Trigonometry starts as the study of triangles and grows into the study of anything that repeats: tides, sound waves, a big wheel, alternating current. In Year 1 you make that jump — from "opposite over hypotenuse" to functions defined for every angle.
sin, cos and tan for any angle · graphs, symmetry and transformations
$\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta}$ and $\sin^2\theta + \cos^2\theta \equiv 1$ · proving identities · solving equations in an interval
| Word | Meaning | Example |
|---|---|---|
| Period | The length of one complete cycle of a repeating graph. | $y = \sin x$ has period $360°$ |
| Amplitude | Half the distance from minimum to maximum. | $y = 3\cos x$ has amplitude $3$ |
| Asymptote | A line the graph approaches but never meets. | $y = \tan x$ at $x = 90°$ |
| Principal value | The single answer a calculator's inverse function gives. | $\sin^{-1} 0.5 = 30°$ |
| Reference angle | The acute angle to the nearest part of the $x$-axis. | For $210°$ it is $30°$ |
| Identity | True for every value of the variable; written $\equiv$. | $\sin^2\theta + \cos^2\theta \equiv 1$ |
| Included angle | The angle between two given sides. | $C$ in $\tfrac12 ab\sin C$ |
| Bearing | Angle clockwise from north, three figures. | $045°$ is north-east |
In triangle $ABC$, $AB = 5$ cm, $AC = 8$ cm and the area is $10\sqrt3$ cm$^2$. Find the two possible sizes of angle $A$, and the length $BC$ in the case where $A$ is obtuse.
The Year 1 content of DfE area E, and the page that covers it.
- ESine and Cosine RulesThe sine rule and the ambiguous case, the cosine rule, area $= \tfrac12 ab\sin C$, and bearings problems.
- ETrigonometric Graphssin, cos and tan for any angle; symmetry and periodicity; exact values; transformations.
- ETrigonometric Identities$\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta}$, $\sin^2\theta + \cos^2\theta \equiv 1$, and how to prove identities.
- ETrigonometric EquationsSolving in an interval, quadratics in sin or cos, and $\sin(n\theta + \alpha) = k$.
Sine rule
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$
Cosine rule
$a^2 = b^2 + c^2 - 2bc\cos A$
Area
$\tfrac12 ab\sin C$
Exact values
$\sin 30° = \tfrac12$, $\sin 45° = \tfrac{\sqrt2}{2}$, $\sin 60° = \tfrac{\sqrt3}{2}$
CAST
Which ratios are positive in each quadrant.
Periods
sin, cos $360°$; tan $180°$; $\sin bx$ has $\dfrac{360°}{b}$.
Identities
$\tan \equiv \dfrac{\sin}{\cos}$; $\sin^2 + \cos^2 \equiv 1$
Second solution
sin: $180° - \alpha$; cos: $360° - \alpha$; tan: $\alpha + 180°$.
Range
$-1 \leqslant \sin x \leqslant 1$, same for cos.
Ten questions drawing on the whole Trigonometry topic. Work each one through before opening the solution.
Write down the exact value of $\cos 210°$.
▶ Show solution
$210° = 180° + 30°$, third quadrant, where cos is negative.
$\cos 210° = -\cos 30° = -\dfrac{\sqrt3}{2}$
In triangle $ABC$, $A = 30°$, $B = 45°$ and $a = 10$ cm. Find the exact length of $b$.
▶ Show solution
$b = \dfrac{10\sin 45°}{\sin 30°} = \dfrac{10 \times \frac{\sqrt2}{2}}{\frac12} = 10\sqrt2$ cm.
In this question you must show detailed reasoning. A triangle has sides $7$, $8$ and $13$. Find its largest angle.
▶ Show solution
$\cos\theta = \dfrac{49 + 64 - 169}{2(7)(8)} = \dfrac{-56}{112} = -\dfrac12$
$\theta = 120°$.
Find the area of a triangle with sides $6$ cm and $10$ cm and included angle $150°$.
▶ Show solution
$\tfrac12(6)(10)\sin 150° = 30 \times \tfrac12 = 15$ cm$^2$.
State the period of $y = \cos 4x$ and the coordinates of its first minimum point for $x > 0$.
▶ Show solution
Period $\dfrac{360°}{4} = 90°$.
$\cos u$ has its first minimum at $u = 180°$, so $4x = 180°$, giving $(45°, -1)$.
Given that $\sin\theta = \tfrac13$ and $\theta$ is acute, find the exact values of $\cos\theta$ and $\tan\theta$.
▶ Show solution
$\cos^2\theta = 1 - \tfrac19 = \tfrac89$; acute so $\cos\theta = \dfrac{\sqrt8}{3} = \dfrac{2\sqrt2}{3}$.
$\tan\theta = \dfrac{1/3}{2\sqrt2/3} = \dfrac{1}{2\sqrt2} = \dfrac{\sqrt2}{4}$.
Prove that $\cos^2\theta - \sin^2\theta \equiv 2\cos^2\theta - 1$.
▶ Show solution
LHS $\equiv \cos^2\theta - (1 - \cos^2\theta)$
$\equiv 2\cos^2\theta - 1 \equiv$ RHS ■
Solve $4\sin x = 3\cos x$ for $0° \leqslant x \leqslant 360°$.
▶ Show solution
$\tan x = \tfrac34$, so $x = 36.9°$ or $36.87° + 180° = 216.9°$.
Solve $2\sin^2 x = 1 + \cos x$ for $0° \leqslant x \leqslant 360°$.
▶ Show solution
$2(1 - \cos^2 x) = 1 + \cos x \;\Rightarrow\; 2\cos^2 x + \cos x - 1 = 0$
$(2\cos x - 1)(\cos x + 1) = 0$
$\cos x = \tfrac12$: $x = 60°, 300°$. $\cos x = -1$: $x = 180°$.
$x = 60°, 180°, 300°$
A triangular field $ABC$ has $AB = 50$ m, $BC = 70$ m and $\angle ABC = 60°$.
(a) Show that $AC = 10\sqrt{39}$ m. (b) Find the exact area of the field. (c) Find $\angle BAC$. (d) A straight path runs from $B$ to the nearest point of $AC$. Find its length.
▶ Show solution
(a) $AC^2 = 50^2 + 70^2 - 2(50)(70)\cos 60° = 2500 + 4900 - 3500 = 3900$
$AC = \sqrt{3900} = \sqrt{100 \times 39} = 10\sqrt{39}$ m ■
(b) Area $= \tfrac12(50)(70)\sin 60° = 1750 \times \tfrac{\sqrt3}{2} = 875\sqrt3$ m$^2$ ($\approx 1520$ m$^2$).
(c) $\cos\angle BAC = \dfrac{50^2 + 3900 - 70^2}{2(50)(10\sqrt{39})} = \dfrac{1500}{1000\sqrt{39}} = 0.2402\ldots$
$\angle BAC = 76.1°$ (1 d.p.)
(d) The path is the perpendicular height $h$ from $B$ to $AC$. Area $= \tfrac12 \times AC \times h$:
$h = \dfrac{2 \times 875\sqrt3}{10\sqrt{39}} = \dfrac{175\sqrt3}{\sqrt{39}} = \dfrac{175}{\sqrt{13}} \approx 48.5$ m.