IGCSE · Practice questions · Trigonometric graphs: amplitude and period

Three numbers, one curve.

Six original Cambridge-style questions on the graphs of \(y = a\sin bx + c\), \(y = a\cos bx + c\) and \(y = a\tan bx + c\): reading amplitude and period, sketching over a domain, labelling asymptotes, working back from a graph and counting the solutions of \(f(x) = k\) and \(|f(x)| = k\).

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Read \(a\), \(b\), \(c\), then sketch and count.

Amplitude \(a\); centre line \(y = c\); maximum \(c + a\), minimum \(c - a\).
Period \(\dfrac{360^{\circ}}{b}\) or \(\dfrac{2\pi}{b}\) for sin and cos; \(\dfrac{180^{\circ}}{b}\) or \(\dfrac{\pi}{b}\) for tan.
Sketch in quarter-periods; label intercepts, maximum and minimum points and the ends of the domain.
Tan: asymptotes where \(bx = 90^{\circ}, 270^{\circ}, \ldots\); label each with its equation.
Count solutions of \(f(x) = k\) as meetings with \(y = k\); for \(|f(x)| = k\) count \(f(x) = k\) and \(f(x) = -k\).

Recall

[3 marks]

For \(y = 4\cos 3x + 2\), \(0^{\circ} \le x \le 360^{\circ}\), state the amplitude, the period and the greatest and least values of \(y\).

Amplitude \(4\). ✓
Period \(\dfrac{360^{\circ}}{3} = 120^{\circ}\). ✓
Greatest value \(2 + 4 = 6\), least value \(2 - 4 = -2\). ✓

Calculation

[4 marks]

The curve \(y = 2\sin\dfrac{x}{2} - 1\) is drawn for \(0 \le x \le 4\pi\).

(a) State the period in radians.

(b) Find the coordinates of the maximum and minimum points.

(a) Period \(2\pi \div \dfrac{1}{2} = 4\pi\). ✓
(b) \(\sin\dfrac{x}{2} = 1\) at \(\dfrac{x}{2} = \dfrac{\pi}{2}\): maximum \((\pi, 1)\). \(\sin\dfrac{x}{2} = -1\) at \(\dfrac{x}{2} = \dfrac{3\pi}{2}\): minimum \((3\pi, -3)\). ✓
(c) \(\sin\dfrac{x}{2} = \dfrac{1}{2}\), so \(\dfrac{x}{2} = \dfrac{\pi}{6}, \dfrac{5\pi}{6}\) (for \(0 \le \dfrac{x}{2} \le 2\pi\)). ✓
\(x = \dfrac{\pi}{3}\) and \(x = \dfrac{5\pi}{3}\). ✓

Calculation

[4 marks]

Sketch the graph of \(y = 2\tan 3x + 2\) for \(0^{\circ} \le x \le 120^{\circ}\). Label the asymptotes with their equations and give the coordinates of the points where the curve meets the axes.

Period \(\dfrac{180^{\circ}}{3} = 60^{\circ}\). Asymptotes where \(3x = 90^{\circ}, 270^{\circ}\): \(x = 30^{\circ}\) and \(x = 90^{\circ}\). ✓
\(y\)-axis: \(y = 2\tan 0 + 2 = 2\), the point \((0^{\circ}, 2)\); the curve is back at \(y = 2\) at \(x = 60^{\circ}\) and \(x = 120^{\circ}\). ✓
\(x\)-axis: \(\tan 3x = -1\), so \(3x = 135^{\circ}, 315^{\circ}\), giving \((45^{\circ}, 0)\) and \((105^{\circ}, 0)\). ✓
Sketch: a rising tan branch from \((0^{\circ}, 2)\) up to the asymptote \(x = 30^{\circ}\), a full branch between \(x = 30^{\circ}\) and \(x = 90^{\circ}\) through \((45^{\circ}, 0)\) and \((60^{\circ}, 2)\), and a branch from the asymptote \(x = 90^{\circ}\) through \((105^{\circ}, 0)\) to \((120^{\circ}, 2)\). ✓

Analysis

[3 marks]

The graph of \(y = a\sin bx + c\), where \(a\) and \(b\) are positive integers, has a maximum point at \((30^{\circ}, 3)\) and the next minimum point at \((90^{\circ}, -5)\). Find \(a\), \(b\) and \(c\).

