At GCSE, $\sin\theta$ was "opposite over hypotenuse" — which only makes sense for angles between $0°$ and $90°$. At A Level we need $\sin 200°$ and $\cos(-45°)$, so we redefine the functions using a circle. This page is part of DfE content area E · Trigonometry; in Year 1 all angles are in degrees.
| Feature | $y = \sin\theta$ | $y = \cos\theta$ | $y = \tan\theta$ |
|---|---|---|---|
| Period | $360°$ | $360°$ | $180°$ |
| Range | $-1 \leqslant y \leqslant 1$ | $-1 \leqslant y \leqslant 1$ | All real numbers |
| Zeros | $0°, \pm180°, \pm360°, \ldots$ | $\pm90°, \pm270°, \ldots$ | $0°, \pm180°, \pm360°, \ldots$ |
| Maximum $1$ at | $90°, 450°, \ldots$ | $0°, \pm360°, \ldots$ | — |
| Asymptotes | None | None | $\theta = \pm90°, \pm270°, \ldots$ |
Every useful fact about related angles can be read from the graphs.
| Property | Sine | Cosine | Tangent |
|---|---|---|---|
| Periodicity | $\sin(\theta + 360°) = \sin\theta$ | $\cos(\theta + 360°) = \cos\theta$ | $\tan(\theta + 180°) = \tan\theta$ |
| Negative angles | $\sin(-\theta) = -\sin\theta$ | $\cos(-\theta) = \cos\theta$ | $\tan(-\theta) = -\tan\theta$ |
| Supplement | $\sin(180° - \theta) = \sin\theta$ | $\cos(180° - \theta) = -\cos\theta$ | $\tan(180° - \theta) = -\tan\theta$ |
| Half turn | $\sin(180° + \theta) = -\sin\theta$ | $\cos(180° + \theta) = -\cos\theta$ | $\tan(180° + \theta) = \tan\theta$ |
| Complement | $\sin(90° - \theta) = \cos\theta$ and $\cos(90° - \theta) = \sin\theta$ | — | |
Given $\sin 20° \approx 0.342$, write down $\sin 160°$, $\sin 200°$, $\sin(-20°)$ and $\cos 70°$.
You must know these without a calculator. They come from two triangles: half of an equilateral triangle of side $2$ (angles $30°, 60°, 90°$, sides $1, \sqrt3, 2$) and an isosceles right-angled triangle with shorter sides $1$ (hypotenuse $\sqrt2$).
| $\theta$ | $0°$ | $30°$ | $45°$ | $60°$ | $90°$ |
|---|---|---|---|---|---|
| $\sin\theta$ | $0$ | $\tfrac12$ | $\tfrac{\sqrt2}{2}$ | $\tfrac{\sqrt3}{2}$ | $1$ |
| $\cos\theta$ | $1$ | $\tfrac{\sqrt3}{2}$ | $\tfrac{\sqrt2}{2}$ | $\tfrac12$ | $0$ |
| $\tan\theta$ | $0$ | $\tfrac{1}{\sqrt3} = \tfrac{\sqrt3}{3}$ | $1$ | $\sqrt3$ | undefined |
Find the exact values of (a) $\sin 240°$ (b) $\cos 135°$ (c) $\tan 330°$ (d) $\sin(-150°)$.
The rules from Transformations of Graphs apply unchanged. With trig graphs, watch what each one does to the period and the amplitude.
| Equation | Transformation of $y = \sin x$ | Effect |
|---|---|---|
| $y = a\sin x$ | Stretch parallel to the $y$-axis, scale factor $a$ | Amplitude $a$; range $-a \leqslant y \leqslant a$ |
| $y = \sin bx$ | Stretch parallel to the $x$-axis, scale factor $\tfrac1b$ | Period $\dfrac{360°}{b}$ |
| $y = \sin(x + c)$ | Translation by $\begin{pmatrix} -c \\ 0 \end{pmatrix}$ | Graph moves $c°$ left |
| $y = \sin x + d$ | Translation by $\begin{pmatrix} 0 \\ d \end{pmatrix}$ | Graph moves up $d$; centre line $y = d$ |
Sketch $y = 2\cos(x - 30°) + 1$ for $0° \leqslant x \leqslant 360°$, giving the coordinates of the maximum and minimum points and of the $y$-intercept.
State the period of $y = \tan 2x$ and the equations of its asymptotes for $0° \leqslant x \leqslant 360°$.
A curve $y = a\sin bx$ ($a, b > 0$) has its first maximum at $(30°, 5)$. Find $a$ and $b$.
How many solutions does $\sin 2x = 0.4$ have for $0° \leqslant x \leqslant 360°$?
Unit circle
$P = (\cos\theta, \sin\theta)$; $\tan\theta$ = gradient of $OP$.
CAST
Positive: All (Q1), Sin (Q2), Tan (Q3), Cos (Q4).
Periods
sin, cos: $360°$. tan: $180°$.
tan asymptotes
$\theta = 90° + 180°n$.
