Exam-style questionsSmall Angle Approximations and Radians questions with mark schemes
Easiest first. Try each question on paper, then open its mark scheme: M marks are for method, A marks for accurate answers that follow from the method, B marks for independent results.
Question 1easy2 marksYear 13
Given that \(\theta\) is small and measured in radians,
(a) show that \(\dfrac{1 - \cos2\theta}{\theta\tan\theta} \approx 2\). (2)
Mark scheme and answer
(a) \(\cos2\theta \approx 1 - \tfrac12(2\theta)^2 = 1 - 2\theta^2\) M1
\(\tan\theta\approx\theta\): \(\dfrac{2\theta^2}{\theta\times\theta} = 2\) A1
Answer: 2
Board notes: Small-angle approximations \(\sin\theta\approx\theta\), \(\cos\theta\approx1-\tfrac12\theta^2\), \(\tan\theta\approx\theta\) are printed in the Edexcel formulae book; candidates for every board should know them and when they are valid. All require \(\theta\) in radians.
Question 2easy4 marksYear 13
Use small-angle approximations to estimate the following, showing your working.
(a) \(\sin 0.04\) (1)
(b) \(\cos 0.2\) (2)
(c) Explain why the approximation in (b) is less accurate, in absolute terms, than using \(\cos 0.02 \approx 0.9998\). (1)
Mark scheme and answer
(a) \(\approx 0.04\) B1
(b) \(1 - \tfrac12(0.2)^2 = 0.98\) M1A1
(c) The error grows with \(\theta\) (next term \(\theta^4/24\)); 0.2 is larger than 0.02 B1
Answer: (a) 0.04; (b) 0.98; (c) error grows with θ
Board notes: Small-angle approximations \(\sin\theta\approx\theta\), \(\cos\theta\approx1-\tfrac12\theta^2\), \(\tan\theta\approx\theta\) are printed in the Edexcel formulae book; candidates for every board should know them and when they are valid. All require \(\theta\) in radians.
Question 3easy5 marksYear 13
A sector \(OAB\) of a circle has centre \(O\), radius 8 cm and angle \(AOB = 1.2\) radians.
(a) Find the length of the arc \(AB\). (2)
(b) Find the area of the sector. (2)
(c) Find the perimeter of the sector. (1)
Mark scheme and answer
(a) \(s = 8\times1.2 = 9.6\) cm M1A1
(b) \(\tfrac12\times64\times1.2 = 38.4\) cm² M1A1
(c) \(9.6 + 16 = 25.6\) cm B1
Answer: (a) 9.6 cm; (b) 38.4 cm²; (c) 25.6 cm
Board notes: All boards: \(s = r\theta\), \(A = \tfrac12r^2\theta\) with \(\theta\) in radians; exact answers in terms of \(\pi\) are expected when angles are given as multiples of \(\pi\).
Question 4easy4 marksYear 13
A sector has radius 9 cm and angle 150°.
(a) Write 150° in radians as a multiple of \(\pi\). (1)
(b) Find the exact arc length and the exact area of the sector. (3)
Mark scheme and answer
(a) \(\tfrac{5\pi}{6}\) B1
(b) Arc \(= 9\times\tfrac{5\pi}{6} = \frac{15 \pi}{2}\) cm B1
Area \(= \tfrac12\times81\times\tfrac{5\pi}{6} = \frac{135 \pi}{4}\) cm² M1A1
Answer: (a) 5π/6; (b) arc 15π/2 cm, area 135π/4 cm²
Board notes: All boards: \(s = r\theta\), \(A = \tfrac12r^2\theta\) with \(\theta\) in radians; exact answers in terms of \(\pi\) are expected when angles are given as multiples of \(\pi\).
Question 5medium4 marksYear 13
The function \(f\) is defined by \(f(\theta) = \cos\theta - 2\sin3\theta\).
