Graphs of sine, cosine and tangent

AS · P1 · 14 min

The sine, cosine and tangent functions repeat themselves for ever, and their graphs are the quickest way to understand them. Paper 1 asks you to sketch them for angles of any size, in degrees or radians, including the forms $y = a\sin(bx) + c$ , $y = a\cos(bx) + c$ and $y = a\tan(bx) + c$ ; to find the constants from a graph or from given values; and to use a sketch to count the solutions of an equation. These sketches also underpin every trigonometric equation you will solve.

Sine and cosine for any angle

At GCSE, sine and cosine are ratios of sides in a right-angled triangle, so they only make sense for angles between $0^\circ$ and $90^\circ$ . To extend them to any angle, use a circle of radius $1$ centred at the origin, the unit circle.

Start at $(1, 0)$ and turn anticlockwise through an angle $\theta$ (clockwise for a negative angle). The point you reach on the unit circle is defined to be

(\cos\theta,\ \sin\theta), \qquad \text{and} \qquad \tan\theta = \frac{\sin\theta}{\cos\theta}

For an acute angle this agrees with the triangle definitions (the hypotenuse is $1$ ). For any other angle it simply carries on: as $\theta$ increases, the point goes round and round, and its height $\sin\theta$ and its horizontal position $\cos\theta$ oscillate between $-1$ and $1$ . The details, including signs in each quadrant and related angles, are in Exact values and angles of any size.

The three basic graphs

Plotting the height of the point against the angle gives the sine wave; plotting its horizontal position gives the cosine wave.

y = sin x y = cos x

y = tan x x = pi/2 x = 3pi/2

Key result

	$y = \sin x$	$y = \cos x$	$y = \tan x$
Period	$360^\circ$ or $2\pi$	$360^\circ$ or $2\pi$	$180^\circ$ or $\pi$
Range	$-1 \le y \le 1$	$-1 \le y \le 1$	all real numbers
Value at $0$	$0$	$1$	$0$
Maximum	$1$ at $90^\circ$ ( $\tfrac{\pi}{2}$ )	$1$ at $0^\circ$ , $360^\circ$	none
Minimum	$-1$ at $270^\circ$ ( $\tfrac{3\pi}{2}$ )	$-1$ at $180^\circ$ ( $\pi$ )	none
Zeros	$0^\circ, 180^\circ, 360^\circ$ ( $0, \pi, 2\pi$ )	$90^\circ, 270^\circ$ ( $\tfrac{\pi}{2}, \tfrac{3\pi}{2}$ )	$0^\circ, 180^\circ, 360^\circ$
Asymptotes	none	none	$x = 90^\circ, 270^\circ$ ( $\tfrac{\pi}{2}, \tfrac{3\pi}{2}$ )

The period is the length of one complete cycle: $\sin(x + 360^\circ) = \sin x$ for every $x$ . The amplitude of a sine or cosine wave is the distance from its middle line to a peak; for $\sin x$ and $\cos x$ it is $1$ .

Tangent has vertical asymptotes where $\cos x = 0$ , since $\tan x = \dfrac{\sin x}{\cos x}$ is undefined there. Between asymptotes it increases through every real value.

Symmetries you can read off the graphs

The sine curve is symmetrical about $x = 90^\circ$ : $\sin(180^\circ - x) = \sin x$ .
The cosine curve is symmetrical about the $y$ -axis: $\cos(-x) = \cos x$ .
The cosine curve is the sine curve shifted left by $90^\circ$ : $\cos x = \sin(x + 90^\circ)$ .
Sine and tangent have rotational symmetry about the origin: $\sin(-x) = -\sin x$ , $\tan(-x) = -\tan x$ .

These symmetries are what you use to find the second solution of a trigonometric equation.