\(a = \dfrac{3 - (-5)}{2} = 4\), \(c = \dfrac{3 + (-5)}{2} = -1\). ✓
Maximum to the next minimum is half a period, \(60^{\circ}\), so the period is \(120^{\circ}\) and \(b = \dfrac{360^{\circ}}{120^{\circ}} = 3\). ✓
Check: \(4\sin 90^{\circ} - 1 = 3\) at \(x = 30^{\circ}\) and \(4\sin 270^{\circ} - 1 = -5\) at \(x = 90^{\circ}\). So \(a = 4\), \(b = 3\), \(c = -1\). ✓

Analysis

[5 marks]

(a) Sketch \(y = 3\cos 2x - 1\) for \(0^{\circ} \le x \le 180^{\circ}\), labelling the end points and the minimum point.

(b) State the number of solutions of \(3\cos 2x - 1 = 1\) in this interval.

(c) Find the value of \(k\) for which \(3\cos 2x - 1 = k\) has exactly one solution, and the set of values of \(k\) for which it has exactly two.

(a) Period \(180^{\circ}\): one wave from \((0^{\circ}, 2)\) down to the minimum \((90^{\circ}, -4)\) and back up to \((180^{\circ}, 2)\). ✓
(b) \(y = 1\) lies between \(-4\) and \(2\), so it cuts the wave twice: \(2\) solutions (\(x = 24.1^{\circ}\) and \(155.9^{\circ}\)). ✓
(c) Exactly one solution: the line only touches the minimum, \(k = -4\). ✓
For \(-4 < k < 2\) the line cuts the wave twice; for \(k = 2\) it meets the two end points \(x = 0^{\circ}\) and \(x = 180^{\circ}\). ✓
Exactly two solutions: \(-4 < k \le 2\). ✓

Analysis

[6 marks]

The function \(f\) is given by \(f(x) = 4\sin x - 2\) for \(0^{\circ} \le x \le 360^{\circ}\).

(a) State the range of \(f\).

(b) Sketch \(y = |f(x)|\), giving the coordinates of the points where it meets the \(x\)-axis and of its maximum points.

(d) Find the values of \(k\) for which \(|f(x)| = k\) has exactly two solutions.

(a) \(4\sin x\) runs from \(-4\) to \(4\), so \(-6 \le f(x) \le 2\). ✓
(b) \(f(x) = 0\) when \(\sin x = \dfrac{1}{2}\): \((30^{\circ}, 0)\) and \((150^{\circ}, 0)\). Reflect the part below the axis: maximum points \((90^{\circ}, 2)\) and \((270^{\circ}, 6)\); the ends are \((0^{\circ}, 2)\) and \((360^{\circ}, 2)\). ✓
(c) \(|f(x)| = 1\): \(f(x) = 1\) gives \(\sin x = \dfrac{3}{4}\), \(2\) solutions; \(f(x) = -1\) gives \(\sin x = \dfrac{1}{4}\), \(2\) solutions. Total \(4\). ✓
\(|f(x)| = 3\): \(f(x) = 3\) is impossible (above \(2\)); \(f(x) = -3\) gives \(\sin x = -\dfrac{1}{4}\), \(2\) solutions. Total \(2\). ✓
(d) For \(0 < k \le 2\) the line \(y = k\) meets the graph \(4\) times; for \(2 < k < 6\) it cuts only the tall hump, twice; \(k = 6\) touches once; \(k = 0\) gives the two zeros. ✓
Exactly two solutions: \(k = 0\) or \(2 < k < 6\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.