MathematicsCore20 min read

Trigonometric Functions and their Graphs

Amplitude, period and phase — three letters, three separate jobs

This topic appears in:

1st Year Mathematics · ch.11
01

The three basic shapes

Before any transformation, know the parent curves cold. Sine and cosine are the same wave shifted by 90°: both run between −1 and 1, both repeat every 360°, and both are continuous everywhere. Tangent is a different animal — it has no maximum, it repeats every 180°, and it is undefined wherever cosine is zero.

FunctionDomainRangePeriodNotable points
y = sin xall real x−1 ≤ y ≤ 1360°starts at 0, peak at 90°
y = cos xall real x−1 ≤ y ≤ 1360°starts at 1, zero at 90°
y = tan xx ≠ 90°, 270°, …all real y180°asymptote wherever cos x = 0
y = cosec xx ≠ 0°, 180°, …|y| ≥ 1360°reciprocal of sin
y = sec xx ≠ 90°, 270°, …|y| ≥ 1360°reciprocal of cos
y = cot xx ≠ 0°, 180°, …all real y180°reciprocal of tan

Why tan has asymptotes and sin does not

tan x = sin x / cos x. Wherever cos x = 0 the fraction has a zero denominator, so the function is undefined and the graph runs off to infinity — at 90°, 270° and every 180° after that. Sine and cosine have no denominators, so nothing can break, which is why they are defined for every value of x.

02

Even, odd, and the symmetries worth using

Cosine is an even function: cos(−x) = cos x, so its graph is symmetrical about the y-axis. Sine and tangent are odd: sin(−x) = −sin x, so their graphs have rotational symmetry about the origin.

These are not decoration. They halve the work in any question that involves a negative angle, and they explain why sin(−30°) = −0.5 while cos(−30°) = +0.866 — a distinction that costs marks when guessed.

cos(−x) = cos x(even)sin(−x) = −sin x(odd)tan(−x) = −tan x(odd)sin(x + 360°) = sin x(periodic)periodicity means every trigonometric equation has infinitely many solutions unless a range is given
03

y = A sin(Bx + C) + D, one letter at a time

Every transformed trigonometric graph on the syllabus fits this template, and each constant does exactly one job. Learn them separately and no combination is confusing.

  • A — amplitude. The curve now runs between −|A| and +|A|: a vertical stretch. A negative A also flips the graph upside down.
  • B — frequency. The period becomes 360°/B for sine and cosine, or 180°/B for tangent. Bigger B means more waves squeezed into the same width.
  • C — phase shift. The graph moves horizontally by −C/B. A positive C moves it to the left, which is the opposite of what most students expect.
  • D — vertical shift. The whole curve moves up by D, so the centre line becomes y = D instead of the x-axis.

Change one slider at a time. A stretches the curve vertically without touching where it crosses the axis; B squeezes it horizontally; C slides it sideways. The read-out gives the period and the phase shift as numbers.

Why the shift is −C/B and not −C

Factorise the bracket: sin(2x + 60°) = sin[2(x + 30°)]. The transformation acting on x is a shift of 30°, not 60°, because the B has already stretched the horizontal axis. Always factorise B out before reading off the phase shift — quoting −C is the most reliable way to lose a mark in this chapter.

04

Sketching one of these in an exam

The marks are for a curve with the right shape in the right place, not for artistic quality. Work through the constants in a fixed order and the sketch takes ninety seconds.

Worked example

Sketch y = 3 sin(2x − 60°) + 1 for 0° ≤ x ≤ 360°, stating the amplitude, period and phase shift.

  1. Amplitude |A| = 3, and the vertical shift D = 1, so the curve oscillates between 1 − 3 = −2 and 1 + 3 = 4.Draw the centre line y = 1 and the two bounds first. They frame everything else.
  2. Period = 360°/2 = 180°, so exactly two complete waves fit into the range.Knowing how many waves to draw prevents the usual over- or under-crowded sketch.
  3. Factorise: 3 sin[2(x − 30°)] + 1, so the phase shift is +30° to the right.C is negative here, so the shift is to the right — the sign works out opposite to C.
  4. The curve therefore starts its cycle at x = 30° on the centre line, rising.A plain sine starts at its centre going up; the shift just moves that starting point.
  5. Mark the peak at x = 30 + 45 = 75° and the trough at x = 30 + 135 = 165°, then repeat 180° later.A peak occurs a quarter of a period after the start, a trough three quarters after.

Amplitude 3, period 180°, phase shift 30° right, centre line y = 1

05

Solving trigonometric equations from the graph

A trigonometric equation has infinitely many solutions, because the graph repeats for ever. A question therefore always states a range, and the number of solutions inside it is decided by how many times the horizontal line cuts the curve.

