# Trigonometry

Here we will learn about trigonometry including how to use SOHCAHTOA, inverse trigonometric functions, exact trigonometric values and the hypotenuse. We’ll also learn about the sine rule, the cosine rule, how to find the area of a triangle using ½abSinC, 3D trigonometry and how to use the sine, cosine and tangent graphs.

There are also trigonometry worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

## What is trigonometry?

Trigonometry is the relationship between angles and side lengths within triangles; it is derived from the greek words “trigōnon” meaning triangle and “metron” meaning measure.

Trigonometry was originally used by the Babylonians, over 1500 years before the Greek form that we use today. It is used widely in science and engineering and product design.

The higher GCSE curriculum expands the use of trigonometric functions for non right-angle triangles, developing from the fundamental knowledge of the three trigonometric ratios (expressed as the mnemonic SOHCAHTOA) and exact trigonometric values in right angle triangles.

### What is trigonometry ## SOHCAHTOA

### What is SOHCAHTOA?

SOHCAHTOA is the abbreviation used to describe the three trigonometric ratios for the sine, cosine and tangent functions.

To determine which trigonometric function you need to use to answer a question, it depends on the location of the angle and the sides of the triangle that will be used.
The trigonometric functions apply to right-angle triangles.

• If you know the hypotenuse and the opposite side of the angle, you would use the sine function.
• If you know the hypotenuse and the adjacent side (next to) the angle, you would use the cosine function.
• If you know the opposite and adjacent sides to the angle, you use the tangent function.

We can use SOHCAHTOA to calculate lengths and angles in 2D and 3D shapes by recognising right-angle triangles.

E.g.
We can find the length AC of a parallelogram or the length AH in the cuboid below.

Step-by-step guide: SOHCAHTOA

## Hypotenuse

### What is the hypotenuse?

The hypotenuse is the longest side of a right angle triangle. It is the side opposite the right angle.

The hypotenuse does not occur for other types of triangles unless we know more information (such as an isosceles triangle can be made from 2 identical right angle triangles, back-to-back).

### Labelling the other sides of the triangle

Once we know which angle we are using, we can label the sides opposite (O), adjacent (A) and hypotenuse (H).
We know the hypotenuse is opposite the right angle.
The opposite side is opposite the angle we are using.
The adjacent side is next to the angle we are using.

The triangle below is labelled based on using the angle θ .

Step-by-step guide: Hypotenuse (coming soon)

### Example 1: find a side given the angle and the hypotenuse

ABC is a right angle triangle. The size of angle ACB = 60º and the length BC = 16cm.

Calculate the value of x.

Labelling the sides OAH in relation to the angle 60º, we can use the hypotenuse, and we need to find the adjacent side. We therefore need to use the cosine function.

\begin{aligned} cos(\theta)&=\frac{A}{H}\\\\ cos(60)&=\frac{x}{16} \\\\ 16\times cos(60&)=x\\\\ x&=8cm. \end{aligned}

## Inverse trigonometric functions

### What are inverse trigonometric functions?

Inverse trigonometric functions allow us to calculate the size of angle θ for a right-angle triangle.
The inverse trigonometric functions look like this:

\begin{aligned} &\quad \;\; \text{sine} \quad \quad \quad \quad \quad \quad \quad \text{cosine} \quad \quad \quad \quad \quad \;\; \quad \text{tangent} \\\\ \text{trigonometric} \quad \quad \quad &\ sin(\theta) = \frac{O}{H} \quad \quad \quad \quad \;\; \cos(\theta)=\frac{A}{H} \quad \quad \quad \quad \;\; \tan(\theta)=\frac{O}{A} \\ \text{function} \quad \quad \quad \quad &\\\\\\ \text{other} \quad \quad \quad \quad \quad &H=\frac{O}{\sin(\theta)} \quad \quad \quad \quad \;\; H=\frac{A}{\cos(\theta)} \quad \quad \quad \quad \;\; A=\frac{O}{\tan(\theta)} \\ \text{forms} \quad \quad \quad \quad \quad &\\ \quad \quad \quad \quad &O=H\times\sin(\theta) \quad \quad \quad A=H\times\cos(\theta) \quad \quad \quad O=A\times\tan(\theta) \\\\\\ \text{inverse} \;\; \quad \quad \quad \quad & \\ \text{trigonometric} \quad \quad \quad &\theta=\sin^{-1}(\frac{O}{H}) \quad \quad \quad \;\; \theta=\cos^{-1}(\frac{A}{H}) \quad \quad \quad \;\;\; \theta=\tan^{-1}(\frac{O}{A})\\ \text{function} \quad \quad \quad \quad & \end{aligned}