Transformed graphs

The transformation rules from Transformations of graphs apply directly. For $y = a\sin(bx) + c$ :

Constant	Transformation	Effect on the wave
$a$	stretch parallel to the $y$ -axis, factor $a$	amplitude becomes $\lvert a\rvert$ ; if $a < 0$ the wave is also reflected in the $x$ -axis
$b$	stretch parallel to the $x$ -axis, factor $\tfrac{1}{b}$	period becomes $\dfrac{360^\circ}{b}$ or $\dfrac{2\pi}{b}$ ( $b$ cycles in $360^\circ$ )
$c$	translation by $\begin{pmatrix} 0 \\ c \end{pmatrix}$	middle line moves to $y = c$

The same table holds for cosine. For $y = a\tan(bx) + c$ , the period is $\dfrac{180^\circ}{b}$ or $\dfrac{\pi}{b}$ , and the asymptotes move with the stretch: they are where $bx = 90^\circ, 270^\circ, \ldots$

Key result

For $y = a\sin(bx) + c$ or $y = a\cos(bx) + c$ with $a > 0$ :

\text{maximum} = c + a, \qquad \text{minimum} = c - a, \qquad \text{period} = \frac{360^\circ}{b} = \frac{2\pi}{b}

So $c = \dfrac{\max + \min}{2}$ and $a = \dfrac{\max - \min}{2}$ .

Sketching y = a sin(bx) + c or y = a cos(bx) + c

Find the period from $b$ , and mark the quarter-periods along the $x$ -axis: the key points of the wave fall at these.
Draw the middle line $y = c$ , and the lines $y = c + a$ and $y = c - a$ .
Start the wave correctly: $\sin$ starts on the middle line going up, $\cos$ starts at a maximum. If $a < 0$ , $\sin$ starts going down and $\cos$ starts at a minimum.
Sketch smooth waves through the key points, covering exactly the interval asked for.
Label the axis intercepts and the maximum and minimum points, or at least their values.

y = 1 - 2cos(2x) y = 1

$y = 1 - 2\cos 2x$ : period $\pi$ , amplitude $2$ , reflected, middle line $y = 1$ , range $-1 \le y \le 3$ .

Using graphs to count solutions

The solutions of $f(x) = k$ are the $x$ -coordinates where the graph of $y = f(x)$ meets the horizontal line $y = k$ . So a sketch tells you how many solutions an equation has in an interval before you solve it, and is the main check that you have found them all. Watch the ends of the interval: a curve that starts and ends at the same height can meet a line at both endpoints.

Worked examples

Sketching a transformed cosine graph

Sketch the graph of $y = 1 - 2\cos 2x$ for $0^\circ \le x \le 360^\circ$ , and state the range of $y$ .

Solution

$\cos 2x$ has period $\tfrac{360^\circ}{2} = 180^\circ$ , so there are two complete waves. The factor $-2$ gives amplitude $2$ and reflects the wave, so it starts at a minimum. Adding $1$ puts the middle line at $y = 1$ .

Key points: $(0^\circ, -1)$ , $(45^\circ, 1)$ , $(90^\circ, 3)$ , $(135^\circ, 1)$ , $(180^\circ, -1)$ , then the same again up to $(360^\circ, -1)$ .

Range: $-1 \le y \le 3$ .

y = 1 - 2cos(2x*pi/180) y = 1

Constants from the maximum, minimum and period

A curve has equation $y = a\sin bx + c$ , where $a$ , $b$ and $c$ are positive constants. The curve has maximum value $5$ , minimum value $-1$ and period $120^\circ$ . Find $a$ , $b$ and $c$ .

Solution

c = \frac{5 + (-1)}{2} = 2, \qquad a = \frac{5 - (-1)}{2} = 3, \qquad \frac{360^\circ}{b} = 120^\circ \Rightarrow b = 3

So $y = 3\sin 3x + 2$ .

Constants from two values

The function $f$ is defined by $f(x) = a + b\sin x$ for $0 \le x \le 2\pi$ . Given that $f\left(\tfrac{\pi}{2}\right) = 5$ and $f\left(\tfrac{7\pi}{6}\right) = 2$ , find $a$ and $b$ , and state the range of $f$ .