(a) Show that, for small values of \(\theta\), \(f(\theta) \approx 1 - 6\theta - \tfrac12\theta^2\). (2)
(b) Find the percentage error when this approximation is used for \(\theta = 0.05\). (2)
Mark scheme and answer
(a) \(\cos\theta\approx1-\tfrac12\theta^2\), \(\sin3\theta\approx3\theta\) M1
\(1 - \tfrac12\theta^2 - 6\theta\) A1
(b) Exact \(f(0.05) = 0.699874\), approximation \(= 0.698750\) M1
Percentage error \(= 0.16\%\) A1
Answer: (a) shown; (b) 0.16%
Board notes: Small-angle approximations \(\sin\theta\approx\theta\), \(\cos\theta\approx1-\tfrac12\theta^2\), \(\tan\theta\approx\theta\) are printed in the Edexcel formulae book; candidates for every board should know them and when they are valid. All require \(\theta\) in radians.
Question 6medium3 marksYear 13
Given that \(\theta\) is small,
(a) find the approximate value of \(\dfrac{\sin\theta\tan2\theta}{1-\cos3\theta}\). (3)
Mark scheme and answer
(a) \(\sin\theta\approx\theta\), \(\tan2\theta\approx2\theta\) M1
\(1 - \cos3\theta \approx \tfrac92\theta^2\) M1
\(\dfrac{2\theta^2}{\tfrac92\theta^2} = \tfrac49\) A1
Answer: 4/9
Board notes: Small-angle approximations \(\sin\theta\approx\theta\), \(\cos\theta\approx1-\tfrac12\theta^2\), \(\tan\theta\approx\theta\) are printed in the Edexcel formulae book; candidates for every board should know them and when they are valid. All require \(\theta\) in radians.
Question 7medium4 marksYear 13
In modelling a pendulum, the approximation \(\sin\theta\approx\theta\) is used.
(a) Find the percentage error in using this approximation when the angle is 10°. (3)
(b) Explain why the approximation \(\sin\theta \approx \theta\) would give a meaningless answer if \(\theta\) were used in degrees. (1)
Mark scheme and answer
(a) \(10° = \tfrac{\pi}{18} = 0.17453\) rad B1
\(\sin(\tfrac{\pi}{18}) = 0.17365\) M1
Error \(= \dfrac{0.17453 - 0.17365}{0.17365}\times100 = 0.51\%\) A1
(b) The approximation comes from \(\dfrac{\sin\theta}{\theta}\to1\) only when \(\theta\) is in radians; in degrees \(\sin10° \approx 0.17\), not 10 B1
Answer: (a) 0.51%; (b) only valid in radians
Board notes: Small-angle approximations \(\sin\theta\approx\theta\), \(\cos\theta\approx1-\tfrac12\theta^2\), \(\tan\theta\approx\theta\) are printed in the Edexcel formulae book; candidates for every board should know them and when they are valid. All require \(\theta\) in radians.
Question 8medium3 marksYear 13
Given that \(\theta\) is small,
(a) show that \((2 + \sin2\theta)\cos\theta \approx 2 + 2\theta - \theta^2\). (3)
Mark scheme and answer
(a) \((2 + 2\theta)(1 - \tfrac12\theta^2)\) M1
\(= 2 + 2\theta - \theta^2 - \theta^3\) M1
Ignoring \(\theta^3\): \(2 + 2\theta - \theta^2\) A1
Answer: shown
Board notes: Small-angle approximations \(\sin\theta\approx\theta\), \(\cos\theta\approx1-\tfrac12\theta^2\), \(\tan\theta\approx\theta\) are printed in the Edexcel formulae book; candidates for every board should know them and when they are valid. All require \(\theta\) in radians.
Question 9medium4 marksYear 13
A sector of a circle of radius 6 cm has perimeter 20 cm.