Step-by-step guide: Trigonometric functions

### Example 2: find the angle using inverse trigonometric functions

Calculate the size of angle θ correct to 2 decimal places.

The two sides that can be used to calculate the value of θ are the opposite and the hypotenuse and so we apply the sine function to θ to get

$sin(\theta)=\frac{8}{10}.$

In order to calculate θ, we rearrange the equation by using the inverse sine function.

We therefore have:

\begin{aligned} \sin^{-1}(\sin(\theta))&=\sin^{-1}(\frac{8}{10})\\ \\\theta&=\sin^{-1}(\frac{8}{10})\\ \\\theta&=53.13^{\circ}\quad(2dp).\\ \end{aligned}

## Exact trigonometric values

### What are exact trigonometric values?

Exact trigonometric values are found when the relationship between the sides and the angles in a triangle have a specific relationship.
Summarising these values, we obtain the exact trigonometric values for sine, cosine and tangent for 0 ≤ θ ≤ 90º.

\begin{aligned} & 0^{\circ} \quad \quad \quad \quad 30^{\circ} \quad \quad \quad \quad 45^{\circ} \quad \quad \quad \quad 60^{\circ} \quad \quad \quad \quad 90^{\circ} \\\\ \text{sin}(x) \quad\quad \quad \quad &0 \; \quad \quad \quad \quad \frac{1}{2} \quad \quad \quad \quad \frac{1}{\sqrt{2}}\quad \quad \quad \quad \frac{\sqrt{3}}{2} \quad \quad \quad \quad 1 \\\\ \text{cos}(x) \quad\quad \quad \quad &1 \; \quad \quad \quad \;\; \frac{\sqrt{3}}{2} \; \; \quad \quad \quad \frac{1}{\sqrt{2}}\;\; \quad \quad \quad \quad \frac{1}{2} \quad \quad \quad \quad \; 0 \\\\ \text{tan}(x) \quad\quad \quad \quad &0 \; \quad \quad \quad \;\; \frac{1}{\sqrt{3}} \; \quad \quad \quad \quad 1 \;\;\; \quad \quad \quad \quad \sqrt{3} \quad \quad \quad \text{Undefined} \end{aligned}

### Example 3: using exact trigonometric values

ABC is an equilateral triangle. M is the midpoint of AC. Calculate the exact size of the angle θ.

AC = 6cm so MC = 3cm

We therefore have the triangle:

\begin{aligned} \sin(\theta)&=\frac{O}{H}\\ \\\sin(\theta)&=\frac{3}{6}\\ \\\sin(\theta)&=\frac{1}{2}\\ \end{aligned}

Using the table above,

\begin{aligned} \text{if } \sin(\theta) = \frac{1}{2}, \text{ then } \theta=30^{\circ}\\ \end{aligned}

### Pythagoras or trigonometry?

We need to be able to interpret problems and recognise whether we need to use Pythagoras’ Theorem in 2D, 3D, or one of the three trigonometric ratios.

This flow chart describes the information you need to know about a shape in order to solve the problem.

It is important to recognise that with most of these problems, you may need to use the Pythagorean Theorem, or trigonometry, or both within the same question so you must be confident with these topics individually to access this topic fully.

