Trigonometric functions and their graphsEdexcel International A Level Maths: Revision notes

Section 1

The sine and cosine graphs

For $x$ in radians, $y=\sin x$ and $y=\cos x$ are periodic with period $2\pi$ and range $-1\le y\le1$ . The sine curve passes through the origin, rises to $1$ at $x=\frac{\pi}{2}$ , crosses zero at $\pi$ and reaches $-1$ at $\frac{3\pi}{2}$ . The cosine curve starts at $1$ when $x=0$ , crosses zero at $\frac{\pi}{2}$ and reaches $-1$ at $\pi$ . The cosine graph is the sine graph translated $\frac{\pi}{2}$ to the left. Both repeat: $\sin(x+2\pi)=\sin x$ .

Key termsperiodicperiodamplitude

Section 2

The tangent graph

$y=\tan x=\frac{\sin x}{\cos x}$ has period $\pi$ and range all real numbers. It passes through the origin and is undefined where $\cos x=0$ , so it has vertical asymptotes at $x=\pm\frac{\pi}{2},\pm\frac{3\pi}{2},\dots$ . Between asymptotes it increases from $-\infty$ to $\infty$ . Key values: $\tan\frac{\pi}{4}=1$ , $\tan\pi=0$ .

Key termsasymptote

Common mistake

Giving a period of $2\pi$ for tangent. Its period is $\pi$ .

Section 3

Symmetries

The graphs have symmetry that produces extra solutions: $\sin(-x)=-\sin x,\quad \cos(-x)=\cos x,\quad \tan(-x)=-\tan x,$ $\sin(\pi-x)=\sin x,\quad \cos(2\pi-x)=\cos x,\quad \tan(\pi+x)=\tan x.$ Sine is odd (rotational symmetry about the origin), cosine is even (reflection in the $y$ -axis). These explain why $\sin x=\frac12$ has the solutions $\frac{\pi}{6}$ and $\pi-\frac{\pi}{6}=\frac{5\pi}{6}$ in $[0,2\pi]$ .

Key termsodd functioneven function

Section 4

Transformations of trigonometric graphs

$y=a\sin x$ : stretch parallel to the $y$ -axis, scale factor $a$ (amplitude $a$ ). Example: $y=3\sin x$ has range $-3\le y\le3$ .
$y=\sin(x+c)$ : translation of $c$ to the left. So $y=\sin\left(x+\frac{\pi}{6}\right)$ moves the curve $\frac{\pi}{6}$ left.
$y=\sin(bx)$ : stretch parallel to the $x$ -axis, scale factor $\frac1b$ , so the period is $\frac{2\pi}{b}$ . $y=\sin2x$ has period $\pi$ .

The same rules hold for $\cos$ and $\tan$ (with period $\frac{\pi}{b}$ for $\tan bx$ ). Combining them, $y=2\sin\left(x-\frac{\pi}{4}\right)$ is a stretch of factor $2$ in $y$ then a translation $\frac{\pi}{4}$ to the right.

Key termsstretchtranslation

Common mistake

Moving $y=\sin(x+c)$ to the right. The sign is opposite to the shift: $+c$ moves it left.

Section 5

Solving equations using the graph

To solve $\sin(bx+c)=k$ in an interval: (1) convert the interval, for example $0\le x\le2\pi$ with $x-\frac{\pi}{4}$ becomes $-\frac{\pi}{4}\le x-\frac{\pi}{4}\le\frac{7\pi}{4}$ ; (2) find the principal value and the symmetric second value ( $\pi-\alpha$ for sine); (3) add or subtract the period to find all values in the converted interval; (4) rearrange for $x$ . Example: $\sin2x=\frac12$ on $0\le x\le2\pi$ uses $0\le2x\le4\pi$ , giving four solutions $\frac{\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12}$ .

Key termsprincipal value

Exam tip

Always convert the interval for the whole bracket before listing solutions, so no solutions are missed.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Trigonometric functions and their graphs

The function $f$ is defined by $f(x)=3\sin x$ , where $x$ is in radians.
Describe the single transformation that maps the curve $y=\sin x$ onto the curve $y=f(x)$ .2 marks
The function $g$ is defined by $g(x)=\sin\left(x+\frac{\pi}{6}\right)$ , where $x$ is in radians.
Solve $g(x)=1$ for $0\le x\le2\pi$ .2 marks
The function $h$ is defined by $h(x)=\sin2x$ , where $x$ is in radians.
State the period of $h$ and find all solutions of $h(x)=0$ in the interval $0\le x\le\pi$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).