A12AlgebraHigherGrades 6–8

Trigonometric Graphs

Q: Why does tan have asymptotes?

x = x x, and x = 0 at 90° and 270°.

In right-angled triangles, angles never exceed 90°. But $\sin x$ , $\cos x$ and $\tan x$ are defined for any angle, and their graphs repeat forever. Knowing these three shapes lets you sketch them, read values, and find every solution of an equation like $\sin x = -0.6$ between 0° and 360°.

45 min AQA GCSE Maths

The big picture

Sine and cosine are waves that repeat every 360° and stay between −1 and 1. Tangent repeats every 180° and has vertical asymptotes where it is undefined.

A calculator gives only one angle for $\sin^{-1}$ , $\cos^{-1}$ or $\tan^{-1}$ . The graph’s symmetry supplies the others — this is the key skill AQA tests.

What you'll learn

Sketch $y = \sin x$ , $y = \cos x$ and $y = \tan x$ for $0^{\circ} \le x \le 360^{\circ}$
Recall key values and the period of each graph
Use symmetry to solve trigonometric equations in a given interval
Apply translations and reflections to trig graphs

The three graphs

$y = \sin x$ : starts at $(0, 0)$ , maximum $(90^{\circ}, 1)$ , back to 0 at $180^{\circ}$ , minimum $(270^{\circ}, -1)$ , 0 at $360^{\circ}$ .

$y = \cos x$ : starts at $(0, 1)$ , 0 at $90^{\circ}$ , minimum $(180^{\circ}, -1)$ , 0 at $270^{\circ}$ , back to 1 at $360^{\circ}$ .

$y = \tan x$ : passes through $(0, 0)$ , $(180^{\circ}, 0)$ and $(360^{\circ}, 0)$ , with asymptotes at $x = 90^{\circ}$ and $x = 270^{\circ}$ .

Why this works

The cosine graph is the sine graph shifted 90° to the left: $\cos x = \sin(x + 90^{\circ})$ . Learning one shape well gives you both.

Tip — Sketch with the axes marked every 90° and label ±1. That is enough for almost every exam question.

Solving equations with symmetry

Sine is symmetrical about $90^{\circ}$ (and $270^{\circ}$ ): if one solution is $\alpha$ , another is $180^{\circ} - \alpha$ .

Cosine is symmetrical about $180^{\circ}$ : the second solution is $360^{\circ} - \alpha$ .

Tangent repeats every $180^{\circ}$ : the second solution is $\alpha + 180^{\circ}$ .

Example 1Solve

\cos x = 0.25

for

0^{\circ} \le x \le 360^{\circ}

x = \cos^{-1}(0.25) = 75.5^{\circ}

2Other solution:

360^{\circ} - 75.5^{\circ}

Answer

x = 75.5^{\circ}

x = 284.5^{\circ}

Your turnSame method, new numbers — try it before moving on.

Solve $\cos x = 0.6$ for $0^\circ \le x \le 360^\circ$ . Give your answers to 1 decimal place.

Your answer

Separate with a comma, to 1 d.p.

Show worked solution

Solution

1 $\cos^{-1}(0.6) = 53.1^\circ$ .
2By symmetry about $180^\circ$ : $360 - 53.1 = 306.9^\circ$ .

Answer: $x = 53.1^\circ$ or $x = 306.9^\circ$

Example 2Solve

\sin x = -0.6

for

0^{\circ} \le x \le 360^{\circ}

1Calculator:

\sin^{-1}(-0.6) = -36.9^{\circ}

, outside the range.

2The graph is negative between

180^{\circ}

and

360^{\circ}

. Using symmetry:

180^{\circ} + 36.9^{\circ}

and

360^{\circ} - 36.9^{\circ}

Answer

x = 216.9^{\circ}

x = 323.1^{\circ}

Rearranging first

Make the trig function the subject before using the inverse.

Example 1Solve

4\cos x + 2 = 0

for

0^{\circ} \le x \le 360^{\circ}

\cos x = -0.5

\cos^{-1}(-0.5) = 120^{\circ}

3Other solution:

360^{\circ} - 120^{\circ}

Answer

x = 120^{\circ}

x = 240^{\circ}

Your turnSame method, new numbers — try it before moving on.

Solve $2\sin x - 1 = 0$ for $0^\circ \le x \le 360^\circ$ .

Your answer

Separate with a comma.

Show worked solution

Solution

1 $\sin x = \frac{1}{2}$ , so the principal value is $30^\circ$ .
2By symmetry about $90^\circ$ : $180 - 30 = 150^\circ$ .