Below is a summary of methods that can be used for right angled triangles:

Name

Pythagoras’ Theorem in 2D

Sine function

Cosine function

Tangent function

Inverse sine function

Inverse cosine function

Inverse tangent function

Pythagoras’ Theorem in 3D

Used to find…

The hypotenuse       (c)

A shorter side (a)

If the triangle       contains a right       angle or not

Missing side:

hypotenuse (H)

or opposite (O)

Missing side:

hypotenuse (H)

Missing side:

or opposite (O)

Missing angle
(\theta)

Missing angle
(\theta)

Missing angle
(\theta)

The diagonal (D)

Rule or formula

### Example 4: find the hypotenuse using trigonometry

Calculate the length of the hypotenuse of a right triangle, x, to 1 decimal place.

The two important sides in this question are the opposite side (O) to the angle and the hypotenuse (H) so we need to use the sine function to calculate the value of x.

\begin{aligned} H&=\frac{O}{\sin(\theta)}\\ \\H&=\frac{56}{\sin(63)}\\ \\H&=62.9m\quad(1dp)\\ \end{aligned}

## Sine rule (the law of sines)

### What is the sine rule?

The sine rule (or the law of sines) is a relationship between the size of an angle in a triangle and the opposite side. There are three relationships in a triangle as there are 3 angles with their opposing sides but you will only need to use two.

Pythagoras’ Theorem cannot be used to find the third side of a non-right angled triangle. Instead we can use the sine rule or the cosine rule, depending on the information we know about the triangle.

To find a missing angle:

\frac{\sin (A)}{a}=\frac{\sin (B)}{b}

To find a missing side:

\frac{a}{\sin (A)}=\frac{b}{\sin (B)}

Step-by-step guide: Sine rule

### Example 5: Finding a missing side of a triangle using the sine rule

Label each angle A, B and C and each side a, b and c:

Here we know side a and we want to find the length of c, therefore we can state:

\begin{array}{l} \frac{a}{\sin (A)}=\frac{c}{\sin (C)}\\ \\\frac{6}{\sin (55)}=\frac{c}{\sin (73)}\\ \\c=\frac{6}{\sin (55)}\times\sin(73)\\ \\c=\frac{6\sin(73)}{\sin (55)}\\ \\c=7.00\quad(2dp) \end{array}

Here, the length AB = 7.00cm (2dp).

## Cosine rule (the law of cosines)

### What is the cosine rule?

The cosine rule (or the law of cosines) is a formula which can be used to calculate the missing sides of a triangle or to find a missing angle. To do this we need to know the two arrangements of the formula and what each variable represents.

To find a missing side:

a^{2}=b^{2}+c^{2}-2bc\cos(A)

To find a missing angle:

A=\cos^{-1}(\frac{b^2+c^2-a^2}{2bc})

Step-by-step guide: Cosine rule

### Example 6: find the missing side using the cosine rule

Find the length of x for triangle ABC, correct to 2 decimal places.

The vertices are already labelled with A located on the angle we are using so we only need to label the opposite sides of a, b, and c.

Here, we need to find the missing side a, therefore we need to state the cosine rule with a2 as the subject:

\begin{array}{l} a^{2}=b^{2}+c^{2}-2bc\cos(A)\\ \\x^{2}=7.1^{2}+6.5^{2}-2\times7.1\times6.5\times\cos(32)\\ \\x^{2}=50.41+42.25-92.3\times\cos(32)\\ \\x^{2}=92.66-78.27483928…\\ \\x^{2}=14.38516072…\\ \\x=\sqrt{14.38516072…}\\ \\x=3.79cm\quad(2dp)\\ \end{array}

## 1/2abSin(C) (area of a triangle)

### What is 1/2abSin(C)?

\frac{1}{2}abSin(C) is a formula to calculate the area of any triangle.

$Area =\frac{1}{2}ab\sin(C).$

Step-by-step guide: Area of a triangle trig

### Example 7: area using A=1/2abSin(C)

Calculate the area of the triangle ABC. Write your answer to 2 decimal places.

