This handout covers Topic 6, Trigonometry. Parts marked (Extended) are only tested on the Extended papers; everything else is for both levels. Give angle answers to one decimal place.
Trigonometry
IGCSE Mathematics · Topic 6
6.1
Pythagoras' theorem
Syllabus
| Subject content | Notes and examples |
|---|---|
| Know and use Pythagoras’ theorem. |
Source: Cambridge International syllabus
Pythagoras' theorem 勾股定理 links the three sides of a right-angled triangle 直角三角形 (a triangle 三角形 with one $90^{\circ}$ angle). If the longest side (the hypotenuse 斜边, opposite the right angle) is $c$, then
$$a^{2} + b^{2} = c^{2}.$$
Use it to find a missing side.
Pythagoras as areas: the squares on the two shorter sides ($9+16$) add up to the square on the hypotenuse ($25$).
Worked example. A right-angled triangle has a hypotenuse of $13\text{ cm}$ and one short side of $5\text{ cm}$. Find the other short side.
$$b^{2} = 13^{2} - 5^{2} = 169 - 25 = 144, \qquad b = \sqrt{144} = 12\text{ cm}.$$
Explore
Pythagoras' theorem
Change the two short sides and see $a^2 + b^2 = c^2$ — the squares on the sides really do add up.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| Pythagoras' theorem | 勾股定理 | gōu gǔ dìng lǐ |
| right-angled triangle | 直角三角形 | zhí jiǎo sān jiǎo xíng |
| triangle | 三角形 | sān jiǎo xíng |
| hypotenuse | 斜边 | xié biān |
6.2
Trigonometry in right-angled triangles
Syllabus
| Subject content | Notes and examples |
|---|---|
| 1 Know and use the sine, cosine and tangent ratios for acute angles in calculations involving sides and angles of a right-angled triangle. | Angles will be given in degrees and answers should be written in degrees, with decimals correct to one decimal place. |
| 2 Solve problems in two dimensions using Pythagoras’ theorem and trigonometry. | Knowledge of bearings may be required. |
| Subject content | Notes and examples |
|---|---|
| 1 Know and use the sine, cosine and tangent ratios for acute angles in calculations involving sides and angles of a right-angled triangle. | Angles will be given in degrees and answers should be written in degrees, with decimals correct to one decimal place. |
| 2 Solve problems in two dimensions using Pythagoras’ theorem and trigonometry. | Knowledge of bearings may be required. |
| 3 Know that the perpendicular distance from a point to a line is the shortest distance to the line. | |
| 4 Carry out calculations involving angles of elevation and depression. |
Source: Cambridge International syllabus
Label the sides from the angle you are using: the opposite 对边 (across from the angle), the adjacent 邻边 (next to the angle), and the hypotenuse. The three ratios are the sine 正弦, cosine 余弦 and tangent ratio 正切 (sin, cos, tan):
$$\sin\theta = \frac{\text{opp}}{\text{hyp}}, \qquad \cos\theta = \frac{\text{adj}}{\text{hyp}}, \qquad \tan\theta = \frac{\text{opp}}{\text{adj}}.$$
Remember them as SOH-CAH-TOA. To find an angle, use the inverse ($\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$).
Name the sides from the angle $\theta$: the opposite is across from it, the adjacent next to it, and the hypotenuse opposite the right angle (SOH-CAH-TOA).
Worked example (find a side). In a right-angled triangle the hypotenuse is $10\text{ cm}$ and the angle is $30^{\circ}$. Find the opposite side.
$$\text{opp} = 10 \times \sin 30^{\circ} = 10 \times 0.5 = 5\text{ cm}.$$
Worked example (find an angle). The opposite side is $4\text{ cm}$ and the adjacent side is $3\text{ cm}$.
$$\tan\theta = \frac{4}{3}, \qquad \theta = \tan^{-1}\!\left(\frac{4}{3}\right) = 53.1^{\circ}.$$
Explore
Sine, cosine and tangent
Drag the angle on the unit circle to see where sin, cos and tan come from.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| opposite | 对边 | duì biān |
| adjacent | 邻边 | lín biān |
| sine | 正弦 | zhèng xián |
| cosine | 余弦 | yú xián |
| tangent ratio | 正切 | zhèng qiē |
6.2
Angles of elevation and depression (Extended)
Syllabus
| Subject content | Notes and examples |
|---|---|
| 1 Know and use the sine, cosine and tangent ratios for acute angles in calculations involving sides and angles of a right-angled triangle. | Angles will be given in degrees and answers should be written in degrees, with decimals correct to one decimal place. |
| 2 Solve problems in two dimensions using Pythagoras’ theorem and trigonometry. | Knowledge of bearings may be required. |
| 3 Know that the perpendicular distance from a point to a line is the shortest distance to the line. | |
| 4 Carry out calculations involving angles of elevation and depression. |
Source: Cambridge International syllabus
Looking up at a tower involves an angle of elevation.