Answer: $x = 30^\circ$ or $x = 150^\circ$

Example 2Solve

\tan x = -1

for

0^{\circ} \le x \le 360^{\circ}

\tan^{-1}(-1) = -45^{\circ}

2Add

180^{\circ}

repeatedly:

135^{\circ}

315^{\circ}

Answer

x = 135^{\circ}

x = 315^{\circ}

Tip — Sketch the graph and draw the horizontal line $y = k$ . Counting the crossings tells you how many solutions to find.

Transforming trig graphs

$y = \sin x + 2$ moves the wave up 2, so it runs between 1 and 3.

$y = \cos(x - 90^{\circ})$ moves the cosine graph 90° right, which gives the sine graph.

$y = -\sin x$ flips the wave in the $x$ -axis, so its first turning point is a minimum at $(90^{\circ}, -1)$ .

Your turnSame method, new numbers — try it before moving on.

What is the maximum value of $y = \sin x + 2$ ?

Your answer

Show worked solution

Solution

1The maximum of $\sin x$ is 1.
2Adding 2 moves the graph up: maximum $3$ .

Answer: $3$

Think like an examiner

Give all solutions in the stated range.
Round angles to 1 decimal place unless told otherwise.
Mark asymptotes on tangent sketches as dashed lines.

Remember these

\sin x = \sin(180^{\circ} - x)

Sine symmetry.

\cos x = \cos(360^{\circ} - x)

Cosine symmetry.

\tan x = \tan(x + 180^{\circ})

Tangent period.

Watch out for these

Giving only the calculator value.

Use the graph to find the second solution.

Using

180^{\circ} - \alpha

for cosine.

Cosine uses

360^{\circ} - \alpha

Including a negative calculator angle as an answer.

Convert it into the required range.

Drawing

\tan x

as a wave between −1 and 1.

It is unbounded with asymptotes at 90° and 270°.

Stretch yourself

Solve $5\sin x + 3 = 1$ for $0^{\circ} \le x \le 360^{\circ}$ , giving answers to 1 decimal place.

Hint — Rearrange to $\sin x = -0.4$ ; both solutions are between 180° and 360°.

Reveal solution

Solution

\sin x = -0.4

. The acute angle with

\sin = 0.4

23.6^{\circ}

. Sine is negative in

180^{\circ} < x < 360^{\circ}

, so

x = 180^{\circ} + 23.6^{\circ} = 203.6^{\circ}

and

x = 360^{\circ} - 23.6^{\circ} = 336.4^{\circ}

\blacksquare

Questions students ask

What is the period of a graph?

The horizontal length of one complete repeat: 360° for sine and cosine, 180° for tangent.

Why does tan have asymptotes?

\tan x = \dfrac{\sin x}{\cos x}

, and

\cos x = 0

at 90° and 270°.

How many solutions does

\sin x = 2

have?

None — sine never exceeds 1.

Key takeaways

Sine and cosine: period 360°, range −1 to 1.
Tangent: period 180°, asymptotes at 90° and 270°.
Sine: $180^{\circ} - \alpha$ . Cosine: $360^{\circ} - \alpha$ . Tangent: $\alpha + 180^{\circ}$ .
Rearrange before using inverse functions.
Translations and reflections work as for any graph.

Exam-style questions

Practise it the way it is examined

3 original questions · 7 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 62 marks

(a)Write down the maximum value of $y = 3\cos x$ .[1]
Show mark scheme (a)
- B1 $3$
(b)Write down the period of $y = \tan x$ .[1]
Show mark scheme (b)
- B1 $180^\circ$

Question 2Grade 73 marks

Solve $5\sin x + 2 = 0$ for $0^\circ \le x \le 360^\circ$ . Give your answers to 1 decimal place.[3]
Show mark scheme
- M1 $\sin x = -0.4$
- M1Principal value $-23.6^\circ$ or one correct solution
- A1 $203.6^\circ$ and $336.4^\circ$
Answer: $203.6^\circ$ , $336.4^\circ$

Question 3Grade 82 marks

Given that $\sin 30^\circ = 0.5$ , use the graph of $y = \sin x$ to find all solutions of $\sin x = -0.5$ for $0^\circ \le x \le 360^\circ$ .[2]
Show mark scheme
- M1Uses symmetry, e.g. $180 + 30$
- A1 $210^\circ$ and $330^\circ$
Answer: $210^\circ$ , $330^\circ$

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

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