Trigonometric graphs, exact values and small angles - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 17 exam-style questions on Trigonometric graphs, exact values and small angles for Edexcel A-Level Maths (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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17 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

The angle \(\alpha\) satisfies \(0 < \alpha < \dfrac{\pi}{2}\) and \(\cos\alpha = c\).

Using the symmetry and periodicity of the graph of \(y = \cos x\), find, in terms of \(\alpha\), all the values of \(x\) in the interval \(0 \leqslant x \leqslant 4\pi\) for which \(\cos x = c\).

Model answer

The graph of \(y = \cos x\) is symmetrical about \(x = \pi\), so \(\cos(2\pi - \alpha) = \cos\alpha\).
In \(0 \leqslant x \leqslant 2\pi\): \(x = \alpha\) and \(x = 2\pi - \alpha\)
The period is \(2\pi\), so adding \(2\pi\) gives the solutions in \(2\pi \leqslant x \leqslant 4\pi\): \(x = 2\pi + \alpha\) and \(x = 4\pi - \alpha\)
\(x = \alpha,\ 2\pi - \alpha,\ 2\pi + \alpha,\ 4\pi - \alpha\)

Explanation

The cosine graph is symmetrical about \(x = \pi\) (and about the \(y\)-axis), so in one cycle \(\cos x = \cos\alpha\) at \(\alpha\) and \(2\pi - \alpha\); because the period is \(2\pi\), adding \(2\pi\) gives the two solutions in the second cycle. The interval contains two whole cycles, so there are four values. The common error is to use the sine symmetry and give \(\pi - \alpha\).

Mark scheme

1 mark
B1 \(x = \alpha\) and \(x = 2\pi - \alpha\)
Correct answers with no working score 3 marks. Both values are needed for this mark. \(\pi - \alpha\) is a solution of \(\sin x = \sin\alpha\), not of \(\cos x = \cos\alpha\): if it is given, the final mark is lost.
1 mark
B1 \(x = 2\pi + \alpha\)
1 mark
B1 \(x = 4\pi - \alpha\), with no extra values in the interval
Values outside the interval (e.g. \(-\alpha\), \(4\pi + \alpha\)) are B0 for this mark if given as answers.

Question 2

4 marks

Solutions relying entirely on calculator technology are not acceptable.

Show that
\[\dfrac{\sin\left(\dfrac{2\pi}{3}\right)}{1 - \cos\left(\dfrac{5\pi}{6}\right)} = 2\sqrt{3} - 3\]

Model answer

\(\sin\dfrac{2\pi}{3} = \sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}\)
\(\cos\dfrac{5\pi}{6} = -\cos\dfrac{\pi}{6} = -\dfrac{\sqrt{3}}{2}\)
\(\dfrac{\frac{\sqrt{3}}{2}}{1 + \frac{\sqrt{3}}{2}} = \dfrac{\sqrt{3}}{2 + \sqrt{3}}\)
\(= \dfrac{\sqrt{3}(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} = \dfrac{2\sqrt{3} - 3}{4 - 3}\)
\(= 2\sqrt{3} - 3\) as required

Explanation

The exact values come from the reference angles \(\dfrac{\pi}{3}\) and \(\dfrac{\pi}{6}\): \(\dfrac{2\pi}{3}\) is in the second quadrant, where sine is positive and cosine negative, so \(1 - \cos\dfrac{5\pi}{6} = 1 + \dfrac{\sqrt{3}}{2}\). Multiplying top and bottom by 2 and then by the conjugate \(2 - \sqrt{3}\) rationalises the denominator. The common errors are taking \(\cos\dfrac{5\pi}{6}\) as positive and checking the result only with decimals.