The angle of elevation 仰角 is the angle up from the horizontal to an object above you. The angle of depression 俯角 is the angle down from the horizontal to an object below you. The shortest distance from a point to a line is the perpendicular 垂直 distance.
The angle of elevation looks up from the horizontal; the angle of depression looks down.
Worked example. From a point $50\text{ m}$ from the foot of a tower, the angle of elevation to the top is $40^{\circ}$. Find the height of the tower.
$$\text{height} = 50 \times \tan 40^{\circ} = 50 \times 0.839 = 42.0\text{ m}.$$
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| angle of elevation | 仰角 | yǎng jiǎo |
| angle of depression | 俯角 | fǔ jiǎo |
| perpendicular | 垂直 | chuí zhí |
6.3
Exact trigonometric values (Extended)
Syllabus
| Subject content | Notes and examples |
|---|---|
| Know the exact values of: 1 $\sin x$ and $\cos x$ for $x = 0^\circ, 30^\circ, 45^\circ, 60^\circ$ and $90^\circ$. 2 $\tan x$ for $x = 0^\circ, 30^\circ, 45^\circ$ and $60^\circ$. |
Source: Cambridge International syllabus
You must know these exact values without a calculator.
| $x$ | $0^{\circ}$ | $30^{\circ}$ | $45^{\circ}$ | $60^{\circ}$ | $90^{\circ}$ |
|---|---|---|---|---|---|
| $\sin x$ | $0$ | $\tfrac{1}{2}$ | $\tfrac{\sqrt{2}}{2}$ | $\tfrac{\sqrt{3}}{2}$ | $1$ |
| $\cos x$ | $1$ | $\tfrac{\sqrt{3}}{2}$ | $\tfrac{\sqrt{2}}{2}$ | $\tfrac{1}{2}$ | $0$ |
| $\tan x$ | $0$ | $\tfrac{1}{\sqrt{3}}$ | $1$ | $\sqrt{3}$ | — |
These two special triangles are where the exact values come from — worth memorising.
6.4
Graphs and trigonometric equations (Extended)
Syllabus
| Subject content | Notes and examples |
|---|---|
| 1 Recognise, sketch and interpret the following graphs for $0^\circ \leqslant x \leqslant 360^\circ$: • $y = \sin x$ • $y = \cos x$ • $y = \tan x$. | |
| 2 Solve trigonometric equations involving $\sin x$, $\cos x$ or $\tan x$, for $0^\circ \leqslant x \leqslant 360^\circ$. | e.g. solve: • $\sin x = \frac{\sqrt{3}}{2}$ for $0^\circ \leqslant x \leqslant 360^\circ$ • $2 \cos x + 1 = 0$ for $0^\circ \leqslant x \leqslant 360^\circ$. |
Source: Cambridge International syllabus
A Ferris wheel: a point on the rim traces a sine curve as it turns.
For $0^{\circ} \leqslant x \leqslant 360^{\circ}$:
- $y = \sin x$ is a wave starting at $0$, peaking at $90^{\circ}$, back to $0$ at $180^{\circ}$, down to $-1$ at $270^{\circ}$.
- $y = \cos x$ is the same wave but starting at $1$.
- $y = \tan x$ rises steeply and repeats every $180^{\circ}$.
$y=\sin x$ and $y=\cos x$ are smooth waves between $-1$ and $1$; cosine is sine shifted left by $90^\circ$.
A trigonometric equation 三角方程 often has more than one answer in this range. Use the graph (or the symmetry of the wave) to find them all.
Worked example. Solve $\sin x = \tfrac{\sqrt{3}}{2}$ for $0^{\circ} \leqslant x \leqslant 360^{\circ}$.
One answer is $x = 60^{\circ}$. The sine wave is also $\tfrac{\sqrt{3}}{2}$ at $180^{\circ} - 60^{\circ} = 120^{\circ}$. So $x = 60^{\circ}$ or $120^{\circ}$.
