Trigonometric Graphs
Trigonometric graphs represent the periodic wave patterns created by sine, cosine, and tangent functions. The standard function y = sin x produces a smooth wave that oscillates between -1 and 1 with a period of 2π. Transformations of these basic functions create variations in amplitude, frequency, and position that appear throughout advanced mathematics.
Why it matters
Trigonometric graphs model countless periodic phenomena in science and engineering. Sound waves follow sine patterns with specific frequencies and amplitudes — a 440 Hz note creates 440 complete cycles per second. Alternating current electricity oscillates sinusoidally at 50 Hz in the UK power grid. Ocean tides, seasonal temperature variations, and even population cycles in biology exhibit trigonometric patterns. In GCSE and A-Level mathematics, these graphs appear in mechanics problems involving simple harmonic motion, where objects oscillate with predictable periods and amplitudes. Engineering students use these concepts to analyse vibrations in structures, while medical equipment like ECG machines interpret the periodic patterns of heartbeats. Weather forecasting relies on trigonometric models to predict seasonal changes, and economists use similar wave functions to analyse cyclical market trends.
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.
Worked examples
What is the period of y = sin(3x)?
Answer: 2π/3
- Identify the period → period = 2π/3 — For sin(Bx), the period is 2π divided by the coefficient B. Here B = 3, so period = 2π/3 = 2π/3.
Find the amplitude and period of y = 3 sin(2x).
Answer: amplitude = 3, period = π
- Amplitude is the leading coefficient → amplitude = 3 — |A| in y = A sin(Bx) gives the amplitude. Here A = 3.
- Period is 2π divided by the coefficient of x → period = 2π/2 = π — For sin, one full cycle spans 2π when the argument increases by 2π. With B = 2, the argument reaches 2π when x reaches 2π/2.
Find the amplitude, period, and phase shift of y = 4 sin(2x + π).
Answer: amplitude = 4, period = π, phase shift = π/2 to the left
- Amplitude from the leading coefficient → amplitude = 4 — |A| = 4
- Period = 2π / |B| → period = π — B = 2, so period = 2π/2 = π.
- Phase shift = −C / B → phase shift = π/2 to the left — The argument is B x + C with B = 2 and C = π. Phase shift is −C/B, which moves the graph horizontally. Positive shift = right; negative = left.
Common mistakes
- Confusing amplitude with period when A = 2 and B = 2, incorrectly stating that both the amplitude and period equal 2, when the amplitude is 2 but the period is π.
- Calculating phase shift incorrectly for y = sin(2x + π) by using C/B instead of -C/B, giving π/2 to the right instead of π/2 to the left.
- Finding the period of y = tan(3x) as 2π/3 instead of π/3, forgetting that tangent has a fundamental period of π rather than 2π.
- Identifying the vertical shift as A instead of D in y = 3 sin x + 5, stating the vertical shift is 3 instead of 5.