theta )=\displaystyle \frac{{-3}}{{25}}$.

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• The length of the sides of similar triangles are always in the same ratios.
• We define the three basic trig ratios in a right-angled triangle as:$\scriptsize \sin \theta =\displaystyle \frac{{\text{opp}}}{{\text{hyp}}}$ $\scriptsize \cos \theta =\displaystyle \frac{{\text{adj}}}{{\text{hyp}}}$ $\scriptsize \tan \theta =\displaystyle \frac{{\text{opp}}}{{\text{adj}}}$.
• We can define the three basic trig ratios on the Cartesian plane as:$\scriptsize \sin \alpha =\displaystyle \frac{y}{r}$ $\scriptsize \cos \alpha =\displaystyle \frac{x}{r}$ $\scriptsize \tan \alpha =\displaystyle \frac{y}{x}$
• All three basic trig ratios are positive in the first quadrant. Only sine is positive in the second quadrant. Only tangent is positive in the third quadrant. Only cosine is positive in the fourth quadrant.

# Unit 1: Assessment

### Suggested time to complete: 35 minutes

1. Given $\scriptsize \Delta \text{XYZ}$ as shown.
1. Determine the value of:
1. $\scriptsize \sin \theta$
2. $\scriptsize {{\cos }^{2}}\alpha$
2. Show that $\scriptsize \left( {1+\tan \theta } \right)\left( {1-\tan \theta } \right)=1-{{\tan }^{2}}\theta$
2. $\scriptsize \cos \theta =\displaystyle \frac{7}{{25}}$ and $\scriptsize 90{}^\circ \le \theta \le 360{}^\circ$. Determine the value of:
1. $\scriptsize \sin \theta$
2. $\scriptsize \tan \theta -\cos \theta$
3. $\scriptsize \displaystyle \frac{{{{{\tan }}^{2}}\theta }}{{{{{\sin }}^{2}}\theta }}$

The full solutions are at the end of the unit.

# Unit 1: Solutions

### Exercise 1.1

1. $\scriptsize \sin x=\displaystyle \frac{{\sqrt{3}}}{2}$ $\scriptsize \cos x=\displaystyle \frac{1}{2}$ $\scriptsize \tan x=\displaystyle \frac{{\sqrt{3}}}{1}=\sqrt{3}$
2. .
1. $\scriptsize \cos \beta =\displaystyle \frac{{\text{adj}}}{{\text{hyp}}}=\displaystyle \frac{3}{5}\ne \displaystyle \frac{4}{5}$ Therefore, this is not correct.
2. $\scriptsize \sin \beta =\displaystyle \frac{{\text{opp}}}{{\text{hyp}}}=\displaystyle \frac{4}{5}$ Therefore, this is correct.
3. $\scriptsize \cos \left( {\displaystyle \frac{3}{5}} \right)=\beta$ is not correctly stated. The trig ratios are defined in terms of the angle. In this case the angle is $\scriptsize \beta$ and the ratio is $\scriptsize \displaystyle \frac{3}{5}$.
4. $\scriptsize \tan \beta =\displaystyle \frac{{\text{opp}}}{{\text{adj}}}=\displaystyle \frac{4}{3}\ne \displaystyle \frac{3}{4}$ or $\scriptsize 0.75$. Therefore, this is not correct.
Therefore only option b. is correct.