(a) Find the angle of the sector in radians. (2)
(b) Find the area of the sector. (2)
Mark scheme and answer
(a) \(6\theta + 12 = 20\) M1
\(\theta = \tfrac43\) A1
(b) \(\tfrac12\times36\times\tfrac43 = 24\) cm² M1A1
Answer: (a) 4/3 rad; (b) 24 cm²
Board notes: All boards: \(s = r\theta\), \(A = \tfrac12r^2\theta\) with \(\theta\) in radians; exact answers in terms of \(\pi\) are expected when angles are given as multiples of \(\pi\).
Question 10medium5 marksYear 13
The points \(A\) and \(B\) lie on a circle with centre \(O\) and radius 10 cm, with angle \(AOB = \dfrac{2\pi}{5}\).
(a) Find the area of the minor segment bounded by the chord \(AB\) and the minor arc \(AB\). (3)
(b) Find the length of the chord \(AB\). (2)
Mark scheme and answer
(a) Sector \(= \tfrac12(100)\tfrac{2\pi}{5} = 20\pi\) M1
Triangle \(= \tfrac12(100)\sin\tfrac{2\pi}{5} = 47.553\) M1
Segment \(= 15.279\) cm² A1
(b) \(2\times10\sin\tfrac{\pi}{5} = 11.756\) cm M1A1
Answer: (a) 15.28 cm²; (b) 11.76 cm
Board notes: All boards: \(s = r\theta\), \(A = \tfrac12r^2\theta\) with \(\theta\) in radians; exact answers in terms of \(\pi\) are expected when angles are given as multiples of \(\pi\).
Question 11medium4 marksYear 13
A windscreen wiper arm of length 50 cm rotates about a fixed point \(O\) through an angle of 2.2 radians. The rubber blade is attached to the outer part of the arm, from 15 cm to 50 cm from \(O\).
(a) Find the area of windscreen wiped by the blade. (3)
(b) Find the distance travelled by the tip of the blade in one sweep. (1)
Mark scheme and answer
(a) \(\tfrac12(2.2)(50^2 - 15^2)\) M1A1
\(= 2502.5\) cm² A1
(b) \(50\times2.2 = 110\) cm B1
Answer: (a) 2502.5 cm²; (b) 110 cm
Board notes: All boards: \(s = r\theta\), \(A = \tfrac12r^2\theta\) with \(\theta\) in radians; exact answers in terms of \(\pi\) are expected when angles are given as multiples of \(\pi\).
Question 12hard6 marksYear 13
It is given that \(\theta\) is small and satisfies \(4\cos\theta + 3\sin\theta = 3.9\).
(a) Use small-angle approximations to show that \(20\theta^2 - 30\theta - 1 = 0\). (2)
(b) Hence find an approximate value of \(\theta\), explaining why you reject the other root. (3)
(c) The exact solution near zero is \(\theta = -0.03263\). Comment on the accuracy of your answer. (1)
Mark scheme and answer
(a) \(4(1 - \tfrac12\theta^2) + 3\theta = 3.9\) M1
\(4 - 2\theta^2 + 3\theta = 3.9\), multiply by \(-10\): \(20\theta^2 - 30\theta - 1 = 0\) A1
(b) \(\theta = \dfrac{30 \pm\sqrt{980}}{40}\) M1
\(\theta = -0.0326\) or \(1.53\) A1
\(\theta \approx -0.0326\); the root \(1.53\) is not small, so the approximations are invalid for it A1
(c) Approximation -0.0326 agrees with -0.03263 to 3 s.f. – good because \(\theta\) is very small B1
Answer: (a) shown; (b) θ ≈ -0.0326; (c) exact -0.03263; close
Board notes: Small-angle approximations \(\sin\theta\approx\theta\), \(\cos\theta\approx1-\tfrac12\theta^2\), \(\tan\theta\approx\theta\) are printed in the Edexcel formulae book; candidates for every board should know them and when they are valid. All require \(\theta\) in radians.
