Reciprocal and inverse trigonometric functions and identities - Edexcel A-Level Maths (Paper 1) Exam Questions

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

18 questions
2–10 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

3 marks

A student is asked to find the exact value of \(\arcsin\left(\sin\dfrac{5\pi}{6}\right)\). The student writes

\(\arcsin\left(\sin\dfrac{5\pi}{6}\right) = \dfrac{5\pi}{6}\), because arcsin is the inverse function of sin.

Explain the error made by the student, and find the correct exact value of \(\arcsin\left(\sin\dfrac{5\pi}{6}\right)\).

Find also the exact value of \(\arctan\left(\tan\dfrac{5\pi}{6}\right)\).

Model answer

The range of arcsin is \(-\dfrac{\pi}{2} \leqslant \arcsin x \leqslant \dfrac{\pi}{2}\). Because sin is not one-to-one, \(\arcsin(\sin\theta) = \theta\) only when \(-\dfrac{\pi}{2} \leqslant \theta \leqslant \dfrac{\pi}{2}\); \(\dfrac{5\pi}{6} > \dfrac{\pi}{2}\), so the student's answer is outside the range of arcsin.
\(\sin\dfrac{5\pi}{6} = \sin\dfrac{\pi}{6} = \dfrac{1}{2}\), so \(\arcsin\left(\sin\dfrac{5\pi}{6}\right) = \arcsin\dfrac{1}{2} = \dfrac{\pi}{6}\)
\(\tan\dfrac{5\pi}{6} = -\tan\dfrac{\pi}{6} = -\dfrac{1}{\sqrt{3}}\), so \(\arctan\left(\tan\dfrac{5\pi}{6}\right) = \arctan\left(-\dfrac{1}{\sqrt{3}}\right) = -\dfrac{\pi}{6}\), since \(-\dfrac{\pi}{2} < \arctan x < \dfrac{\pi}{2}\)

Explanation

Inverse trigonometric functions return only their principal value: arcsin gives an angle from \(-\dfrac{\pi}{2}\) to \(\dfrac{\pi}{2}\), and arctan an angle strictly between \(-\dfrac{\pi}{2}\) and \(\dfrac{\pi}{2}\), so \(\arcsin(\sin\theta)\) equals \(\theta\) only when \(\theta\) is already in that range. Working out \(\sin\dfrac{5\pi}{6} = \dfrac{1}{2}\) and \(\tan\dfrac{5\pi}{6} = -\dfrac{1}{\sqrt{3}}\) first, then taking the principal values, gives \(\dfrac{\pi}{6}\) and \(-\dfrac{\pi}{6}\). The common error is to assume that the inverse function always cancels the function, as the student did.

Mark scheme

1 mark
B1 Explains that the range of arcsin is \(-\dfrac{\pi}{2} \leqslant \arcsin x \leqslant \dfrac{\pi}{2}\), so \(\arcsin(\sin\theta) = \theta\) only for \(-\dfrac{\pi}{2} \leqslant \theta \leqslant \dfrac{\pi}{2}\), and \(\dfrac{5\pi}{6}\) is not in this range (sin is not one-to-one, so arcsin gives only the angle in this range)
Correct values (\(\dfrac{\pi}{6}\) and \(-\dfrac{\pi}{6}\)) with no working score 2 marks; this first mark needs the explanation. Accept "\(\dfrac{5\pi}{6} > \dfrac{\pi}{2}\), which is outside the range of arcsin". Do not accept vague comments such as "arcsin and sin do not cancel" with no reference to the range of arcsin (or to sin not being one-to-one).
1 mark
B1 \(\arcsin\left(\sin\dfrac{5\pi}{6}\right) = \dfrac{\pi}{6}\)
From \(\sin\dfrac{5\pi}{6} = \dfrac{1}{2}\). A decimal (0.524) only is B0. \(\dfrac{5\pi}{6}\) is B0.
1 mark
B1 \(\arctan\left(\tan\dfrac{5\pi}{6}\right) = -\dfrac{\pi}{6}\)
From \(\tan\dfrac{5\pi}{6} = -\dfrac{1}{\sqrt{3}}\) and the range \(-\dfrac{\pi}{2} < \arctan x < \dfrac{\pi}{2}\). \(\dfrac{5\pi}{6}\) or \(\dfrac{11\pi}{6}\) is B0. A decimal (\(-0.524\)) only is B0.

