Trigonometric graphs, exact values and small angles - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 17 exam-style questions on Trigonometric graphs, exact values and small angles for Edexcel A-Level Maths (Paper 2). 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.

17 questions
2–8 marks each
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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

In this question you must show all stages of your working.

Given that \(x\) is small and measured in radians, use the small-angle approximations to find the approximate value of

\[\dfrac{3\sin^2 x + 4\cos x - 4}{x\tan 3x}\]

[Small-angle approximations: \(\sin\theta \approx \theta\), \(\cos\theta \approx 1 - \dfrac{\theta^2}{2}\), \(\tan\theta \approx \theta\), where \(\theta\) is measured in radians]

Model answer

\(3\sin^2 x \approx 3x^2\)
\(4\cos x - 4 \approx 4\left(1 - \dfrac{x^2}{2}\right) - 4 = -2x^2\)
\(x\tan 3x \approx x \times 3x = 3x^2\)
\(\dfrac{3x^2 - 2x^2}{3x^2} = \dfrac{x^2}{3x^2} = \dfrac{1}{3}\)

Explanation

Each function is replaced by its small-angle approximation: the 4s cancel in \(4\cos x - 4\), leaving \(-2x^2\), and with \(3\sin^2 x \approx 3x^2\) the numerator is \(x^2\). The denominator is \(x\tan 3x \approx 3x^2\), so the \(x^2\) cancels to leave \(\dfrac{1}{3}\). Common errors are dropping the "\(1 -\)" in the cosine approximation and not squaring \(x\) in \(\sin^2 x\).

Mark scheme

1 mark
M1 Uses at least two of \(\sin x \approx x\), \(\cos x \approx 1 - \dfrac{x^2}{2}\), \(\tan 3x \approx 3x\), with correct brackets. An answer with no working scores 0 marks.
An answer of \(\dfrac{1}{3}\) with no working scores 0 marks: the question asks for all stages of working. Writing \(\sin^2 x \approx x\) (not squared) is an error, not condoned for the A marks.
1 mark
A1 Numerator \(\approx 3x^2 + 4\left(1 - \dfrac{x^2}{2}\right) - 4 = x^2\) and denominator \(\approx 3x^2\)
Both needed.
1 mark
A1 \(\dfrac{1}{3}\) cao
Depends on the M mark. Accept 0.333 (recurring). Do not accept answers that still contain \(x\).

Question 2

4 marks

A camera with a telephoto lens has a horizontal field of view of 0.035 radians. The camera is pointed at a flat cliff face 1.2 km away, so that its line of sight is perpendicular to the cliff face and bisects the field of view.

Use a small-angle approximation to estimate the width of cliff face that appears in a photograph, giving your answer in metres.

Explain why your estimate is very close to the true width.

[The small-angle approximation for tangent: \(\tan\theta \approx \theta\), where \(\theta\) is in radians]

Model answer

The line of sight splits the view into two right-angled triangles, each with angle \(\dfrac{0.035}{2} = 0.0175\) radians at the camera and adjacent side 1200 m.
Width \(= 2 \times 1200 \times \tan 0.0175\)
\(\tan 0.0175 \approx 0.0175\), so width \(\approx 2 \times 1200 \times 0.0175 = 42\) m
The angle 0.0175 radians is very small, so \(\tan\theta \approx \theta\) is extremely accurate: the true width is 42.004 m to 3 decimal places.

Explanation

Splitting the field of view into two right-angled triangles gives a half-width of \(1200\tan 0.0175\), and for such a small angle \(\tan\theta \approx \theta\), so the width is about \(1200 \times 0.035\). The approximation is excellent because the angle is tiny: the error in \(\tan\theta \approx \theta\) is roughly \(\dfrac{\theta^3}{3}\). A common error is to use the whole angle 0.035 in each triangle, which doubles the width.

Mark scheme

1 mark
M1 Width \(= 2 \times 1200 \times \tan 0.0175\) (two right-angled triangles, each with angle \(\dfrac{0.035}{2}\) at the camera); or a correct final answer of 42 m with no working
An answer of 42 m with no working scores 3 marks (all except the B mark for the explanation). Writing the width directly as \(1200 \times 0.035\) (or \(2 \times 1200 \times 0.0175\)) earns both M marks. Using 1.2 (km) is fine if the final answer is converted.
1 mark
M1 Uses \(\tan 0.0175 \approx 0.0175\)
Or width \(\approx 1200 \times 0.035\) directly.
1 mark
A1 Width \(\approx 42\) m
Accept 0.042 km. The exact value 42.004... m from a calculator without the approximation scores M1 M0 A0.
1 mark
B1 Explains that 0.0175 radians is a very small angle, so \(\tan\theta \approx \theta\) is very accurate (the error is of order \(\theta^3\), much smaller than \(\theta\))
Accept "the angle is very small" or "the true value is 42.004 m, only about 0.01% more". Do not accept "because it is an approximation".

