P5.2Small angle approximations

Edexcel A level Maths (9MA0) · Pure mathematics › Trigonometry

Practise Small angle approximations. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.

Build a paper on this topic

▶ Watch videos on Small angle approximations (TLMaths on YouTube) · Practise all of Trigonometry

Sample questions

Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.

Question 1Easy4 marks

(a) Given that \(\theta\) is small, use the small angle approximations for \(\cos\theta\) and \(\sin\theta\) to show that \[6\cos 2\theta - 5\sin 2\theta \approx p + q\theta + r\theta^2,\] where \(p\), \(q\) and \(r\) are integers to be found.[3]

(b) Hence find an approximate value of \(6\cos 2\theta - 5\sin 2\theta\) when \(\theta = 0.1\).[1]

Show the answer and mark scheme

(a) Answer: \(6\cos 2\theta - 5\sin 2\theta \approx 6 - 10\theta - 12\theta^2\)

M1 for using \(\cos 2\theta \approx 1 - 2\theta^2\) and \(\sin 2\theta \approx 2\theta\)
A1 for two of \(p = 6\), \(q = -10\), \(r = -12\) correct
A1 for \(6 - 10\theta - 12\theta^2\)

Worked solution: \(\cos 2\theta \approx 1 - 2\theta^2\) and \(\sin 2\theta \approx 2\theta\), so
\(6\cos 2\theta - 5\sin 2\theta \approx 6\left(1 - 2\theta^2\right) - 10\theta = 6 - 10\theta - 12\theta^2\)

(b) Answer: \(4.88\)

B1ft for \(4.88\) (follow through their \(p + 0.1q + 0.01r\))

Worked solution: \(6 - 10(0.1) - 12(0.1)^2 = 4.88\) (the true value is \(4.88705\ldots\)).

Question 2Medium6 marks

(a) Show that, when \(\theta\) is small and measured in radians,
\[\frac{\sin 3\theta \tan 2\theta}{1 - \cos 4\theta} \approx \frac{3}{4}\][3]

(b) Find the percentage error when this approximation is used for the value of the expression when \(\theta = 0.1\).[2]

Show the answer and mark scheme

(a) Answer: \(\dfrac{3\theta \times 2\theta}{1 - \left(1 - \frac{(4\theta)^2}{2}\right)} = \dfrac{6\theta^2}{8\theta^2} = \dfrac{3}{4}\)

M1 for \(\sin 3\theta \approx 3\theta\) and \(\tan 2\theta \approx 2\theta\)
M1 for \(\cos 4\theta \approx 1 - \frac{(4\theta)^2}{2} = 1 - 8\theta^2\)
A1* for \(\frac{6\theta^2}{8\theta^2} = \frac{3}{4}\)

Worked solution: For small \(\theta\): \(\sin 3\theta \approx 3\theta\), \(\tan 2\theta \approx 2\theta\), \(\cos 4\theta \approx 1 - \frac{1}{2}(4\theta)^2 = 1 - 8\theta^2\).
So the expression \(\approx \dfrac{6\theta^2}{8\theta^2} = \dfrac{3}{4}\).

(b) Answer: 1.17% (awrt 1.2%) %

M1 for evaluating the expression at \(\theta = 0.1\): 0.7588…
A1 for awrt 1.2%

Worked solution: \(\dfrac{\sin 0.3 \tan 0.2}{1 - \cos 0.4} = 0.75888\ldots\). Percentage error \(= \dfrac{0.75888 - 0.75}{0.75888} \times 100 = 1.17\%\).

(c) Answer: 1.5 radians is not a small angle, so the small angle approximations are not accurate (the true value is about 3.5).

B1 for \(\theta = 1.5\) (radians) is not small, so the approximations do not hold

Worked solution: The approximations \(\sin x \approx x\), \(\tan x \approx x\) and \(\cos x \approx 1 - \frac{x^2}{2}\) are only accurate for small angles in radians. At \(\theta = 1.5\) the angles \(3\theta\), \(2\theta\) and \(4\theta\) are large (the expression is actually about 3.5).

Question 3Hard6 marks

(a) Given that \(\theta\) is small, use the small angle approximations to find the value of the constant \(k\) such that \[\frac{3\sec 2\theta - 3}{\theta\sin 2\theta} \approx k.\][4]

(b) Find the percentage error when your value of \(k\) is used as an estimate for the value of the expression when \(\theta = 0.1\). Give your answer to 2 significant figures.[2]

Show the answer and mark scheme

(a) Answer: \(k = 3\)

M1 for \(\sec 2\theta \approx \left(1 - 2\theta^2\right)^{-1}\)
M1 for a binomial expansion giving \(1 + 2\theta^2\)
M1 for \(\sin 2\theta \approx 2\theta\) and forming \(\frac{a\theta^2}{b\theta^2}\)
A1 for \(k = 3\) oe

Worked solution: \(\sec 2\theta = \frac{1}{\cos 2\theta} \approx \left(1 - 2\theta^2\right)^{-1} \approx 1 + 2\theta^2\) (binomial expansion)
\(3\sec 2\theta - 3 \approx 6\theta^2\) and \(\theta\sin 2\theta \approx 2\theta^2\)
So the expression \(\approx 3\)

(b) Answer: \(2.3\%\)

M1 for evaluating the expression at \(\theta = 0.1\) (\(= 3.07126\ldots\)) and using \(\frac{|k - \text{true value}|}{|\text{true value}|} \times 100\)
A1 for awrt \(2.3\%\)

Worked solution: At \(\theta = 0.1\) the expression equals \(3.07126\ldots\), so the percentage error is \(\frac{\left|3 - 3.07126\ldots\right|}{3.07126\ldots} \times 100 = 2.3\%\)

Related subtopics

Stuck? Get 1-to-1 help. Chhetri Academy tutors GCSE and A level Maths and Science online, with a free 30-minute trial lesson.

Book a free trial