§ Trigonometry

Trigonometric Graphs

CCSS.HSF.TF.B.53 min readApr 13, 2026

Trigonometric graphs form the backbone of Year 12 A-level mathematics, yet many students struggle to identify transformations like amplitude and period correctly. When teaching y = A sin(Bx + C) + D, the key is helping students visualise how each parameter changes the parent function's shape and position.

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Why it matters

Trigonometric graphs model real-world phenomena from sound waves to tidal patterns. Engineers use sine curves to design bridges that withstand wind oscillations — the Millennium Bridge's infamous wobble was caused by resonant frequency matching pedestrian footsteps. Sound engineers adjust amplitude (volume) and frequency (pitch) parameters when mixing music, whilst marine engineers predict tide heights using cosine functions with periods of approximately 12.5 hours. In medicine, ECG readings show heartbeats as periodic waves, and understanding transformations helps diagnose irregular rhythms. GCSE students encounter these concepts at Foundation level, but A-level mathematics demands mastery of compound transformations. Students who grasp y = 2 sin(3x - π/2) + 1 can later tackle Fourier analysis in university engineering courses, making this topic essential for STEM progression.

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How to solve trigonometric graphs

Trig Graphs — A sin(Bx + C) + D

Amplitude = |A|. Vertical stretch/compression.
Period = 2π/|B| (π/|B| for tan).
Phase shift = −C/B (horizontal shift; + is left, − is right).
Vertical shift = D; midline y = D; max = D + |A|, min = D − |A|.

Example: y = 2 sin(3x − π) + 1: amp=2, period=2π/3, shift=π/3 right, midline y=1.

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Worked examples

Beginner§ 01

What is the amplitude of y = 3 cos(x)?

Answer: 3

Identify the amplitude → amplitude = 3 — The amplitude is the coefficient in front of cos, which is 3.

Easy§ 02

Find the amplitude and period of y = 4 cos(4x).

Answer: amplitude = 4, period = π/2

Amplitude is the leading coefficient → amplitude = 4 — |A| in y = A cos(Bx) gives the amplitude. Here A = 4.
Period is 2π divided by the coefficient of x → period = 2π/4 = π/2 — For cos, one full cycle spans 2π when the argument increases by 2π. With B = 4, the argument reaches 2π when x reaches 2π/4.

Medium§ 03

Find the amplitude, period, and phase shift of y = 4 cos(3x + π).

Answer: amplitude = 4, period = 2π/3, phase shift = π/3 to the left

Amplitude from the leading coefficient → amplitude = 4 — |A| = 4
Period = 2π / |B| → period = 2π/3 — B = 3, so period = 2π/3 = 2π/3.
Phase shift = −C / B → phase shift = π/3 to the left — The argument is B x + C with B = 3 and C = π. Phase shift is −C/B, which moves the graph horizontally. Positive shift = right; negative = left.

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Common mistakes

Confusing period calculation by writing 2π × B instead of 2π ÷ B. Students often calculate the period of y = 3 sin(2x) as 2π × 2 = 4π instead of 2π ÷ 2 = π.
Mixing up phase shift direction, claiming y = sin(x + π/2) shifts π/2 to the right instead of π/2 to the left. The formula −C/B determines direction, so positive C values shift left.
Forgetting amplitude is |A|, writing amplitude = −2 for y = −2 cos(x) instead of amplitude = 2. Negative coefficients create reflections, not negative amplitudes.
Calculating maximum and minimum values incorrectly for vertical shifts. For y = 3 sin(x) + 2, students write max = 3 and min = −3 instead of max = 5 and min = −1.

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Frequently asked questions

How do I remember whether phase shifts go left or right?

Use the formula phase shift = −C/B. If the result is positive, shift right; if negative, shift left. For y = sin(x + π/2), C = π/2 and B = 1, so shift = −π/2 ÷ 1 = −π/2 (left). Think 'negative C pulls the graph leftward'.

Why is tan's period π/|B| instead of 2π/|B|?

The tangent function completes one full cycle every π radians, not 2π like sine and cosine. This shorter period reflects tan's vertical asymptotes occurring at odd multiples of π/2. For y = tan(2x), the period becomes π/2, creating asymptotes every π/4 units.

What's the difference between amplitude and vertical stretch factor?

Amplitude specifically measures the maximum displacement from the midline and is always positive (|A|). Vertical stretch factor includes the sign, so y = −3 sin(x) has amplitude 3 but stretch factor −3, indicating both scaling and reflection across the x-axis.

How do I find maximum and minimum values when there's a vertical shift?

Add the vertical shift D to the amplitude |A| for the maximum, and subtract |A| from D for the minimum. For y = 2 sin(x) − 3, the midline is y = −3, so maximum = −3 + 2 = −1 and minimum = −3 − 2 = −5.

Can I have negative periods or amplitudes?

No, periods and amplitudes are always positive by definition. If B is negative in y = sin(Bx), use |B| for period calculations. Negative A values create reflections but the amplitude remains |A|. Think of these as magnitudes — they measure size, not direction.

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