Trigonometric Graphs & Graph Transformations — AQA GCSE Mathematics

SPEC A12 (H) + A13 • HIGHER TIER ONLY

Trig Graphs & Graph Transformations

Sine, cosine, tangent — and how to translate or reflect any function sketch.

y = sin x HIGHER

The sine graph is a smooth wave. Amplitude 1, period 360°.

Key values for sine

sin 0° = 0 · sin 90° = 1 · sin 180° = 0 · sin 270° = −1 · sin 360° = 0

Maximum value +1, minimum −1. Repeats every 360°.

y = cos x HIGHER

The cosine graph has the same wave shape as sine — but it starts at y = 1 (shifted 90° left of sine).

Key values for cosine

cos 0° = 1 · cos 90° = 0 · cos 180° = −1 · cos 270° = 0 · cos 360° = 1

Starts at maximum. Period 360°. Same wave shape as sine — just shifted.

y = tan x HIGHER

The tangent graph has a very different character — period 180° (not 360°) and vertical asymptotes at 90°, 270° … (wherever cos = 0).

Key features of tan x

Period 180° — shape repeats every 180° (half of sin/cos period).

Vertical asymptotes at x = 90°, 270°, 450°, … (wherever cos x = 0, since tan = sin/cos).

Crosses x-axis at 0°, 180°, 360° (where sin x = 0).

Not bounded — tan x can take any real value (unlike sin and cos, which are between −1 and 1).

Key Values to Memorise

Two sets of values anchor every trig graph sketch and every trig-solve question. Know them cold.

Quadrantal values (the corners of each graph)

Anglesin xcos xtan x0°01090°10undefined (asymptote)180°0−10270°−10undefined (asymptote)360°010

Exact values (spec G21 — required at F + H)

Anglesincostan30°¹⁄₂^√3⁄₂¹⁄_√345°^√2⁄₂^√2⁄₂160°^√3⁄₂¹⁄₂√3You need to reproduce these without a calculator. In particular: sin 30° = ¹⁄₂ and cos 60° = ¹⁄₂ come up often in solve-from-graph questions.

Pattern to remember

sin starts at 0 and rises; cos starts at 1 and falls. At 90° sin peaks, cos is zero. At 180° sin is zero, cos bottoms out. The cycle flips and repeats for 180°–360°.

Transformation 1: y = f(x) + a HIGHER

Adding a constant outside the function moves the graph vertically.

y = f(x) + a → translate by (⁰⁄ₐ)

Rules (outside the function)

y = f(x) + 3 → graph moves UP 3 units.

y = f(x) − 3 → graph moves DOWN 3 units.

Every y-coordinate increases/decreases by the same amount. Shape and position on x-axis unchanged.

Transformation 2: y = f(x + a) HIGHER

Adding a constant inside the brackets moves the graph horizontally — but the direction is the opposite of what you'd expect.

y = f(x + a) → translate by (⁻ᵃ⁄₀)

⚠️ Sign reversal! Inside the brackets, the sign of the constant is opposite to the direction of the shift.

y = f(x + 3) → translate LEFT 3 (not right).

y = f(x − 3) → translate RIGHT 3 (not left).

Reflections: y = −f(x) and y = f(−x) HIGHER

A minus sign outside reflects in the x-axis; a minus sign inside reflects in the y-axis.

y = −f(x): reflect in x-axis

y = f(−x): reflect in y-axis

Rules for the minus sign

Outside the function (y = −f(x)): flip up/down. Each y becomes −y. Reflection in the x-axis.

Inside the function (y = f(−x)): flip left/right. Each x becomes −x. Reflection in the y-axis.

All Four Transformations HIGHER

RuleEffectMnemonic

y = f(x) + aTranslate UP by a (outside → vertical, same sign)"Outside same"

y = f(x + a)Translate LEFT by a (inside → horizontal, opposite sign)"Inside opposite"

y = −f(x)Reflect in x-axis (minus outside → flip up/down)"Minus outside, flip over x"

y = f(−x)Reflect in y-axis (minus inside → flip left/right)"Minus inside, flip through y"

The universal rule

Anything outside f(·) affects the graph in the y-direction and behaves normally. Anything inside f(·) affects the graph in the x-direction and behaves opposite to what you'd expect.

Worked Example — Vertical Translation

HIGHER

Given y = f(x) = x², sketch y = f(x) + 3 on the same axes. [2 marks]

M

Method: adding 3 outside the function translates the graph vertically. Outside → same sign → UP by 3.

C

Take each point (x, y) on y = x² and move it to (x, y + 3):

(−2, 4) → (−2, 7) · (0, 0) → (0, 3) · (2, 4) → (2, 7).

Vertex moves from (0, 0) to (0, 3).

E

Evaluate: the new graph is y = x² + 3 — parabola with vertex at (0, 3), same shape, 3 units higher. B1 correct translation, B1 through (0, 3).

THE SKETCH

Worked Example — Horizontal Translation

HIGHER

Given y = f(x) = x², sketch y = f(x − 3) on the same axes. [2 marks]

M

Method: subtracting 3 inside the function translates horizontally. Inside → opposite sign → RIGHT by 3 (not left!).

C

Each point (x, y) on y = f(x) moves to (x + 3, y). The new function is y = (x − 3)².

(0, 0) → (3, 0) · (−2, 4) → (1, 4) · (2, 4) → (5, 4).

Vertex moves from (0, 0) to (3, 0).

E

Evaluate: translated graph is y = (x − 3)² with vertex at (3, 0). Same parabola shape, shifted right by 3. B1 correct direction, B1 through (3, 0).

THE SKETCH

Worked Example — Reflection

HIGHER

The graph of y = f(x) passes through (0, 1), (−2, 5) and (2, 5). Sketch the reflection of y = f(x) in the x-axis. [2 marks]

(Given: f(x) = x² + 1.)