Solution

$\sin\tfrac{\pi}{2} = 1$ and $\sin\tfrac{7\pi}{6} = -\tfrac{1}{2}$ , so

a + b = 5, \qquad a - \tfrac{1}{2}b = 2

Subtracting: $\tfrac{3}{2}b = 3$ , so $b = 2$ and $a = 3$ . Then $f(x) = 3 + 2\sin x$ , and since $-1 \le \sin x \le 1$ , the range is $1 \le f(x) \le 5$ .

A transformed tangent graph

Sketch the graph of $y = \tan 2x$ for $0 \le x \le \pi$ , showing the asymptotes and the points where the graph meets the $x$ -axis.

Solution

The period of $\tan 2x$ is $\tfrac{\pi}{2}$ . Asymptotes occur where $2x = \tfrac{\pi}{2}$ or $\tfrac{3\pi}{2}$ , i.e. $x = \tfrac{\pi}{4}$ and $x = \tfrac{3\pi}{4}$ . Zeros occur where $2x = 0, \pi, 2\pi$ , i.e. $x = 0, \tfrac{\pi}{2}, \pi$ .

y = tan(2x) x = pi/4 x = 3pi/4

The graph rises from $(0, 0)$ towards the asymptote $x = \tfrac{\pi}{4}$ , reappears from below, passes through $\left(\tfrac{\pi}{2}, 0\right)$ , rises towards $x = \tfrac{3\pi}{4}$ , and reappears to end at $(\pi, 0)$ .

Counting solutions with a parameter

(a) Sketch $y = 2\cos 2x$ for $0^\circ \le x \le 180^\circ$ .

(b) Find the set of values of $k$ for which the equation $2\cos 2x = k$ has exactly two solutions in $0^\circ \le x \le 180^\circ$ .

Solution

(a) Period $180^\circ$ , so one complete wave: it starts at $(0^\circ, 2)$ , falls through $(45^\circ, 0)$ to $(90^\circ, -2)$ , and rises through $(135^\circ, 0)$ to $(180^\circ, 2)$ .

y = 2cos(2x*pi/180) y = 1

(b) Slide a horizontal line $y = k$ across the sketch.

$-2 < k < 2$ : the line cuts the wave once on the way down and once on the way up, two solutions.
$k = 2$ : the line meets the curve at both ends, $x = 0^\circ$ and $x = 180^\circ$ , two solutions.
$k = -2$ : the line touches only at $x = 90^\circ$ , one solution.
$|k| > 2$ : no solutions.

So there are exactly two solutions for $-2 < k \le 2$ .

Range, sketch and number of roots

The function $f$ is defined by $f(x) = 2 - 3\sin x$ for $0 \le x \le 2\pi$ .

(a) State the range of $f$ .

(b) Sketch the graph of $y = f(x)$ .

Solution

(a) $-1 \le \sin x \le 1$ , so $-3 \le -3\sin x \le 3$ and $-1 \le f(x) \le 5$ .

(b) The sine wave is reflected and stretched by factor $3$ , then moved up $2$ . It starts at $(0, 2)$ , falls to a minimum $\left(\tfrac{\pi}{2}, -1\right)$ , rises through $(\pi, 2)$ to a maximum $\left(\tfrac{3\pi}{2}, 5\right)$ and returns to $(2\pi, 2)$ .

y = 2 - 3sin(x) y = 2

once if $k = -1$ or $k = 5$ (touching at a turning point);
twice if $-1 < k < 2$ or $2 < k < 5$ ;
three times if $k = 2$ (at $x = 0$ , $\pi$ and $2\pi$ ).

Exactly two solutions: $-1 < k < 2$ or $2 < k < 5$ .

Watch out

Period the wrong way round. $y = \sin 3x$ has period $120^\circ$ , not $1080^\circ$ . The factor $b$ squeezes the graph, so the period is divided by $b$ .