Here, we label each side a, b, and c.

\begin{array}{l} A=\frac{1}{2}ab\sin(C)\\ \\A=\frac{1}{2}(12\times7\times\sin(77))\\ \\A=\frac{81.84708544}{2}\\ \\A=40.92cm^{2}\quad(2dp)\\ \end{array}

## 3D trigonometry

### What is 3D trigonometry?

3D trigonometry is the application of the trigonometric skills developed for 2 dimensional triangles.

To find missing sides or angles in 3 dimensional shapes, we need to be very clear with the rules and formulae to find these different angles and side lengths.
The flowchart below can help determine which function you need to use:

Once you can justify which rule or formulae you need to use, you may need to carry out this process again for another triangle in the question.

Top tip: Look out for common angles or common sides.

Step-by-step guide: 3D trigonometry

### Example 8: find the missing angle in a triangular prism

Calculate the size of the angle in the triangular prism ABCDEF.

We can see that triangle ABF and triangle ACF share the side AF. We can use triangle ACF to calculate the length of AF, which will then help us calculate the size of angle θ .

This triangle contains no information about the angles so we need to use Pythagoras Theorem.

\begin{aligned} a^{2}+b^{2}&=c^{2}\\ 4^{2}+8^{2}&=c^{2}\\ 80&=c^{2}\\ c&=\sqrt{80}\\ c&=8.94427191 \end{aligned}

This is a right angled triangle involving angles so we need to use SOHCAHTOA.

Since we know O and A, we need to use tan.

\begin{aligned} \tan(\theta)&=\frac{O}{A}\\ \tan(\theta) &= \frac{5}{8.94427191}\\ \theta &= \tan^{-1}(\frac{5}{8.94427191})\\ \theta &= 29.21^{\circ} \end{aligned}

## Trigonometric graphs

The trigonometric functions sine, cosine and tangent can be represented by graphs.

For example, as the angle changes, so does the value of sine. This can be plotted on a graph.

Angle0 153045607590
Sine (to 3 decimal places) 0 0.259 0.5 0.707 0.866 0.966 1

Let’s look at this in more detail below.

## Sine, cosine and tangent graphs

### What are sine, cosine, and tangent graphs?

Trigonometric graphs are a visual representation of the sine, cosine and tangent functions. The horizontal axis represents the angle, usually written as θ, and the vertical axis is the trig function.

See below for all three trigonometric graphs for all angles of θ between -360º and 360º (-360 < θ < 360).

The graph of y = sin(θ)

Step-by-step guide: Sin graph

The graph of y = cos(θ)

Step-by-step guide: Cos graph

The graph of y = tan(θ)

Step-by-step guide: Tan graph

### Example 9: state the value of tan(θ) with θ known

Use the graph of y = tan(θ) to estimate the value of y when θ = 120º.

Here we draw the vertical line at 120º until it reaches the tangent curve and then a horizontal line towards the y-axis.

As the scale for each mark on the y-axis is 0.25, the value for tan(120) is approximately equal to -1.7 (1dp).

### Common misconceptions

• Labelling a triangle incorrectly

E.g.
This triangle has been incorrectly labelled with the side next to the angle.

This will have an impact on the formula for the sine rule, the cosine rule, and the area of the triangle.

• Using the incorrect trigonometric function

If the triangle is incorrectly labelled it can lead to the use of the incorrect standard or inverse trigonometric function.

• Rounding the decimal too early

This can lose accuracy marks. Always use as many decimal places as possible throughout the calculation, then round the solution.

• Pythagoras’ Theorem or trigonometry?

Use the flowchart to help you to recognise when to use Pythagoras’ Theorem and when to use trigonometry. Remember, you may need to use both.

• Using the sine rule instead of cosine rule

In order to use the sine rule we need to have pairs of opposite angles and sides.

• Not using the included angle

For the cosine rule and the area of a triangle using A=1/2absin(C), the angle is included between the two sides. Using any other angle will result in an incorrect solution.