The reliable method: find the principal value from the calculator, then use the symmetry of the graph to find every other solution in range. For sine, the second solution in each cycle is 180° − θ; for cosine it is 360° − θ; for tangent, solutions simply repeat every 180°.

sin x = k → x = θ and 180° − θ, then add 360° repeatedlycos x = k → x = θ and 360° − θ, then add 360° repeatedlytan x = k → x = θ, then add 180° repeatedlyθ is the principal value from the calculator; a sketch shows how many of the rest fall inside the range

Before you leave this chapter

  1. sin and cos have period 360° and range [−1, 1]; tan has period 180°, no bound, and asymptotes where cos x = 0.
  2. cos is even, sin and tan are odd.
  3. Amplitude |A|; period 360°/B (or 180°/B for tan); phase shift −C/B; centre line y = D.
  4. Factorise B out of the bracket before reading the phase shift.
  5. Every equation has infinitely many solutions — use the stated range and the graph's symmetry to find exactly the ones asked for.

Practice questions

6 questions · 20 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
State the period and range of y = 4 cos 3x.
Model answer

Period = 360°/3 = 120°. Amplitude 4, so the range is −4 ≤ y ≤ 4.

Examiner tip. The 4 affects the range only, and the 3 affects the period only. Mixing them up is what this question is designed to catch.

SQ2[2 marks]
Explain why y = tan x has vertical asymptotes.
Model answer

tan x = sin x / cos x, so wherever cos x = 0 — at 90°, 270° and every 180° thereafter — the function is undefined and the graph rises or falls without limit. Those values of x are excluded from the domain.

Examiner tip. Name the actual values, not just "where cos is zero". The mark scheme wants at least one of 90° or 270°.

SQ3[2 marks]
Is sin x even or odd? Give the identity that shows it.
Model answer

Odd, since sin(−x) = −sin x. Its graph has rotational symmetry of order 2 about the origin.

Examiner tip. Cos is the even one. Remembering that cos(0) = 1 sits on the axis of symmetry makes it easy to keep them apart.

Solved numericals

2 · 8 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
For y = 2 sin(3x + 90°) − 1, state the amplitude, the period, the phase shift and the range.
Full working
  1. Amplitude = |2| = 2[1]
  2. Period = 360°/3 = 120°[1]
  3. Factorising, 3x + 90° = 3(x + 30°), so the phase shift is 30° to the leftshift = −C/B = −90/3 = −30, i.e. 30° left[1]
  4. Centre line y = −1, so the range is −3 ≤ y ≤ 1D ± amplitude[1]

Amplitude 2, period 120°, shift 30° left, range −3 ≤ y ≤ 1

Examiner tip. Answering 90° for the phase shift is the standard error. Factorise the B out of the bracket first, every time.

N2[4 marks]
Solve 2 sin x = 1 for 0° ≤ x ≤ 360°.
Full working
  1. sin x = 0.5[1]
  2. Principal value x = 30°from the calculator or from the exact value[1]
  3. Second solution 180° − 30° = 150°sine is positive in the first and second quadrants[1]
  4. Both lie in range, and the next would be 390°, outside it: x = 30°, 150°a check that no further solutions fit[1]

x = 30° and x = 150°

Examiner tip. Sketch y = sin x and the line y = 0.5. The number of crossings inside the range tells you how many answers to expect before you calculate any of them.

Long questions

1 · 6 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[6 marks]
Consider the function f(x) = 3 cos(2x) + 1.
  1. State the amplitude, period, maximum and minimum values.
  2. Find the values of x in 0° ≤ x ≤ 360° at which f(x) takes its maximum.
  3. Sketch the graph over that interval.
Mark scheme
  1. Amplitude 3; period 360/2 = 180°[1]
  2. Maximum = 1 + 3 = 4; minimum = 1 − 3 = −2centre line plus and minus the amplitude[1]
  3. Maximum occurs when cos 2x = 1, i.e. 2x = 0°, 360°, 720°the range for 2x is 0° to 720°, which is where the extra solutions come from[1]
  4. x = 0°, 180°, 360°dividing the whole solution set by 2[1]
  5. Sketch shows two complete cycles across the intervalbecause the period is half the interval[1]
  6. Curve oscillates between −2 and 4 about the line y = 1, starting at its maximumcosine starts at its peak[1]

(a) amplitude 3, period 180°, max 4, min −2 (b) x = 0°, 180°, 360° (c) two cycles about y = 1

Examiner tip. When the argument is 2x, widen the search range to 0°–720° before solving, then halve every answer. Solving in the original range first is how solutions go missing.