Symmetry
$\sin(180° - \theta) = \sin\theta$; $\cos(-\theta) = \cos\theta$.
Exact values
From the $1, \sqrt3, 2$ and $1, 1, \sqrt2$ triangles.
$a\sin x$
Amplitude $a$.
$\sin bx$
Period $360° \div b$.
$\sin(x + c)$
Shift $c°$ left.
$\sin x + d$
Shift up $d$.
Write down the exact value of $\sin 300°$.
▶ Show solution
$300° = 360° - 60°$, fourth quadrant, where sin is negative.
$\sin 300° = -\sin 60° = -\dfrac{\sqrt3}{2}$
Find the exact values of $\cos 225°$ and $\tan 150°$.
▶ Show solution
$225° = 180° + 45°$ (third quadrant, cos negative): $\cos 225° = -\dfrac{\sqrt2}{2}$.
$150° = 180° - 30°$ (second quadrant, tan negative): $\tan 150° = -\dfrac{\sqrt3}{3}$.
Given that $\cos 40° \approx 0.766$, write down the values of $\cos 320°$, $\cos 140°$ and $\sin 50°$.
▶ Show solution
$\cos 320° = \cos(360° - 40°) = \cos 40° \approx 0.766$
$\cos 140° = \cos(180° - 40°) = -\cos 40° \approx -0.766$
$\sin 50° = \cos(90° - 50°) = \cos 40° \approx 0.766$
State the amplitude and the period of $y = 5\sin 3x$.
▶ Show solution
Amplitude $5$. Period $\dfrac{360°}{3} = 120°$.
Describe fully the single transformation that maps $y = \cos x$ onto $y = \cos(x + 45°)$.
▶ Show solution
A translation by the vector $\begin{pmatrix} -45 \\ 0 \end{pmatrix}$, i.e. $45°$ in the negative $x$-direction.
Find the maximum and minimum values of $y = 3 - 2\sin x$, and the values of $x$ in $0° \leqslant x \leqslant 360°$ where they occur.
▶ Show solution
$y$ is largest when $\sin x$ is smallest: $\sin x = -1$ at $x = 270°$, giving $y = 3 + 2 = 5$.
$y$ is smallest when $\sin x = 1$ at $x = 90°$, giving $y = 3 - 2 = 1$.
Maximum $5$ at $x = 270°$; minimum $1$ at $x = 90°$.
For $y = \tan(x - 45°)$, $0° \leqslant x \leqslant 360°$, find the asymptotes, the $x$-intercepts and the $y$-intercept.
▶ Show solution
The graph of $y = \tan x$ moves $45°$ right.
Asymptotes: $x - 45° = 90°, 270°$, so $x = 135°$ and $x = 315°$.
$x$-intercepts: $x - 45° = 0°, 180°$, so $x = 45°$ and $x = 225°$.
$y$-intercept: $\tan(-45°) = -1$, so $(0, -1)$.
The graph of $y = a\cos bx$ ($a, b > 0$) has a maximum at $(0°, 4)$ and its next minimum at $(60°, -4)$. Find $a$ and $b$.
▶ Show solution
Amplitude $a = 4$.
Maximum to next minimum is half a period, so the period is $120°$.
$\dfrac{360°}{b} = 120°$, so $b = 3$.
Without solving, state the number of solutions of (a) $\cos 3x = -0.2$ for $0° \leqslant x \leqslant 360°$ and (b) $2\sin x = 3$ for all $x$. Explain each answer.
▶ Show solution
(a) $y = \cos 3x$ has period $120°$, so it completes three cycles. Each cycle crosses $y = -0.2$ twice: $6$ solutions.
(b) $\sin x = 1.5$ is impossible because $-1 \leqslant \sin x \leqslant 1$: no solutions.
The height $h$ metres of a seat on a big wheel, $t$ minutes after it passes the lowest point, is modelled by $h = 20 - 18\cos(12t)°$.
(a) Find the minimum and maximum heights. (b) How long does one revolution take? (c) Find the height after $5$ minutes. (d) Find the first time at which the seat is $29$ m high. (e) For how long in each revolution is the seat higher than $29$ m?
▶ Show solution
(a) $\cos(12t)°$ ranges from $-1$ to $1$, so $h$ ranges from $20 - 18 = 2$ m to $20 + 18 = 38$ m.
(b) Period $= \dfrac{360}{12} = 30$ minutes.
(c) $h = 20 - 18\cos 60° = 20 - 9 = 11$ m.
(d) $20 - 18\cos(12t)° = 29 \;\Rightarrow\; \cos(12t)° = -\tfrac12$.
The first angle with cosine $-\tfrac12$ is $120°$, so $12t = 120$ and $t = 10$ minutes.
(e) The next angle with cosine $-\tfrac12$ is $240°$, at $t = 20$. Between these, $\cos(12t)° < -\tfrac12$, so $h > 29$.
The seat is above $29$ m for $20 - 10 = 10$ minutes per revolution.