• Using A = b × h ÷ 2

If the vertical height of a triangle is not available then we cannot calculate the area by halving the base times the height.

• Using the inverse trig function instead, inducing a mathematical error

If the inverse trig function is used instead of the standard trig function, the calculator may return a maths error as the solution does not exist.

• Sine and cosine graphs switched

The sine and cosine graphs are very similar and can easily be confused with one another. A tip to remember is that you “sine up” from 0 for the sine graph so the line is increasing whereas you “cosine down” from 1 so the line is decreasing for the cosine graph.

• Asymptotes are drawn incorrectly for the graph of the tangent function

The tangent function has an asymptote at 90º because this value is undefined. As the curve repeats every 180º, the next asymptote is at 270º and so on.

• The graphs are sketched using a ruler

Each trigonometric graph is a curve and therefore the only time you are required to use a ruler is to draw a set of axes. Practice sketching each curve freehand and label important values on each axis.

### Practice trigonometry questions

1. Calculate the length of the side BC: 17.87cm 13.96cm 0.06cm 0.21cm This is a right angled triangle involving angles so use SOHCAHTOA.

First we need to label the sides O, A and H. We know A and we want to find H so we need to use cos.

\begin{aligned} H&=\frac{A}{\cos(\theta)}\\\\ H&=\frac{11}{\cos(52)}\\\\ H&=17.87\mathrm{cm} \end{aligned} 2. Using your knowledge of exact trigonometric values, work out the size of the angle marked . \theta 30^{\circ} 45^{\circ} 60^{\circ} 90^{\circ} First we need to label the sides O, A and H. Since we know O and A , we need to use tan.

\begin{aligned} \tan(\theta) &= \frac{O}{A}\\\\ \tan(\theta) &= \frac{\sqrt{3}}{1}\\\\ \tan(\theta) &=\sqrt{3} \end{aligned}

Using the exact trigonometric values,

\tan(\theta)=\sqrt{3} \text{ when } \theta=60^{\circ}

3. Work out the size of angle \theta . 60.67^{\circ} 54.53^{\circ} 46.69^{\circ} 60.07^{\circ} This is not a right angled triangle and we know an angle and its opposite side so we need to use the sine rule. \begin{aligned} \frac{\sin(A)}{a}&=\frac{\sin(B)}{b}\\\\ \frac{\sin(\theta)}{13}&=\frac{sin(70)}{15} \\\\ \sin(\theta)&= \frac{sin(70)}{15} \times 13\\\\ \sin(\theta)&=0.8144002713\\\\ \theta &=\sin^{-1}(0.8144002713)\\\\ \theta&=54.53^{\circ} \end{aligned}

4. Calculate the area of the following triangle: 168cm^{2} 91.50cm^{2} 23.71cm^{2} 140.90cm^{2} We do not know the base or the height so we need to use:

A=\frac{1}{2}ab \sin(C) \begin{aligned} A&=\frac{1}{2} \times 12 \times 28 \times \sin(57)\\\\ A&= 140.90cm^{2} \end{aligned}

5. Calculate the length of AE. 9.11cm 7.81cm 10.39 cm 10.91cm The triangles AGH and AEH share the line AH .
Using the triangle AGH we can calculate the length of the line AH. \begin{aligned} A &= \frac{O}{\tan(\theta)}\\\\ A &= \frac{6}{\tan(30)}\\\\ A &= 6\sqrt{3} \mathrm{cm} \end{aligned}

Using the triangle AEH we can calculate the length of AE . \begin{aligned} a^{2}&+b^{2}=c^{2}\\\\ (AE)^{2}+5^{2}&=(6\sqrt{3})^{2}\\\\ (AE)^{2}&=(6\sqrt{3})^{2}-5^{2}\\\\ (AE)^{2}&=83\\\\ AE&=\sqrt{3}\\\\ AE&=9.11 \mathrm{cm} \end{aligned}

6. Write down the coordinates of a minimum point on the graph of y=cos(\theta) for 0^{\circ} \leqslant \theta \leqslant 360^{\circ}

(90, -1) (180, 0) (180, -1) (90, 0)  The minimum point occurs at (180, -1)

### Trigonometry GCSE questions

1. Below is a sketch of a football pitch ABCD . (a)   Player F is standing exactly 60m perpendicular to Player E on the goal line and 75m from the corner where Player A is standing.