Back to Exercise 1.1

### Exercise 1.2

1. .
1. .
\scriptsize \begin{align*}\text{A}{{\text{B}}^{2}} & =\text{A}{{\text{C}}^{2}}+\text{B}{{\text{C}}^{2}}\\\therefore \text{A}{{\text{C}}^{2}} & =\text{A}{{\text{B}}^{2}}-\text{B}{{\text{C}}^{2}}\\\therefore \text{AC} & =\sqrt{{\text{A}{{\text{B}}^{2}}-\text{B}{{\text{C}}^{2}}}}\\ & =\sqrt{{25-16}}\\ & =3\end{align*}
$\scriptsize \sin \alpha =\displaystyle \frac{3}{5}$
2. .
$\scriptsize \sin \alpha =\displaystyle \frac{3}{5}$
\scriptsize \begin{align*}\cos \theta +\tan \theta &=\displaystyle \frac{3}{5}+\displaystyle \frac{3}{4}\\&=\displaystyle \frac{{15+12}}{{20}}\\&=\displaystyle \frac{{37}}{{20}}\end{align*}
3. .
\scriptsize \begin{align*}1+{{\sin }^{2}}\theta &=1+{{\left( {\displaystyle \frac{4}{5}} \right)}^{2}}\\&=1+\displaystyle \frac{{16}}{{25}}\\&=\displaystyle \frac{{25+16}}{{25}}\\&=\displaystyle \frac{{41}}{{25}}\end{align*}
4. .
\scriptsize \begin{align*}{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha &={{\left( {\displaystyle \frac{3}{5}} \right)}^{2}}+{{\left( {\displaystyle \frac{4}{5}} \right)}^{2}}\\&=\displaystyle \frac{9}{{25}}+\displaystyle \frac{6}{{25}}\\&=\displaystyle \frac{{25}}{{25}}\\&=1\end{align*}

Back to Exercise 1.2

### Exercise 1.3

1. $\scriptsize \cos \alpha =-\displaystyle \frac{{\sqrt{3}}}{2}$ and $\scriptsize 0{}^\circ \le \alpha \le 180{}^\circ$. Therefore, the angle is in the second quadrant, $\scriptsize r=2$ and the x-coordinate of the point on the circle is $\scriptsize -\sqrt{3}$.

\scriptsize \begin{align*}{{r}^{2}} & ={{x}^{2}}+{{y}^{2}}\\\therefore {{y}^{2}} & ={{r}^{2}}-{{x}^{2}}\\\therefore y & =\sqrt{{{{r}^{2}}-{{x}^{2}}}}\\&=\sqrt{{4-3}}\\&=1\end{align*}
1. .
\scriptsize \begin{align*}{{\tan }^{2}}\alpha -1&={{\left( {\displaystyle \frac{1}{{-\sqrt{3}}}} \right)}^{2}}-1\\&=\displaystyle \frac{1}{3}-1\\&=-\displaystyle \frac{2}{3}\end{align*}
2. .
\scriptsize \begin{align*}3\sin \alpha +2(1+\cos \alpha )&=3\left( {\displaystyle \frac{1}{2}} \right)+2\left( {1+\displaystyle \frac{{(-\sqrt{{3)}}}}{2}} \right)\\&=\displaystyle \frac{3}{2}+2\left( {\displaystyle \frac{{2-\sqrt{3}}}{2}} \right)\\&=\displaystyle \frac{3}{2}+2-\sqrt{3}\\&=\displaystyle \frac{7}{2}-\sqrt{3}\end{align*}
2. If $\scriptsize \tan \theta =\displaystyle \frac{6}{8}$ and $\scriptsize 90{}^\circ \le \theta \le 360{}^\circ$,then as tangent is positive, the angle must be in the third quadrant. The coordinates of the point on the circumference of the circle are $\scriptsize \left( {-8,-6} \right)$. Therefore, the radius of the circle is $\scriptsize 10$ (using Pythagoras).\scriptsize \begin{align*}\sin \theta (1+\cos \theta ) & =\displaystyle \frac{{-6}}{{10}}\left( {1-\displaystyle \frac{8}{{10}}} \right)\\&=\displaystyle \frac{{-6}}{{10}}\left( {\displaystyle \frac{2}{{10}}} \right)\\&=\displaystyle \frac{{-12}}{{100}}\\&=\displaystyle \frac{{4\times -3}}{{4\times 25}}\\&=\displaystyle \frac{{-3}}{{25}}\end{align*}