Amplitude and range confused. $y = 4\cos x - 1$ has amplitude $4$ but range $-5 \le y \le 3$ .

Wrong starting point. $y = -\sin x$ starts at the origin going down; $y = -\cos x$ starts at $-1$ .

Tangent drawn like a wave. $\tan x$ has no maximum or minimum. Draw the asymptotes first, dashed, and make each branch approach them.

Degrees and radians mixed. If the interval is $0 \le x \le 2\pi$ , label the axis in radians. Do not mark $90$ on a radian axis.

Exam tip

"Sketch" means the correct shape over the whole interval, with key coordinates labelled: intercepts, maximum and minimum points, asymptotes. It does not mean plot accurately.
Interval endpoints. Start and finish the curve exactly at the ends of the interval given, and check whether the endpoints are included (for counting solutions).
Finding constants. Use maximum, minimum and period, or substitute given points. Cambridge often gives values at angles with exact sines and cosines, so use exact values, not decimals.
"Hence" after a sketch. If the next part says "hence state the number of solutions", count intersections on your sketch; no algebra is required.
Range. For $a\sin(bx) + c$ over a full period, the range is $c - |a| \le y \le c + |a|$ . Over a shorter interval, check the values at the ends.

Summary

On the unit circle, the point at angle $\theta$ is $(\cos\theta, \sin\theta)$ ; this defines sine and cosine for all angles.
$\sin x$ and $\cos x$ : period $360^\circ$ ( $2\pi$ ), range $[-1, 1]$ . $\tan x$ : period $180^\circ$ ( $\pi$ ), asymptotes at $90^\circ + 180^\circ n$ .
$y = a\sin(bx) + c$ : amplitude $|a|$ , period $\tfrac{360^\circ}{b}$ , middle line $y = c$ , range $c - |a|$ to $c + |a|$ .
$y = a\tan(bx) + c$ : period $\tfrac{180^\circ}{b}$ ; asymptotes where $bx = 90^\circ, 270^\circ, \ldots$
Constants: $c = \tfrac{\max + \min}{2}$ , $a = \tfrac{\max - \min}{2}$ , or substitute known points.
Solutions of $f(x) = k$ are the intersections with $y = k$ : count them on a sketch, watching the endpoints.

Practice questions

Question

State the period and the range of $y = 2 - \cos 3x$ .
Sketch $y = 3\sin\left(x + 60^\circ\right)$ for $0^\circ \le x \le 360^\circ$ , stating the coordinates of the maximum and minimum points and the $y$ -intercept.
The curve $y = a\sin bx + c$ , where $a$ , $b$ , $c$ are positive, has maximum value $7$ , minimum value $1$ and period $\pi$ . Find $a$ , $b$ and $c$ .
Sketch $y = \tan\left(x - \tfrac{\pi}{4}\right)$ for $0 \le x \le 2\pi$ , showing the asymptotes and the $x$ -intercepts.
The curve $y = a + b\cos x$ passes through $(0, 7)$ and $\left(\tfrac{\pi}{3}, 4\right)$ . Find $a$ and $b$ , and state the range of $y$ .
Use a sketch to state the number of solutions of $3\sin 2x = 2$ for $0 \le x \le 2\pi$ .
Describe a sequence of transformations that maps $y = \sin x$ onto $y = 2\sin\tfrac{x}{2} - 1$ , and state the period and range of the new curve.
The function $f$ is defined by $f(x) = 3\cos x - 1$ for $0 \le x \le 2\pi$ . (a) Find the range of $f$ . (b) Sketch the graph of $y = f(x)$ . (c) State the set of values of $k$ for which $f(x) = k$ has exactly one solution.
The graph of $y = a\tan(bx) + c$ , for $-\tfrac{\pi}{6} < x < \tfrac{\pi}{6}$ , has asymptotes $x = \pm\tfrac{\pi}{6}$ , passes through $(0, 1)$ and through $\left(\tfrac{\pi}{12}, 3\right)$ . Find $a$ , $b$ and $c$ .