Player A kicks the football directly to Player F . Calculate the bearing of F from A . Write your answer correct to 2 decimal places.

(b)   Player F then passes the ball to Player G (the goal keeper). Player G is standing at the midpoint of BC at a bearing of 060^{\circ} from F .

How far is Player G from Player F ? Write your answer to 2 decimal places. (7 Marks)

(a)

Bearing of F from A = 180 – angle EAF

(1)

Angle AEF = \sin^{-1}(\frac{60}{75})

(1)

Angle AEF = 53.13^{\circ}

(1)

Bearing of F from  A = 180 – 53.13 =126.87^{\circ}

(1)

(b) (1)

FG = \frac{45}{sin(60)}

(1)

FG = 51.96m

(1)

2.  Triangle ABE and ACD are similar with AB:BC = 1:3. Using the information on the diagram, calculate the area of the shaded region BCDE . (8 Marks)

Area (ABE) = ½ × a × b × sin(C) = ½ × 4 × 6 × sin(30)

(1)

Sin(30) = ½

(1)

Area (ABE) = 6cm^2

(1)

As AB:BC=1:3, BC= 4 × 3 = 12cm and DE = 6 × 3 = 18cm

(1)

AC = 12+4 = 16cm and AD = 6+18 = 24cm

(1)

Area (ACD) = ½ × a × b × sin(C) = ½ × 16 × 24 × sin(30)

(1)

Area (ACD) = 96cm^2

(1)

Area BCDE = 96 – 6 = 90cm^2

(1)

3.  (a)  The cuboid ABCDEFGH is shown below. ADEF is a square face with the side length of 2m , and the length AB = 8cm.

Calculate the length of the line AE. Write your answer as a surd in its simplest form. (b)  Given that the point X lies on the line EH so that XH = 3EX and angle XAB = 54.7^{\circ}, calculate the length of the line BX . (10 Marks)

(a)

AE^2 = AF^2 + EF^2

(1)

AE^2 = 2^2 + 2^2 = 8

(1)

AE = \sqrt{8} = 2\sqrt{2}

(1)

(b)

EX = 8 / 4 = 2m

(1)

AX = \sqrt{2^{2}+2\sqrt{2}^{2}}

(1)

AX = 2\sqrt{3}

(1)

Cosine rule stated to find BX :

BX^{2}=AX^{2}+AB^{2}-2\times{AX}\times{AB}\times\cos(A)

(1)

BX^{2}=(2\sqrt{3})^{2}+8^{2}-2\times{2\sqrt{3}}\times{8}\times\cos(54.7)

(1)

BX^{2}=43.97m

(1)

BX=6.63m

(1)

## Learning checklist

You have now learned how to:

• use trigonometric ratios in similar triangles to solve problems involving right-angled triangles
• recognise, sketch and interpret graphs of trigonometric functions (with arguments in degrees) y = sin x, y = cos x and y = tan x for angles of any size
• apply trigonometric ratios to find angles and lengths in right-angled triangles and, where possible, general triangles in 2 and 3 dimensional figures
• know the exact values of sin θ and cos θ for θ = 0°, 30°, 45°, 60° and 90°; know the exact value of tan θ for θ = 0°, 30°, 45°, 60°
• know and apply the sine rule, and cosine rule, to find unknown lengths and angles
• know and apply area = \frac{1}{2}abSinC to calculate the area, sides or angles of any triangle

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#### FREE GCSE Maths Practice Papers - 2022 Topics

Practice paper packs based on the advanced information for the Summer 2022 exam series from Edexcel, AQA and OCR.

Designed to help your GCSE students revise some of the topics that will come up in the Summer exams.