M

Method: reflection in the x-axis is the transformation y = −f(x). Every y-coordinate flips sign.

C

Apply (x, y) → (x, −y) to each point:

(0, 1) → (0, −1) · (−2, 5) → (−2, −5) · (2, 5) → (2, −5).

The new function is y = −(x² + 1) = −x² − 1.

E

Evaluate: reflected graph is a parabola opening downwards, with maximum at (0, −1). Always below the x-axis (whereas original was always above). B1 mark intention for correct reflection through the x-axis.

THE SKETCH

Examiner Alert: Horizontal Direction HIGHER

⚠️ EXAMINER ALERT: The #1 transformation mistake at GCSE. Students see y = f(x + 3) and instinctively shift right by 3. Wrong — it shifts left.

✓ Correct

y = f(x + 3) → LEFT 3

y = f(x − 3) → RIGHT 3

y = f(x + 5) → LEFT 5

✗ Common wrong thinking

"Plus means right" — WRONG for inside-the-bracket shifts.

"Same sign = same direction" — only true for y = f(x) + a, not y = f(x + a).

Why the sign reverses

Think about what makes the output equal: if the new function is f(x + 3), then to get the same y-value that the original had at some input X, we need the new x to satisfy x + 3 = X, so x = X − 3. The new input is 3 less — the graph has moved LEFT.

Examiner Alert: Key Trig Values & Solving HIGHER

⚠️ EXAMINER ALERT: "Solve tan x = 5 for 0° ≤ x ≤ 360°" is a classic AQA-style question that catches many students out. A common wrong answer is −78.69° — students forget to adjust to the required range and leave the negative calculator output as the final answer. Always check your answer lies in the stated range.

Trig solve strategy

Use your calculator in degree mode to find the first solution: arctan(5) ≈ 78.69°.

Use the graph pattern to find the second: tan has period 180°, so the second solution is 78.69° + 180° = 258.69°.

Check: both solutions are in [0°, 360°] ✓. Reject any outside the range (including negative answers).

Memorise these anchor values

sin 90° = 1 (not 0). cos 0° = 1 (not 0). tan 45° = 1. tan 90° is undefined (asymptote — not 1, not infinity on a calculator display, just undefined).

Practice Question 1 HIGHER

Use the graph of y = sin x shown below to find all solutions of sin x = 0.5 in the range 0° ≤ x ≤ 360°. [2 marks]

M

Method: find where the sine curve meets the line y = 0.5. First solution via calculator (arcsin); second solution via symmetry of the sine curve about 90°.

C

First solution: since sin 30° = ¹⁄₂ (exact value from G21), x = 30°.

Second solution: sine is symmetric about x = 90° on the rising-and-falling arch, so if one solution is 30° before the peak, the other is 30° after: 180° − 30° = 150°.

Both within the range [0°, 360°].

E

Evaluate: x = 30° and x = 150°. A1 first solution, A1 second solution. A sketch confirming both intersections with y = 0.5 reassures the examiner.

Practice Question 2 HIGHER

The graph of y = f(x) is shown. On the same axes, sketch y = f(x) − 2. [2 marks]

M

Method: subtracting 2 outside the function means vertical translation DOWN by 2 units. Every y-coordinate decreases by 2; the shape stays the same.

C

Transform key points:

(0, 2) → (0, 0) · (−4, 6) → (−4, 4) · (4, 6) → (4, 4).

Draw the same curve shape, 2 units lower — passes through the origin (where (0, 2) used to be).

E

Evaluate: new graph same shape, now with vertex at (0, 0). B1 vertical translation, B1 passes through (0, 0). A common error is translating up instead of down — watch the sign of the constant.

THE SKETCH

Practice Question 3 HIGHER

The graph of y = f(x) has a local maximum at (2, 7) and passes through (0, 3).

1

Write down the coordinates of the local maximum of y = f(x) + 4 . [1 mark]

2

Write down the coordinates of the local maximum of y = f(x − 1) . [1 mark]

3

Write down the coordinates of the local maximum of y = −f(x) (specify whether it becomes a max or min). [2 marks]

M

Method: apply each transformation rule to the point (2, 7). Outside/inside/sign — be careful.

C

(a) y = f(x) + 4: vertical translation UP 4. (2, 7) → (2, 11).

(b) y = f(x − 1): horizontal translation RIGHT 1 (sign reversal). (2, 7) → (3, 7).

(c) y = −f(x): reflect in x-axis. Every y flips sign. (2, 7) → (2, −7). A maximum flips to become a minimum.

E

Evaluate: (a) (2, 11), (b) (3, 7), (c) (2, −7), now a minimum. B1 each. For (c): reflection doesn't just flip the y-value — it flips max ↔ min.

Summary

Trig graphs (angles in degrees, any size)

y = sin x : wave, amplitude 1, period 360°. Key: sin 0 = 0, sin 90 = 1, sin 180 = 0, sin 270 = −1.

y = cos x : same wave, shifted 90° left. Key: cos 0 = 1, cos 90 = 0, cos 180 = −1.

y = tan x : period 180°, asymptotes at 90° and 270°. Crosses x-axis at 0°, 180°, 360°.

Transformations (translations & reflections)

y = f(x) + a : UP by a. Outside → vertical → normal sign.

y = f(x + a) : LEFT by a. Inside → horizontal → opposite sign.

y = −f(x) : reflect in x-axis. (Max becomes min and vice versa.)

y = f(−x) : reflect in y-axis.

Exam Tip

Always identify inside vs outside first. Inside the function means x-direction and sign-reversal. Outside means y-direction and normal sign. That single rule handles every GCSE transformation question at A13.