Worked example. Solve $2\cos x + 1 = 0$ for $0^{\circ} \leqslant x \leqslant 360^{\circ}$.
$$\cos x = -\tfrac{1}{2} \;\Rightarrow\; x = 120^{\circ} \text{ or } 240^{\circ}.$$
Explore
Trig graphs & equations
(cos θ, sin θ)
As θ turns, sin and cos trace their waves — and repeat every 360°.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| trigonometric equation | 三角方程 | sān jiǎo fāng chéng |
6.5
The sine and cosine rules (Extended)
Syllabus
| Subject content | Notes and examples |
|---|---|
| 1 Use the sine and cosine rules in calculations involving lengths and angles for any triangle. | Includes problems involving obtuse angles and the ambiguous case. |
| 2 Use the formula $\text{area of triangle} = \frac{1}{2} ab \sin C$. | The sine and cosine rules and the formula for area of a triangle are given in the List of formulas. |
Source: Cambridge International syllabus
For any triangle (not just right-angled), with sides $a, b, c$ opposite angles $A, B, C$:
$$\text{sine rule:}\quad \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C},$$
$$\text{cosine rule:}\quad a^{2} = b^{2} + c^{2} - 2bc\cos A.$$
In any triangle, sides $a$, $b$, $c$ lie opposite the angles $A$, $B$, $C$ of the same letter.
Use the sine rule 正弦定理 when you have a side and its opposite angle. Use the cosine rule 余弦定理 when you have two sides and the angle between them, or all three sides. With the sine rule, watch for the ambiguous case 两解情况, where an angle could be acute or obtuse.
The area of any triangle is
$$\text{area} = \tfrac{1}{2}ab\sin C.$$
Worked example. A triangle has $b = 7\text{ cm}$, $c = 8\text{ cm}$ and the angle between them $A = 40^{\circ}$. Find side $a$.
$$a^{2} = 7^{2} + 8^{2} - 2(7)(8)\cos 40^{\circ} = 113 - 112 \times 0.766 = 27.2,$$
$$a = \sqrt{27.2} = 5.2\text{ cm}.$$
Explore
Sine & cosine rule
Two sides and the angle between them fix the triangle: the cosine rule finds the third side, the sine rule the other angles.
Explore
What sin and cos mean
Spin the angle on the unit circle: the horizontal leg is cos θ and the vertical leg is sin θ — the same ratios the sine and cosine rules use.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| sine rule | 正弦定理 | zhèng xián dìng lǐ |
| cosine rule | 余弦定理 | yú xián dìng lǐ |
| ambiguous case | 两解情况 | liǎng jiě qíng kuàng |
6.6
Pythagoras and trigonometry in 3D (Extended)
Syllabus
| Subject content | Notes and examples |
|---|---|
| Carry out calculations and solve problems in three dimensions using Pythagoras' theorem and trigonometry, including calculating the angle between a line and a plane. |
Source: Cambridge International syllabus
In three dimensions, find a right-angled triangle inside the solid, then use Pythagoras or trigonometry on it. A common task is the angle between a line and a flat surface (a plane 平面).
Worked example. A box has a base $6\text{ cm}$ by $8\text{ cm}$ and height $5\text{ cm}$. Find the angle between a space diagonal and the base.
First the base diagonal: $\sqrt{6^{2} + 8^{2}} = \sqrt{100} = 10\text{ cm}$. This diagonal and the height form a right-angled triangle, so the angle $\theta$ with the base is
$$\tan\theta = \frac{5}{10} = 0.5, \qquad \theta = \tan^{-1}(0.5) = 26.6^{\circ}.$$
Explore
Pythagoras' theorem
In a right-angled triangle a² + b² = c². The same idea, applied twice, gives lengths inside 3-D solids.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| plane | 平面 | píng miàn |
6.6
Exam tips
- Use Pythagoras ($a^2 + b^2 = c^2$) when there is no angle; use SOH-CAH-TOA when an angle is involved.
- The hypotenuse is always opposite the right angle. Label the sides (opposite, adjacent, hypotenuse) relative to the angle you are using.
- To find an angle, use the inverse ($\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$), and check your calculator is set to degrees.
- For a triangle with no right angle, use the sine rule or the cosine rule — the cosine rule when you know two sides and the angle between them.