Question 13hard6 marksYear 13
Given that \(\theta\) is small,
(a) show that \(\cos\left(\theta + \dfrac{\pi}{3}\right) \approx \dfrac12 - \dfrac{\sqrt3}{2}\theta - \dfrac14\theta^2\). (4)
(b) Hence estimate \(\cos(1.1)\), giving your answer to 3 decimal places, using \(\tfrac{\pi}{3} \approx 1.0472\). (2)
Mark scheme and answer
(a) \(\cos\theta\cos\tfrac\pi3 - \sin\theta\sin\tfrac\pi3\) M1
\(\tfrac12(1 - \tfrac12\theta^2) - \tfrac{\sqrt3}{2}\theta\) M1A1
\(= \tfrac12 - \tfrac{\sqrt3}{2}\theta - \tfrac14\theta^2\) A1
(b) \(\theta = 1.1 - \tfrac\pi3 = 0.0528\) M1
\(\approx 0.454\) (exact 0.4536) A1
Answer: (a) shown; (b) 0.454
Board notes: Requires the compound angle formulae (all boards, Year 2). Small-angle approximations \(\sin\theta\approx\theta\), \(\cos\theta\approx1-\tfrac12\theta^2\), \(\tan\theta\approx\theta\) are printed in the Edexcel formulae book; candidates for every board should know them and when they are valid. All require \(\theta\) in radians.
Question 14hard5 marksYear 13
In triangle \(ABC\), \(AB = 5\) cm, \(AC = 6\) cm and angle \(BAC = \theta\) radians, where \(\theta\) is small.
(a) Show that \(BC^2 \approx 1 + 30\theta^2\). (3)
(b) Hence estimate \(BC\) when \(\theta = 0.1\), and compare with the exact value. (2)
Mark scheme and answer
(a) \(BC^2 = 25 + 36 - 60\cos\theta\) M1
\(\approx 61 - 60(1 - \tfrac12\theta^2) = 1 + 30\theta^2\) M1A1
(b) \(\sqrt{1 + 0.3} = 1.1402\) cm B1
Exact \(\sqrt{61 - 60\cos0.1} = 1.1401\) cm – very close B1
Answer: (a) shown; (b) ≈ 1.140 cm (exact 1.140 cm)
Board notes: Small-angle approximations \(\sin\theta\approx\theta\), \(\cos\theta\approx1-\tfrac12\theta^2\), \(\tan\theta\approx\theta\) are printed in the Edexcel formulae book; candidates for every board should know them and when they are valid. All require \(\theta\) in radians.
Question 15hard6 marksYear 13
A chord of length 12 cm is drawn in a circle of radius 10 cm.
(a) Find the angle subtended by the chord at the centre, in radians to 4 decimal places. (2)
(b) Find the area of the minor segment cut off by the chord. (3)
(c) Find the perimeter of the minor segment. (1)
Mark scheme and answer
(a) \(\sin\tfrac\theta2 = 0.6\) M1
\(\theta = 1.2870\) A1
(b) \(\tfrac12(100)(\theta - \sin\theta)\) M1
\(\sin\theta = 2(0.6)(0.8) = 0.96\) B1
\(= 16.35\) cm² A1
(c) \(10\theta + 12 = 24.87\) cm B1
Answer: (a) 1.2870 rad; (b) 16.35 cm²; (c) 24.87 cm
Board notes: All boards: \(s = r\theta\), \(A = \tfrac12r^2\theta\) with \(\theta\) in radians; exact answers in terms of \(\pi\) are expected when angles are given as multiples of \(\pi\).
Question 16hard6 marksYear 13
A garden bed is in the shape of a sector of a circle with radius \(r\) m and angle \(\theta\) radians. The perimeter of the bed is 20 m.