Question 3

4 marks

Prove that the curves with equations \(y = \tan 2x\) and \(y = \tan\left(x + \dfrac{\pi}{4}\right)\), where \(x\) is in radians, have no points of intersection.

Model answer

Tangent has period \(\pi\) and takes each value exactly once in each period, so if \(\tan A = \tan B\) (both defined) then \(A = B + n\pi\) for some integer \(n\).
So at an intersection, \(2x = x + \dfrac{\pi}{4} + n\pi\), i.e. \(x = \dfrac{\pi}{4} + n\pi\).
But then \(2x = \dfrac{\pi}{2} + 2n\pi\), and \(\tan 2x\) is undefined there: these are asymptotes of \(y = \tan 2x\).
So there is no value of \(x\) at which both curves are defined and equal: the curves have no points of intersection.

Explanation

Because tangent repeats every \(\pi\) and takes each value once per period, equal tangents force the angles to differ by a multiple of \(\pi\). That gives \(x = \dfrac{\pi}{4} + n\pi\), exactly where \(\tan 2x\) has its asymptotes, so the only candidate points are not on the first curve at all. A complete proof must explain why these values fail and end with a conclusion.

Mark scheme

1 mark
B1 States that \(\tan A = \tan B\) (both defined) only if \(A = B + n\pi\) for some integer \(n\), since tangent has period \(\pi\) and takes each value once in each period; an answer with no working scores 0 marks
An answer with no working scores 0 marks: this is a proof. Alternative method (using \(\tan(A \pm B)\) and \(\tan 2x\)): B1 \(\tan\left(x + \dfrac{\pi}{4}\right) = \dfrac{1 + t}{1 - t}\) and \(\tan 2x = \dfrac{2t}{1 - t^2}\) with \(t = \tan x\); M1 equates and simplifies to \(2t = (1 + t)^2\); A1 \(t^2 + 1 = 0\), which has no real solutions; A1* also deals with \(x = \dfrac{\pi}{2} + n\pi\) (where \(\tan x\) is undefined: \(\tan 2x = 0\) but \(\tan\left(x + \dfrac{\pi}{4}\right) = -1\)), with a conclusion.
1 mark
M1 \(2x = x + \dfrac{\pi}{4} + n\pi\)
1 mark
A1 \(x = \dfrac{\pi}{4} + n\pi\)
1 mark
A1* At these values \(2x = \dfrac{\pi}{2} + 2n\pi\), where \(\tan 2x\) is undefined (an asymptote of \(y = \tan 2x\)), so there are no points of intersection, with a conclusion
Depends on the M mark. Needs the reason that \(\tan 2x\) is undefined there and a conclusion. Checking a few values of \(x\) is not a proof.

Question 4

7 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

The function f is defined by
\[\mathrm{f}(x) = 2\cos x + \sin 2x, \quad x \in \mathbb{R}\]
where \(x\) is in radians.

Using the symmetry and periodicity of the sine and cosine functions, show that \(\mathrm{f}(x + \pi) = \sin 2x - 2\cos x\).

The function g is defined by \(\mathrm{g}(x) = \mathrm{f}(x) + \mathrm{f}(x + \pi)\), \(x \in \mathbb{R}\). Show that \(\mathrm{g}(x) = 2\sin 2x\), and write down the maximum value of \(\mathrm{g}(x)\) and the smallest positive value of \(x\) at which it occurs.

Hence, or otherwise, find the exact value of \(\mathrm{f}\left(\dfrac{\pi}{6}\right) + \mathrm{f}\left(\dfrac{7\pi}{6}\right)\) and the exact value of \(\mathrm{f}\left(\dfrac{\pi}{6}\right) - \mathrm{f}\left(\dfrac{7\pi}{6}\right)\).

