Reciprocal and inverse trigonometric functions and identities - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 18 exam-style questions on Reciprocal and inverse trigonometric functions and identities 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.

18 questions
2–9 marks each
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18 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

4 marks

The function f is defined by

\[\mathrm{f}(x) = 2\operatorname{cosec} x + 1, \quad \dfrac{\pi}{6} \leqslant x \leqslant \dfrac{5\pi}{6}\]

Find the range of f.

State, with a reason, whether f has an inverse function.

Model answer

For \(\dfrac{\pi}{6} \leqslant x \leqslant \dfrac{5\pi}{6}\): \(\dfrac{1}{2} \leqslant \sin x \leqslant 1\), so \(1 \leqslant \operatorname{cosec} x \leqslant 2\).
So \(3 \leqslant 2\operatorname{cosec} x + 1 \leqslant 5\): the range is \(3 \leqslant \mathrm{f}(x) \leqslant 5\).
\(\mathrm{f}\left(\dfrac{\pi}{6}\right) = 2 \times 2 + 1 = 5\) and \(\mathrm{f}\left(\dfrac{5\pi}{6}\right) = 5\): two inputs give the same output.
So f is many-one and does not have an inverse function.

Explanation

On this domain \(\sin x\) runs from \(\dfrac{1}{2}\) up to 1 and back, so \(\operatorname{cosec} x\) runs between 1 and 2, giving the range of f. Because the graph is symmetrical about \(x = \dfrac{\pi}{2}\), different inputs give the same output, so f is not one-one and has no inverse. A common error is to give the range in terms of \(x\).

Mark scheme

1 mark
M1 Uses \(\dfrac{1}{2} \leqslant \sin x \leqslant 1\) on this domain, so \(1 \leqslant \operatorname{cosec} x \leqslant 2\); or a correct final answer of \(3 \leqslant \mathrm{f}(x) \leqslant 5\) with no working
An answer of \(3 \leqslant \mathrm{f}(x) \leqslant 5\) with no working scores 2 marks (M1 A1); the inverse marks need a reason.
1 mark
A1 \(3 \leqslant \mathrm{f}(x) \leqslant 5\)
Accept \(3 \leqslant y \leqslant 5\), \([3, 5]\) or \(\{\mathrm{f}(x) : 3 \leqslant \mathrm{f}(x) \leqslant 5\}\). Not \(3 \leqslant x \leqslant 5\).
1 mark
B1 f does not have an inverse, because f is many-one (not one-one)
"No" with no reason, or "it is not a function", is B0.
1 mark
B1 Supports this with two inputs that give the same output, e.g. \(\mathrm{f}\left(\dfrac{\pi}{6}\right) = \mathrm{f}\left(\dfrac{5\pi}{6}\right) = 5\), or the symmetry of the graph about \(x = \dfrac{\pi}{2}\)
Needs the previous B mark.

Question 2

4 marks

Determine the values of \(k\) for which there is at least one real \(x\) satisfying

\[\operatorname{cosec}^2 x = k\cot x\]

Write your answer using set notation.

Model answer

\(1 + \cot^2 x = k\cot x \Rightarrow \cot^2 x - k\cot x + 1 = 0\)
\(\cot x\) can take any real value, so the equation has real solutions exactly when this quadratic in \(\cot x\) has real roots:
\(k^2 - 4 \geqslant 0 \Rightarrow k \leqslant -2\) or \(k \geqslant 2\)
\(\{k : k \leqslant -2\} \cup \{k : k \geqslant 2\}\)

Explanation

The identity \(\operatorname{cosec}^2 x = 1 + \cot^2 x\) gives a quadratic in \(\cot x\), and since \(\cot x\) takes every real value, there are solutions exactly when the quadratic has real roots, i.e. when the discriminant is at least 0. The boundary values \(\pm 2\) are included, as a repeated root still gives solutions. Common errors are using a strict inequality and writing the answer without set notation.

Mark scheme

1 mark
M1 Uses \(\operatorname{cosec}^2 x = 1 + \cot^2 x\) to form the quadratic \(\cot^2 x - k\cot x + 1 = 0\); a correct final answer of \(\{k : k \leqslant -2\} \cup \{k : k \geqslant 2\}\) with no working scores 1 mark
An answer of \(\{k : k \leqslant -2\} \cup \{k : k \geqslant 2\}\) with no working scores 1 mark (special case: the final A1 only), as "Determine" requires the reasoning to be shown.
1 mark
M1 Uses the discriminant of their quadratic in \(\cot x\), \(b^2 - 4ac \geqslant 0\): \(k^2 - 4 \geqslant 0\), noting that \(\cot x\) takes every real value, so any real root gives a solution
Using \(> 0\) is M1 but leads to A0 for the final mark.
1 mark
A1 Critical values \(k = \pm 2\)
1 mark
A1 \(\{k : k \leqslant -2\} \cup \{k : k \geqslant 2\}\)
Depends on both M marks (except in the special case for an answer with no working). Set notation with \(\cup\) needed; accept \(\{k : \lvert k\rvert \geqslant 2\}\). Strict inequalities or \(\cap\) are A0. \(k \leqslant -2\) or \(k \geqslant 2\) without set notation is A0.