Question 2

4 marks

The functions f and g are defined by
\[\mathrm{f}(x) = \arccos x, \quad -1 \leqslant x \leqslant 1\]
\[\mathrm{g}(x) = \dfrac{x - 3}{2}, \quad x \in \mathbb{R}\]

Find the set of values of \(x\) for which \(\mathrm{fg}(x)\) is defined, giving your answer as an inequality.

Solve the equation \(\mathrm{fg}(x) = \dfrac{3\pi}{4}\), giving your answer in exact form.

Model answer

\(\mathrm{fg}(x) = \arccos\left(\dfrac{x - 3}{2}\right)\)
arccos is defined only for inputs from \(-1\) to 1, so \(-1 \leqslant \dfrac{x - 3}{2} \leqslant 1\)
\(-2 \leqslant x - 3 \leqslant 2\), so \(1 \leqslant x \leqslant 5\)
\(\arccos\left(\dfrac{x - 3}{2}\right) = \dfrac{3\pi}{4} \Rightarrow \dfrac{x - 3}{2} = \cos\dfrac{3\pi}{4} = -\dfrac{\sqrt{2}}{2}\)
\(x - 3 = -\sqrt{2}\), so \(x = 3 - \sqrt{2}\)

Explanation

fg means apply g first, so \(\mathrm{fg}(x) = \arccos\left(\dfrac{x - 3}{2}\right)\), and it is defined only when the input to arccos lies between \(-1\) and 1, giving \(1 \leqslant x \leqslant 5\). Since \(\dfrac{3\pi}{4}\) lies in the range \([0, \pi]\) of arccos, taking cos of both sides gives \(\dfrac{x - 3}{2} = -\dfrac{\sqrt{2}}{2}\) and \(x = 3 - \sqrt{2}\), which is in the domain. Common errors are finding gf instead of fg and using strict inequalities for the domain.

Mark scheme

1 mark
M1 Uses the domain of arccos: \(-1 \leqslant \dfrac{x - 3}{2} \leqslant 1\) (or either inequality); or correct final answers of \(1 \leqslant x \leqslant 5\) and \(x = 3 - \sqrt{2}\) with no working
Answers of \(1 \leqslant x \leqslant 5\) and \(x = 3 - \sqrt{2}\) with no working score 4 marks.
1 mark
A1 \(1 \leqslant x \leqslant 5\)
Accept \([1, 5]\) or \(\{x : 1 \leqslant x \leqslant 5\}\). Strict inequalities are A0, since \(\arccos(\pm 1)\) is defined.
1 mark
M1 Writes \(\arccos\left(\dfrac{x - 3}{2}\right) = \dfrac{3\pi}{4}\) and takes cos of both sides: \(\dfrac{x - 3}{2} = \cos\dfrac{3\pi}{4}\)
Using gf instead of fg, i.e. \(\dfrac{\arccos x - 3}{2}\), is M0.
1 mark
A1 \(x = 3 - \sqrt{2}\)
From \(\cos\dfrac{3\pi}{4} = -\dfrac{\sqrt{2}}{2}\). Accept \(3 - \dfrac{2}{\sqrt{2}}\). A decimal answer (1.59) only is A0, since the exact form is asked for.

Question 3

5 marks

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

Solve, for \(0 < x < 2\pi\), the equation
\[\cot^2 x = 1 + \operatorname{cosec} x\]
Give your answers as exact multiples of \(\pi\).

Model answer

Using \(\cot^2 x = \operatorname{cosec}^2 x - 1\):
\(\operatorname{cosec}^2 x - 1 = 1 + \operatorname{cosec} x\)
\(\operatorname{cosec}^2 x - \operatorname{cosec} x - 2 = 0\)
\((\operatorname{cosec} x - 2)(\operatorname{cosec} x + 1) = 0\)
\(\operatorname{cosec} x = 2 \Rightarrow \sin x = \dfrac{1}{2} \Rightarrow x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}\)
\(\operatorname{cosec} x = -1 \Rightarrow \sin x = -1 \Rightarrow x = \dfrac{3\pi}{2}\)
\(x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}\)

Explanation

The identity \(\cot^2 x = \operatorname{cosec}^2 x - 1\) turns the equation into a quadratic in \(\operatorname{cosec} x\), which factorises to give \(\operatorname{cosec} x = 2\) or \(-1\). These mean \(\sin x = \dfrac{1}{2}\) (two solutions in the interval) and \(\sin x = -1\) (one solution), and none of them makes \(\sin x = 0\), so all three are valid. Common errors are using the identity with the wrong sign and missing the solution \(\dfrac{3\pi}{2}\).