Model answer

Translating the graph of \(y = \cos x\) by \(\pi\) parallel to the x-axis gives the graph of \(y = -\cos x\), so \(\cos(x + \pi) = -\cos x\); and \(\sin\) has period \(2\pi\), so \(\sin(2x + 2\pi) = \sin 2x\).
\(\mathrm{f}(x + \pi) = 2\cos(x + \pi) + \sin(2(x + \pi)) = -2\cos x + \sin 2x = \sin 2x - 2\cos x\)
\(\mathrm{g}(x) = 2\cos x + \sin 2x + \sin 2x - 2\cos x = 2\sin 2x\), whose maximum value is 2, first when \(2x = \dfrac{\pi}{2}\), i.e. \(x = \dfrac{\pi}{4}\).
\(\mathrm{f}\left(\dfrac{\pi}{6}\right) + \mathrm{f}\left(\dfrac{7\pi}{6}\right) = \mathrm{g}\left(\dfrac{\pi}{6}\right) = 2\sin\dfrac{\pi}{3} = \sqrt{3}\)
\(\mathrm{f}(x) - \mathrm{f}(x + \pi) = 4\cos x\), so \(\mathrm{f}\left(\dfrac{\pi}{6}\right) - \mathrm{f}\left(\dfrac{7\pi}{6}\right) = 4\cos\dfrac{\pi}{6} = 2\sqrt{3}\)

Explanation

Moving \(\pi\) along the x-axis changes the sign of \(\cos x\) but leaves \(\sin 2x\) unchanged, because \(\sin 2x\) has period \(\pi\); so adding \(\mathrm{f}(x)\) and \(\mathrm{f}(x + \pi)\) cancels the cosine terms and subtracting them cancels the sine terms. With \(x = \dfrac{\pi}{6}\) these give \(\sqrt{3}\) and \(2\sqrt{3}\) at once. Common errors are writing \(\cos(x + \pi) = \cos x\) and forgetting that the angle in \(\sin 2x\) increases by \(2\pi\), not \(\pi\).

Mark scheme

1 mark
M1 Uses \(\cos(x + \pi) = -\cos x\) and \(\sin(2x + 2\pi) = \sin 2x\) (from the symmetry and periodicity of the graphs, or from the addition formulae). An answer with no working scores 0 marks.
An answer of \(\sqrt{3}\) and \(2\sqrt{3}\) with no working scores 0 marks: solutions relying entirely on calculator technology are not acceptable. Both results are needed for this mark.
1 mark
A1* \(\mathrm{f}(x + \pi) = 2\cos(x + \pi) + \sin(2x + 2\pi) = -2\cos x + \sin 2x\), cso
The step \(2\cos(x + \pi) + \sin(2(x + \pi))\) must be seen.
1 mark
B1* \(\mathrm{g}(x) = (2\cos x + \sin 2x) + (\sin 2x - 2\cos x) = 2\sin 2x\)
1 mark
B1 The maximum value of \(\mathrm{g}(x)\) is 2, first occurring at \(x = \dfrac{\pi}{4}\)
Both the value and \(x = \dfrac{\pi}{4}\) are needed for this mark. Accept 45° for \(x\).
1 mark
M1 Uses \(\mathrm{f}\left(\dfrac{\pi}{6}\right) + \mathrm{f}\left(\dfrac{7\pi}{6}\right) = \mathrm{g}\left(\dfrac{\pi}{6}\right)\) and \(\mathrm{f}(x) - \mathrm{f}(x + \pi) = 4\cos x\), or evaluates \(\mathrm{f}\left(\dfrac{\pi}{6}\right) = \dfrac{3\sqrt{3}}{2}\) and \(\mathrm{f}\left(\dfrac{7\pi}{6}\right) = -\dfrac{\sqrt{3}}{2}\) with exact values
Either route. For direct evaluation, \(\sin\dfrac{7\pi}{3} = \sin\dfrac{\pi}{3}\) and \(\cos\dfrac{7\pi}{6} = -\dfrac{\sqrt{3}}{2}\) must be used.
1 mark
A1 \(\mathrm{f}\left(\dfrac{\pi}{6}\right) + \mathrm{f}\left(\dfrac{7\pi}{6}\right) = 2\sin\dfrac{\pi}{3} = \sqrt{3}\)
Exact.
1 mark
A1 \(\mathrm{f}\left(\dfrac{\pi}{6}\right) - \mathrm{f}\left(\dfrac{7\pi}{6}\right) = 4\cos\dfrac{\pi}{6} = 2\sqrt{3}\)
Depends on the M mark in point 5. Exact; decimals (1.73, 3.46) score A0.

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