Question 3

6 marks

A security camera is fixed at the point \(C\) on the vertical wall of a building, 4 m above horizontal ground. The point \(F\) on the ground is vertically below \(C\). The camera points at a point \(P\) on the ground, where \(FP = x\) metres, \(x > 0\), and the angle of depression of \(P\) from \(C\) is \(\theta\) radians.

Show that \(\theta = \arctan\left(\dfrac{4}{x}\right)\).

Find the value of \(\theta\) when \(x = 6\), giving your answer to 3 significant figures.

State the set of possible values of \(\theta\) as \(x\) takes all positive values, giving a reason for your answer.

Find the exact value of \(x\) for which \(\theta = \dfrac{\pi}{3}\).

Model answer

The horizontal through \(C\) is parallel to the ground, so angle \(CPF = \theta\) (alternate angles), and triangle \(CFP\) is right-angled at \(F\).
\(\tan\theta = \dfrac{CF}{FP} = \dfrac{4}{x}\). As \(\theta\) is acute, \(\theta = \arctan\left(\dfrac{4}{x}\right)\), as required.
When \(x = 6\): \(\theta = \arctan\dfrac{2}{3} = 0.588\) (3 s.f.)
For \(x > 0\), \(\dfrac{4}{x}\) takes every positive value, and arctan of a positive number lies strictly between 0 and \(\dfrac{\pi}{2}\): so \(0 < \theta < \dfrac{\pi}{2}\).
\(\theta = \dfrac{\pi}{3}\): \(\dfrac{4}{x} = \tan\dfrac{\pi}{3} = \sqrt{3}\), so \(x = \dfrac{4}{\sqrt{3}} = \dfrac{4\sqrt{3}}{3}\)

Explanation

The angle of depression from \(C\) equals the angle of elevation of \(C\) from \(P\), so the right-angled triangle \(CFP\) gives \(\tan\theta = \dfrac{4}{x}\), and as \(\theta\) is acute it is \(\arctan\dfrac{4}{x}\). The range of arctan, \(-\dfrac{\pi}{2} < \arctan u < \dfrac{\pi}{2}\), with \(u = \dfrac{4}{x} > 0\), gives \(0 < \theta < \dfrac{\pi}{2}\), and the exact value \(\tan\dfrac{\pi}{3} = \sqrt{3}\) gives \(x\). Common errors are using degree mode and including the end values 0 and \(\dfrac{\pi}{2}\), which are never reached.

Mark scheme

1 mark
B1* Angle \(CPF = \theta\) (alternate angles with the horizontal through \(C\)), so in triangle \(CFP\), right-angled at \(F\), \(\tan\theta = \dfrac{4}{x}\); as \(\theta\) is acute, \(\theta = \arctan\left(\dfrac{4}{x}\right)\), with a conclusion; correct answers of 0.588 and \(\dfrac{4\sqrt{3}}{3}\) with no working score 4 marks
Correct answers of \(\theta = 0.588\) and \(x = \dfrac{4\sqrt{3}}{3}\) with no working score 4 marks (M1 A1 for each; \(\theta =\) awrt 0.588, i.e. in [0.5875, 0.5885)); the show-that needs working and the set of values needs a reason. For this mark the link between the angle of depression and the angle at \(P\), and \(\tan\theta = \dfrac{4}{x}\), must be seen.
1 mark
M1 \(\theta = \arctan\left(\dfrac{4}{6}\right)\), with the calculator in radian mode
Degree mode gives \(33.7\): M1 A0.
1 mark
A1 \(\theta =\) awrt 0.588, i.e. in [0.5875, 0.5885)
From 0.58800...
1 mark
B1 \(0 < \theta < \dfrac{\pi}{2}\), because \(\dfrac{4}{x}\) takes every positive value and arctan of a positive number lies strictly between 0 and \(\dfrac{\pi}{2}\)
Accept a reason based on the limits: as \(x \to 0\), \(\theta \to \dfrac{\pi}{2}\), and as \(x \to \infty\), \(\theta \to 0\), with neither value reached. Strict inequalities needed; set notation is not required. \(0 \leqslant \theta \leqslant \dfrac{\pi}{2}\), or the interval with no reason, is B0.
1 mark
M1 \(\tan\dfrac{\pi}{3} = \dfrac{4}{x}\), i.e. \(\sqrt{3} = \dfrac{4}{x}\)
Uses the exact value \(\tan\dfrac{\pi}{3} = \sqrt{3}\).
1 mark
A1 \(x = \dfrac{4}{\sqrt{3}} = \dfrac{4\sqrt{3}}{3}\)
Accept \(\dfrac{4}{\sqrt{3}}\). A decimal (2.31) alone is A0, as the exact value is asked for.