Mark scheme

1 mark
M1 Uses \(\cot^2 x = \operatorname{cosec}^2 x - 1\) to form an equation in \(\operatorname{cosec} x\) only (or writes the equation in \(\sin x\) and \(\cos x\), multiplies by \(\sin^2 x\) and uses \(\cos^2 x = 1 - \sin^2 x\)). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: the question demands all stages of working. Using \(\cot^2 x = 1 + \operatorname{cosec}^2 x\) or \(\cot^2 x = \operatorname{cosec}^2 x + 1\) is M0.
1 mark
A1 \(\operatorname{cosec}^2 x - \operatorname{cosec} x - 2 = 0\) (or \(2\sin^2 x + \sin x - 1 = 0\))
Any equivalent three-term quadratic, with all the terms on one side.
1 mark
M1 Solves their three-term quadratic: \((\operatorname{cosec} x - 2)(\operatorname{cosec} x + 1) = 0\) (or \((2\sin x - 1)(\sin x + 1) = 0\))
Roots written down from a calculator with no method seen are M0, because of the instruction at the start of the question.
1 mark
A1 \(\operatorname{cosec} x = 2\) or \(\operatorname{cosec} x = -1\), i.e. \(\sin x = \dfrac{1}{2}\) or \(\sin x = -1\)
1 mark
A1 \(x = \dfrac{\pi}{6},\ \dfrac{5\pi}{6},\ \dfrac{3\pi}{2}\) only
Depends on the second M mark. All three values are needed and no others in the interval; decimals (0.524, 2.62, 4.71) only are A0. A common error is to lose \(x = \dfrac{3\pi}{2}\) by solving only \(\sin x = \dfrac{1}{2}\).

Question 4

8 marks

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

The point \(P(\cos\theta, \sin\theta)\), where \(0 < \theta < \dfrac{\pi}{2}\), lies on the circle with equation \(x^2 + y^2 = 1\), which has centre \(O\). The tangent to the circle at \(P\) meets the \(x\)-axis at the point \(A\) and the \(y\)-axis at the point \(B\).

Show that \(OA = \sec\theta\) and \(OB = \operatorname{cosec}\theta\).

Show further that \(AB = \sec\theta\operatorname{cosec}\theta\).

Given that \(AB = 4\), find the possible values of \(\theta\), giving your answers as exact multiples of \(\pi\).

Model answer

\(OP\) has gradient \(\dfrac{\sin\theta}{\cos\theta}\), so the tangent at \(P\) has gradient \(-\dfrac{\cos\theta}{\sin\theta}\):
\(y - \sin\theta = -\dfrac{\cos\theta}{\sin\theta}(x - \cos\theta)\)
\(y\sin\theta - \sin^2\theta = -x\cos\theta + \cos^2\theta \Rightarrow x\cos\theta + y\sin\theta = 1\)
\(y = 0\): \(x = \dfrac{1}{\cos\theta}\), so \(OA = \sec\theta\)
\(x = 0\): \(y = \dfrac{1}{\sin\theta}\), so \(OB = \operatorname{cosec}\theta\)
Angle \(AOB = \dfrac{\pi}{2}\), so \(AB^2 = \sec^2\theta + \operatorname{cosec}^2\theta = \dfrac{1}{\cos^2\theta} + \dfrac{1}{\sin^2\theta} = \dfrac{\sin^2\theta + \cos^2\theta}{\sin^2\theta\cos^2\theta} = \dfrac{1}{\sin^2\theta\cos^2\theta}\)
\(AB > 0\), so \(AB = \dfrac{1}{\sin\theta\cos\theta} = \sec\theta\operatorname{cosec}\theta\)
\(AB = 4\): \(\sin\theta\cos\theta = \dfrac{1}{4} \Rightarrow \dfrac{1}{2}\sin 2\theta = \dfrac{1}{4} \Rightarrow \sin 2\theta = \dfrac{1}{2}\)
\(0 < 2\theta < \pi\), so \(2\theta = \dfrac{\pi}{6}\) or \(\dfrac{5\pi}{6}\)
\(\theta = \dfrac{\pi}{12}\) or \(\theta = \dfrac{5\pi}{12}\)