Mark scheme

1 mark
M1 Uses exact values with the correct signs, at least one correct: \(\sin\dfrac{2\pi}{3} = \dfrac{\sqrt{3}}{2}\), \(\cos\dfrac{5\pi}{6} = -\dfrac{\sqrt{3}}{2}\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: the result is given and calculator technology is not acceptable.
1 mark
A1 A correct single fraction, e.g. \(\dfrac{\sqrt{3}}{2 + \sqrt{3}}\)
Both exact values correct. Accept \(\dfrac{\sqrt{3}/2}{1 + \sqrt{3}/2}\).
1 mark
M1 Rationalises the denominator: multiplies the numerator and denominator by \(2 - \sqrt{3}\) (or by their equivalent conjugate)
1 mark
A1* \(\dfrac{\sqrt{3}(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} = \dfrac{2\sqrt{3} - 3}{4 - 3} = 2\sqrt{3} - 3\) cso
The denominator \(4 - 3 = 1\) must be seen. Showing only that both sides equal 0.464 is A0*.

Question 3

5 marks

The equation
\[\sin 3x = 0.4\]
has exactly six solutions in the interval \(0 \leqslant x \leqslant 2\pi\), where \(x\) is in radians.

Find the smallest of these solutions, giving your answer to 3 significant figures.

By considering the symmetry and periodicity of the graph of \(y = \sin 3x\), or otherwise, find the exact value of the sum of the six solutions.

Model answer

Let \(\beta = \arcsin 0.4 = 0.41152...\)
\(0 \leqslant x \leqslant 2\pi\), so \(0 \leqslant 3x \leqslant 6\pi\)
\(\sin\theta = 0.4\) for \(\theta = \beta,\ \pi - \beta,\ 2\pi + \beta,\ 3\pi - \beta,\ 4\pi + \beta,\ 5\pi - \beta\) in \([0, 6\pi]\)
Smallest solution: \(x = \dfrac{\beta}{3} = 0.13717... = 0.137\) (3 s.f.)
Sum of the six values of \(3x\): \(\beta + (\pi - \beta) + (2\pi + \beta) + (3\pi - \beta) + (4\pi + \beta) + (5\pi - \beta) = 15\pi\)
So the sum of the six solutions is \(\dfrac{15\pi}{3} = 5\pi\)
(Check: \(0.137 + 0.910 + 2.232 + 3.004 + 4.326 + 5.099 = 15.708 = 5\pi\))

Explanation

Since \(0 \leqslant 3x \leqslant 6\pi\), there are three periods of \(\sin 3x\), and in each period \(\sin\theta = 0.4\) has the two solutions \(\theta\) and \(\pi - \theta\) shifted by \(2\pi\). The smallest solution is \(\dfrac{1}{3}\arcsin 0.4\), and when the six values of \(3x\) are added the \(\beta\) terms cancel to leave \(15\pi\), so the six values of \(x\) add up to \(5\pi\). Equivalently, the solutions come in pairs placed symmetrically about the maximum points at \(x = \dfrac{\pi}{6}\), \(\dfrac{5\pi}{6}\) and \(\dfrac{3\pi}{2}\); common errors are working in degrees and using the interval \(0 \leqslant 3x \leqslant 2\pi\).

Mark scheme

1 mark
M1 Finds \(\arcsin 0.4 = 0.4115...\) and divides by 3; or correct final answers of \(x = 0.137\) and a sum of \(5\pi\) with no working
Answers of 0.137 and \(5\pi\) with no working score 2 marks (the M1 A1 for 0.137); the marks for the sum need the six solutions or a symmetry argument. Working in degrees (\(3x = 23.6^\circ\)) and not converting is M1 A0. Accept \(\arcsin 0.4\) as awrt 0.412, i.e. in [0.4115, 0.4125).
1 mark
A1 \(x =\) awrt 0.137, i.e. in [0.1365, 0.1375)
Exact form \(\dfrac{1}{3}\arcsin 0.4\) is also acceptable.
1 mark
M1 Finds all six values of \(3x\) in \([0, 6\pi]\) in terms of \(\beta = \arcsin 0.4\): \(\beta,\ \pi - \beta,\ 2\pi + \beta,\ 3\pi - \beta,\ 4\pi + \beta,\ 5\pi - \beta\); or all six values of \(x\) (0.137, 0.910, 2.232, 3.004, 4.326, 5.099); or uses the symmetry of the graph about the lines \(x = \dfrac{\pi}{6}\), \(x = \dfrac{5\pi}{6}\) and \(x = \dfrac{3\pi}{2}\)
Accept the decimal solutions to 2 or more decimal places, e.g. awrt 0.910, i.e. in [0.9095, 0.9105). The interval for \(3x\) is \(0 \leqslant 3x \leqslant 6\pi\): three periods of \(\sin 3x\), each giving two solutions.
1 mark
M1 (dM1) Adds the solutions so that the \(\beta\) terms cancel: the sum of the values of \(3x\) is \(15\pi\) (or each symmetric pair of values of \(x\) adds up to twice the \(x\)-coordinate of a maximum point, \(2 \times \dfrac{\pi}{6}\), \(2 \times \dfrac{5\pi}{6}\), \(2 \times \dfrac{3\pi}{2}\))
Adding six decimals to get 15.7... and then recognising \(5\pi\) earns this mark only if the six solutions are all correct.
1 mark
A1 Sum of the solutions \(= 5\pi\) cso
A decimal answer (15.7) only is A0, since the exact value is asked for.