Question 4

8 marks

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

Prove that

\[\sec 2x - \sec^2 x \equiv \tan^2 x\sec 2x\]

for all values of \(x\) for which both sides are defined.

Hence solve, for \(0 \leqslant x < \pi\), the equation

\[\sec 2x - \sec^2 x = 2\tan x\sec 2x\]

giving each solution exactly or to 3 significant figures.

Model answer

\(\sec 2x - \sec^2 x = \dfrac{1}{\cos 2x} - \dfrac{1}{\cos^2 x} = \dfrac{\cos^2 x - \cos 2x}{\cos 2x\cos^2 x}\)
\(\cos 2x = \cos^2 x - \sin^2 x\), so the numerator is \(\cos^2 x - \cos^2 x + \sin^2 x = \sin^2 x\)
\(= \dfrac{\sin^2 x}{\cos^2 x} \times \dfrac{1}{\cos 2x} = \tan^2 x\sec 2x\), as required.
Hence \(\tan^2 x\sec 2x = 2\tan x\sec 2x\).
\(\sec 2x ≠ 0\), so \(\tan^2 x - 2\tan x = 0\), i.e. \(\tan x(\tan x - 2) = 0\)
\(\tan x = 0\): \(x = 0\) (in \(0 \leqslant x < \pi\))
\(\tan x = 2\): \(x = \arctan 2 = 1.11\) (3 s.f.)
Both are valid, as \(\cos x\) and \(\cos 2x\) are non-zero there.
\(x = 0\) or \(x = 1.11\)

Explanation

Over the common denominator \(\cos 2x\cos^2 x\) the numerator is \(\cos^2 x - \cos 2x\), which the double-angle formula \(\cos 2x = \cos^2 x - \sin^2 x\) reduces to \(\sin^2 x\), giving \(\tan^2 x\sec 2x\). In the equation, \(\sec 2x\) can be divided out because it is never zero, but \(\tan x\) must be factorised, not cancelled. The commonest loss of marks is cancelling \(\tan x\) and so losing the solution \(x = 0\).

Mark scheme

1 mark
M1 Writes the left-hand side in terms of cosines over a common denominator: \(\dfrac{1}{\cos 2x} - \dfrac{1}{\cos^2 x} = \dfrac{\cos^2 x - \cos 2x}{\cos 2x\cos^2 x}\). An answer with no working scores 0 marks.
An answer of \(x = 0\) and \(x = 1.11\) with no working scores 0 marks: the proof needs working and solutions relying on calculator technology are not acceptable. Alternative for the proof: start from the right-hand side, \(\tan^2 x\sec 2x = \dfrac{\sin^2 x}{\cos^2 x\cos 2x}\), and work towards the left-hand side, marked in the same way, with a conclusion.
1 mark
M1 Uses a correct double-angle formula for \(\cos 2x\) to simplify the numerator: \(\cos^2 x - (\cos^2 x - \sin^2 x) = \sin^2 x\) (or \(\cos^2 x - (2\cos^2 x - 1) = 1 - \cos^2 x = \sin^2 x\))
Using \(\cos 2x = 1 - 2\cos^2 x\) (a sign error) is M0.
1 mark
M1 Writes \(\dfrac{\sin^2 x}{\cos^2 x\cos 2x}\) as \(\dfrac{\sin^2 x}{\cos^2 x} \times \dfrac{1}{\cos 2x} = \tan^2 x\sec 2x\)
Uses \(\tan x = \dfrac{\sin x}{\cos x}\) and \(\sec 2x = \dfrac{1}{\cos 2x}\).
1 mark
A1* \(\sec 2x - \sec^2 x \equiv \tan^2 x\sec 2x\), with a conclusion, cso
Needs all three M marks; no errors seen.
1 mark
M1 Uses the identity: \(\tan^2 x\sec 2x = 2\tan x\sec 2x\)
Solving the original equation without the identity is not using "hence": M0, but if \(x = 0\) and \(x = \arctan 2\) are found correctly that way with full working, award SC 1 for them.
1 mark
M1 (dM1) Divides by \(\sec 2x\), which is never zero, and solves \(\tan^2 x - 2\tan x = 0\) by factorising: \(\tan x(\tan x - 2) = 0\)
Depends on the previous M mark. Cancelling \(\tan x\) instead of factorising still scores this mark for \(\tan x = 2\), but it loses the solution \(x = 0\) (A0 next).
1 mark
A1 \(x = 0\)
From \(\tan x = 0\); \(x = \pi\) is outside the interval.
1 mark
A1 \(x = \arctan 2 =\) awrt 1.11, i.e. in [1.105, 1.115), and no other solutions in the interval
Depends on the dM1 mark. Accept \(\arctan 2\). Both solutions are valid, as \(\cos x\) and \(\cos 2x\) are non-zero there (\(\cos 2x = 1\) and \(-0.6\)). An answer in degrees (\(63.4^\circ\)) is A0.

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