(a) Show that the area \(A\) of the bed is given by \(A = 10r - r^2\). (2)
(b) Find the maximum area and the corresponding value of \(\theta\). (4)
Mark scheme and answer
(a) \(r\theta + 2r = 20 \Rightarrow \theta = \dfrac{20 - 2r}{r}\) M1
\(A = \tfrac12r^2\theta = \tfrac12r(20 - 2r) = 10r - r^2\) A1
(b) \(\dfrac{dA}{dr} = 10 - 2r = 0\), \(r = 5\) M1A1
\(A = 25\) m² A1
\(\theta = \dfrac{10}{5} = 2\) rad A1
Answer: (a) shown; (b) A = 25 m², θ = 2
Board notes: All boards: \(s = r\theta\), \(A = \tfrac12r^2\theta\) with \(\theta\) in radians; exact answers in terms of \(\pi\) are expected when angles are given as multiples of \(\pi\).
Question 17challenging5 marksYear 13
Given that \(\dfrac{\sin h}{h}\to1\) and \(\dfrac{\cos h - 1}{h}\to0\) as \(h\to0\),
(a) prove from first principles that the derivative of \(\sin x\) is \(\cos x\). (4)
(b) Use small-angle approximations to justify the result \(\dfrac{\cos h - 1}{h} \to 0\). (1)
Mark scheme and answer
(a) \(\dfrac{\sin(x+h) - \sin x}{h}\) B1
\(= \dfrac{\sin x\cos h + \cos x\sin h - \sin x}{h}\) M1
\(= \sin x\left(\dfrac{\cos h - 1}{h}\right) + \cos x\left(\dfrac{\sin h}{h}\right)\) M1
As \(h\to0\): \(\sin x\times0 + \cos x\times1 = \cos x\) A1
(b) \(\cos h - 1 \approx -\tfrac12h^2\), so \(\dfrac{\cos h - 1}{h} \approx -\tfrac12h \to 0\) B1
Answer: (a) proof; (b) −h/2 → 0
Board notes: Edexcel explicitly examines this proof; AQA and OCR also list differentiation of sin x and cos x from first principles. Small-angle approximations \(\sin\theta\approx\theta\), \(\cos\theta\approx1-\tfrac12\theta^2\), \(\tan\theta\approx\theta\) are printed in the Edexcel formulae book; candidates for every board should know them and when they are valid. All require \(\theta\) in radians.
Question 18challenging7 marksYear 13
A sector \(OAB\) has centre \(O\), radius \(r\) and angle \(AOB = \theta\) radians, where \(0 \lt \theta \lt \pi\). The area of the sector is three times the area of triangle \(OAB\).
(a) Show that \(\theta = 3\sin\theta\). (2)
(b) Show that the equation has a root between 2 and 2.5. (2)
(c) Use the Newton-Raphson method with \(\theta_0 = 2.3\) to find the root to 3 decimal places. (3)
Mark scheme and answer
(a) \(\tfrac12r^2\theta = 3\times\tfrac12r^2\sin\theta\) M1
\(\theta = 3\sin\theta\) A1
(b) \(f(\theta) = \theta - 3\sin\theta\): \(f(2) = -0.728\), \(f(2.5) = 0.705\) M1
Sign change and \(f\) continuous, so root in \((2, 2.5)\) A1
(c) \(\theta_{n+1} = \theta_n - \dfrac{\theta_n - 3\sin\theta_n}{1 - 3\cos\theta_n}\) M1
\(\theta_1 = 2.27903\) A1
\(\theta = 2.279\) A1
Answer: (a) shown; (b) sign change; (c) θ = 2.279
Board notes: Links radians with Year 2 numerical methods. All boards: \(s = r\theta\), \(A = \tfrac12r^2\theta\) with \(\theta\) in radians; exact answers in terms of \(\pi\) are expected when angles are given as multiples of \(\pi\).
Original questions in the style of Edexcel 9MA0, independently checked before publishing. Not past-paper questions. Covers: Small-angle approximations, Arc length and sector area, Validity of small-angle approximations, Area of a segment, Arc length and sector area in context, Solving equations with small-angle approximations, Small angles with compound angles, Small angles in geometry, Optimisation with sectors, Differentiation from first principles, Sectors and numerical methods.
A Level Math Revision