Answers

Period $\tfrac{360^\circ}{3} = 120^\circ$ (or $\tfrac{2\pi}{3}$ ). $-1 \le \cos 3x \le 1$ , so $1 \le 2 - \cos 3x \le 3$ .
The sine wave is stretched by factor $3$ and translated $60^\circ$ to the left. Maximum when $x + 60^\circ = 90^\circ$ : $(30^\circ, 3)$ . Minimum when $x + 60^\circ = 270^\circ$ : $(210^\circ, -3)$ . $y$ -intercept: $3\sin 60^\circ = \tfrac{3\sqrt{3}}{2}$ . Zeros where $x + 60^\circ = 180^\circ$ or $360^\circ$ : $x = 120^\circ, 300^\circ$ .
$c = \tfrac{7 + 1}{2} = 4$ , $a = \tfrac{7 - 1}{2} = 3$ , and $\tfrac{2\pi}{b} = \pi$ gives $b = 2$ . So $y = 3\sin 2x + 4$ .
The tangent graph translated $\tfrac{\pi}{4}$ to the right. Asymptotes where $x - \tfrac{\pi}{4} = \tfrac{\pi}{2}$ or $\tfrac{3\pi}{2}$ : $x = \tfrac{3\pi}{4}$ and $x = \tfrac{7\pi}{4}$ . Zeros where $x - \tfrac{\pi}{4} = 0$ or $\pi$ : $x = \tfrac{\pi}{4}$ and $\tfrac{5\pi}{4}$ . The $y$ -intercept is $\tan\left(-\tfrac{\pi}{4}\right) = -1$ .
$a + b = 7$ and $a + \tfrac{1}{2}b = 4$ , so $\tfrac{1}{2}b = 3$ , $b = 6$ , $a = 1$ . $y = 1 + 6\cos x$ has range $-5 \le y \le 7$ .
$\sin 2x = \tfrac{2}{3}$ . Over $0 \le x \le 2\pi$ , $y = \sin 2x$ makes two complete waves, and the line $y = \tfrac{2}{3}$ (between $0$ and $1$ ) cuts each wave twice: $4$ solutions.
A stretch parallel to the $x$ -axis with factor $2$ , a stretch parallel to the $y$ -axis with factor $2$ , and a translation by $\begin{pmatrix} 0 \\ -1 \end{pmatrix}$ (the two stretches can be in either order; the translation comes after the $y$ -stretch). Period $4\pi$ (or $720^\circ$ ); range $-3 \le y \le 1$ .
(a) $-1 \le \cos x \le 1$ , so $-4 \le f(x) \le 2$ . (b) A cosine wave with amplitude $3$ and middle line $y = -1$ : from $(0, 2)$ down to $(\pi, -4)$ and back to $(2\pi, 2)$ , crossing $y = -1$ at $\tfrac{\pi}{2}$ and $\tfrac{3\pi}{2}$ . (c) $y = k$ meets the graph once only at the minimum, $k = -4$ . (For $-4 < k < 2$ there are two solutions, and for $k = 2$ there are two, at $x = 0$ and $x = 2\pi$ .) So $k = -4$ .
At $x = 0$ , $\tan 0 = 0$ , so $c = 1$ . The asymptotes of $\tan(bx)$ nearest the origin are at $bx = \pm\tfrac{\pi}{2}$ , i.e. $x = \pm\tfrac{\pi}{2b}$ , so $\tfrac{\pi}{2b} = \tfrac{\pi}{6}$ and $b = 3$ . Then $a\tan\tfrac{\pi}{4} + 1 = 3$ , so $a = 2$ . The curve is $y = 2\tan 3x + 1$ .

Sine and cosine for any angle

The three basic graphs

Symmetries you can read off the graphs

Transformed graphs

Using graphs to count solutions

Worked examples

Practice questions

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