Back to Exercise 1.3

### Unit 1: Assessment

1. .
1. .
\scriptsize \begin{align*}\text{X}{{\text{Z}}^{2}} & =\text{X}{{\text{Y}}^{2}}+\text{Y}{{\text{Z}}^{2}}\\\therefore \text{XZ} & =\sqrt{{\text{X}{{\text{Y}}^{2}}+\text{Y}{{\text{Z}}^{2}}}}\\&=\sqrt{{25+144}}=\sqrt{{169}}\\&=13\end{align*}
1. $\scriptsize \sin \theta =\displaystyle \frac{{\text{opp}}}{{\text{hyp}}}=\displaystyle \frac{{12}}{{13}}$
2. $\scriptsize {{\cos }^{2}}\alpha ={{\left( {\displaystyle \frac{{12}}{{13}}} \right)}^{2}}=\displaystyle \frac{{144}}{{169}}$
2. .
\scriptsize \begin{align*}\left( {1+\tan \theta } \right)\left( {1-\tan \theta } \right)&=\left( {1+\displaystyle \frac{{12}}{5}} \right)\left( {1-\displaystyle \frac{{12}}{5}} \right)\\&=\left( {\displaystyle \frac{{5+12}}{5}} \right)\left( {\displaystyle \frac{{5-12}}{5}} \right)\\&=\left( {\displaystyle \frac{{17}}{5}} \right)\left( {\displaystyle \frac{{-7}}{5}} \right)\\&=-\displaystyle \frac{{119}}{{25}}\end{align*}
.
\scriptsize \begin{align*}1-{{\tan }^{2}}\theta & =1-{{\left( {\displaystyle \frac{{12}}{5}} \right)}^{2}}\\&=1-\displaystyle \frac{{144}}{{25}}\\&=\displaystyle \frac{{25-144}}{{25}}\\&=-\displaystyle \frac{{119}}{{25}}\end{align*}
Therefore, $\scriptsize \left( {1+\tan \theta } \right)\left( {1-\tan \theta } \right)=1-{{\tan }^{2}}\theta$.
2. $\scriptsize \cos \theta =\displaystyle \frac{7}{{25}}$ and $\scriptsize 90{}^\circ \le \theta \le 360{}^\circ$.

\scriptsize \begin{align*}{{r}^{2}} & ={{x}^{2}}+{{y}^{2}}\\\therefore {{y}^{2}} & ={{r}^{2}}-{{x}^{2}}\\\therefore y & =\sqrt{{{{r}^{2}}-{{x}^{2}}}}\\&=\sqrt{{{{{25}}^{2}}-{{7}^{2}}}}\\&=\sqrt{{625-49}}\\&=\sqrt{{576}}\\&=\pm24\end{align*}
But the point is in the fourth quadrant. Therefore, $\scriptsize y=-24$.
1. $\scriptsize \sin \theta =-\displaystyle \frac{{24}}{{25}}$
2. .
\scriptsize \begin{align*}\tan \theta -\cos \theta &=-\displaystyle \frac{{24}}{7}-\displaystyle \frac{7}{{25}}\\&=\displaystyle \frac{{-600-49}}{{175}}\\&=-\displaystyle \frac{{649}}{{175}}\end{align*}
3. .
\scriptsize \begin{align*}\displaystyle \frac{{{{{\tan }}^{2}}\theta }}{{{{{\sin }}^{2}}\theta }} & =\displaystyle \frac{{{{{\left( {-\displaystyle \frac{{24}}{7}} \right)}}^{2}}}}{{{{{\left( {-\displaystyle \frac{{24}}{{25}}} \right)}}^{2}}}}\\&=\displaystyle \frac{{\displaystyle \frac{{576}}{{49}}}}{{\displaystyle \frac{{576}}{{625}}}}\\&=\displaystyle \frac{{576}}{{49}}\times \displaystyle \frac{{625}}{{576}}\\&=\displaystyle \frac{{625}}{{49}}\end{align*}
You could take this result further as follows:
$\scriptsize \displaystyle \frac{{625}}{{49}}=\displaystyle \frac{{{{{25}}^{2}}}}{{{{7}^{2}}}}={{\left( {\displaystyle \frac{{25}}{7}} \right)}^{2}}={{\left( {\displaystyle \frac{1}{{{}^{7}\!\!\diagup\!\!{}_{{25}}\;}}} \right)}^{2}}={{\left( {\displaystyle \frac{1}{{\cos \theta }}} \right)}^{2}}=\displaystyle \frac{1}{{{{{\cos }}^{2}}\theta }}$

Back to Unit 1: Assessment