Explanation

The tangent at \(P\) is perpendicular to the radius \(OP\), so either its equation \(x\cos\theta + y\sin\theta = 1\) or the right-angled triangles \(OPA\) and \(OPB\) give \(OA = \sec\theta\) and \(OB = \operatorname{cosec}\theta\); this is where the names secant and tangent come from. Pythagoras with \(\sin^2\theta + \cos^2\theta = 1\) then gives \(AB = \dfrac{1}{\sin\theta\cos\theta}\), and \(AB = 4\) becomes \(\sin 2\theta = \dfrac{1}{2}\), with two solutions for \(\theta\) between 0 and \(\dfrac{\pi}{2}\). Common errors are using the gradient of \(OP\) for the tangent and giving only \(\dfrac{\pi}{12}\).

Mark scheme

1 mark
M1 A correct method for \(OA\): the tangent is perpendicular to \(OP\), so its gradient is \(-\dfrac{\cos\theta}{\sin\theta}\), giving \(y - \sin\theta = -\dfrac{\cos\theta}{\sin\theta}(x - \cos\theta)\) (i.e. \(x\cos\theta + y\sin\theta = 1\)), and \(y = 0\) is substituted; or uses the right-angled triangle \(OPA\) (right angle at \(P\), angle \(AOP = \theta\)) to write \(\cos\theta = \dfrac{OP}{OA}\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: the results are printed and the question demands all stages of working. Restating \(OA = \sec\theta\) earns nothing.
1 mark
A1* \(OA = \dfrac{1}{\cos\theta} = \sec\theta\) cso
The equation of the tangent, or the triangle and the right angle at \(P\), must be seen.
1 mark
B1* \(OB = \dfrac{1}{\sin\theta} = \operatorname{cosec}\theta\), from \(x = 0\) in the tangent, or from triangle \(OPB\) with angle \(BOP = \dfrac{\pi}{2} - \theta\) and \(\cos\left(\dfrac{\pi}{2} - \theta\right) = \sin\theta\)
cso.
1 mark
M1 Uses Pythagoras: \(AB^2 = OA^2 + OB^2 = \sec^2\theta + \operatorname{cosec}^2\theta\); or uses \(AB = AP + PB\) with \(AP = \tan\theta\) and \(PB = \cot\theta\) (from the right-angled triangles \(OPA\) and \(OPB\))
Either method earns this mark.
1 mark
M1 Combines using an identity: \(\sec^2\theta + \operatorname{cosec}^2\theta = \dfrac{\sin^2\theta + \cos^2\theta}{\sin^2\theta\cos^2\theta} = \dfrac{1}{\sin^2\theta\cos^2\theta}\) (or \(\tan\theta + \cot\theta = \dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \dfrac{1}{\sin\theta\cos\theta}\))
Depends on the previous M mark. The identity \(\sin^2\theta + \cos^2\theta = 1\) must be used.
1 mark
A1* \(AB = \dfrac{1}{\sin\theta\cos\theta} = \sec\theta\operatorname{cosec}\theta\) cso
With the Pythagoras method, the positive square root is taken because \(AB\) is a length and \(0 < \theta < \dfrac{\pi}{2}\).
1 mark
M1 Sets \(\sec\theta\operatorname{cosec}\theta = 4\), i.e. \(\sin\theta\cos\theta = \dfrac{1}{4}\), and uses \(\sin 2\theta = 2\sin\theta\cos\theta\) to reach \(\sin 2\theta = \dfrac{1}{2}\)
A correct alternative is to square: \(\sin^2\theta\cos^2\theta = \dfrac{1}{16}\), leading to a quadratic in \(\sin^2\theta\); this earns the M mark when it is solved.
1 mark
A1 \(\theta = \dfrac{\pi}{12}\) and \(\theta = \dfrac{5\pi}{12}\) only
From \(2\theta = \dfrac{\pi}{6}\) or \(\dfrac{5\pi}{6}\), since \(0 < 2\theta < \pi\). Both values are needed; decimals (0.262, 1.31) only are A0.

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