Question 4

6 marks

The function f is defined by
\[f(x) = 3\cos 2x + 1, \quad 0 \leqslant x < 2\pi\]
where \(x\) is in radians.

State the maximum value of f(\(x\)), the minimum value of f(\(x\)) and the period of f.

Find the set of values of the constant \(k\) for which the equation f(\(x\)) \(= k\) has exactly four solutions. Give your answer in set notation.

Model answer

\(-1 \leqslant \cos 2x \leqslant 1\), so the maximum value is \(3 + 1 = 4\) and the minimum value is \(-3 + 1 = -2\).
\(\cos x\) has period \(2\pi\), so \(\cos 2x\) has period \(\pi\): the period of f is \(\pi\).
For \(0 \leqslant x < 2\pi\), \(2x\) runs through \([0, 4\pi)\): two complete cycles.
In each cycle, every value strictly between \(-2\) and 4 is taken twice, so four times in all.
\(f(x) = 4\): \(\cos 2x = 1\), \(2x = 0, 2\pi\), so only \(x = 0, \pi\) (two solutions).
\(f(x) = -2\): \(\cos 2x = -1\), \(2x = \pi, 3\pi\), so only \(x = \dfrac{\pi}{2}, \dfrac{3\pi}{2}\) (two solutions).
\(\{k : -2 < k < 4\}\)

Explanation

The amplitude 3 and the shift 1 give a maximum of 4 and a minimum of \(-2\), and replacing \(x\) by \(2x\) halves the period to \(\pi\), so the interval \(0 \leqslant x < 2\pi\) contains two complete cycles. A horizontal line \(y = k\) strictly between the minimum and maximum meets each cycle twice, but at the maximum or minimum it touches the curve only once per cycle. Common errors are including the end values and giving the answer without set notation.

Mark scheme

1 mark
B1 Maximum value 4; correct answers with no working score 4 marks
Correct answers (maximum 4, minimum \(-2\), period \(\pi\) and \(\{k : -2 < k < 4\}\)) with no working score 4 marks: B1 B1 B1 for the maximum, minimum and period, and 1 mark (SC) for the set, as the M marks need the reasoning.
1 mark
B1 Minimum value \(-2\)
If the maximum and minimum values are given the wrong way round, award B1 B0.
1 mark
B1 Period \(\pi\)
Accept \(180^\circ\). \(2\pi\) is B0.
1 mark
M1 Recognises that \(2x\) takes values in \([0, 4\pi)\), so the graph of f has two complete cycles and each value strictly between the minimum and the maximum is taken twice in each cycle
May be shown by a sketch with a correct explanation, or by solving \(\cos 2x = c\) for a general \(c\) with \(-1 < c < 1\).
1 mark
M1 Considers the end values: \(f(x) = 4\) has only two solutions (\(x = 0, \pi\)) and \(f(x) = -2\) has only two solutions (\(x = \dfrac{\pi}{2}, \dfrac{3\pi}{2}\))
1 mark
A1 \(\{k : -2 < k < 4\}\)
Needs both M marks. Set notation is required: \(-2 < k < 4\) without the set brackets is A0. Allow \(\{k : k > -2\} \cap \{k : k < 4\}\). Including \(-2\) or 4 is A0.

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