Small Angle Approximations

Builds on: aqa-alevel-maths-pure-2 / further-trigonometry — this lesson is the A2-level deep dive.

This lesson covers the small angle approximations for $\sin\theta$ , $\cos\theta$ , and $\tan\theta$ when $\theta$ is small and measured in radians. These approximations simplify expressions and are used in limit calculations, mechanics, and numerical analysis. They are part of the AQA A-Level Mathematics specification 7357.

Spec Mapping — AQA 7357 Section E Trigonometry. This lesson covers the small-angle approximations $\sin\theta \approx \theta$ , $\cos\theta \approx 1 - \theta^2/2$ , $\tan\theta \approx \theta$ content of Section E, at the depth required for A2-level synoptic questions. Refer to the official AQA specification document for exact wording.

The Three Approximations

When $\theta$ is small and measured in radians:

\begin{aligned} \sin \theta \approx \theta \\ \cos \theta \approx 1 - \theta ^{2}/2 \\ \tan \theta \approx \theta \end{aligned}

These approximations become more accurate as $\theta$ gets closer to zero, and they are exact in the limit as $\theta \to 0$ .

Crucially, $\theta$ must be in radians. The approximations do not work with degrees.

Why Do These Work?

The $\sin\theta \approx \theta$ Approximation

Consider the unit circle. For a small angle $\theta$ (in radians), the arc length subtended is $\theta$ (since $s = r\theta = 1 \times \theta$ ), and the opposite side of the triangle (which equals $\sin\theta$ ) is approximately equal to the arc length for small angles. As the angle shrinks, the chord and the arc become indistinguishable.

More formally, from the Maclaurin series (Taylor series at 0):

\sin\theta = \theta - \frac{\theta^{3}}{3!} + \frac{\theta^{5}}{5!} - \cdots

For small $\theta$ , the higher-order terms ( $\theta^3, \theta^5, \ldots$ ) are negligibly small, so $\sin\theta \approx \theta$ .

The $\cos\theta \approx 1 - \theta^2/2$ Approximation

From the Maclaurin series:

\cos\theta = 1 - \frac{\theta^{2}}{2!} + \frac{\theta^{4}}{4!} - \cdots

For small $\theta$ , the terms from $\theta^4$ onwards are negligible, so $\cos\theta \approx 1 - \theta^2/2$ .

Alternatively, using $\cos\theta = 1 - 2\sin^2(\theta/2)$ and the approximation $\sin(\theta/2) \approx \theta/2$ :

\cos \theta \approx 1 - 2( \theta /2)^{2} = 1 - \theta ^{2}/2

The $\tan\theta \approx \theta$ Approximation

Since $\tan\theta = \sin\theta / \cos\theta$ , and for small $\theta$ :

\tan\theta \approx \frac{\theta}{1 - \frac{\theta^{2}}{2}} \approx \theta\left(1 + \frac{\theta^{2}}{2} + \cdots\right) \approx \theta

The error term is of order $\theta^3$ , just as for $\sin\theta$ .

From the Maclaurin series:

\tan\theta = \theta + \frac{\theta^{3}}{3} + \frac{2\theta^{5}}{15} + \cdots

For small $\theta$ , $\tan\theta \approx \theta$ .

Numerical Verification

Let us check the approximations for $\theta = 0.1$ radians (about $5.7°$ ):

Function	Exact Value	Approximation	Error
$\sin(0.1)$	$0.0998334\ldots$	$0.1$	$0.0001666\ldots$
$\cos(0.1)$	$0.9950042\ldots$	$1 - 0.005 = 0.995$	$0.0000042\ldots$
$\tan(0.1)$	$0.1003347\ldots$	$0.1$	$0.0003347\ldots$

The approximations are very accurate for small angles. For $\theta = 0.01$ radians, the errors would be even smaller (roughly 1000 times smaller).

For $\theta = 0.5$ radians (about $28.6°$ ):

Function	Exact Value	Approximation	Error
$\sin(0.5)$	$0.4794\ldots$	$0.5$	$0.0206\ldots$
$\cos(0.5)$	$0.8776\ldots$	$1 - 0.125 = 0.875$	$0.0026\ldots$
$\tan(0.5)$	$0.5463\ldots$	$0.5$	$0.0463\ldots$

The errors are larger but the approximations are still reasonable. As a rule of thumb, the approximations are good for $\theta \lesssim 0.2$ radians and become progressively worse beyond that.

Using Small Angle Approximations to Simplify Expressions

Example 1: Use small angle approximations to estimate $\dfrac{\sin 3\theta}{2\theta}$ when $\theta$ is small.

\begin{aligned} \sin 3\theta &\approx 3\theta \quad \text{(since } 3\theta \text{ is also small when } \theta \text{ is small)} \\ \frac{\sin 3\theta}{2\theta} &\approx \frac{3\theta}{2\theta} = \tfrac{3}{2} \end{aligned}

Example 2: Simplify $\dfrac{1 - \cos\theta}{\theta^2}$ for small $\theta$ .

\begin{aligned} \cos \theta \approx 1 - \theta ^{2}/2 \\ 1 - \cos \theta \approx \theta ^{2}/2 \\ (1 - \cos \theta )/ \theta ^{2} \approx ( \theta ^{2}/2)/ \theta ^{2} = 1/2 \end{aligned}

Example 3: Simplify $\dfrac{\tan 2\theta - \sin 2\theta}{\theta^3}$ for small $\theta$ .

\begin{aligned} \tan 2 \theta \approx 2 \theta \\ \sin 2 \theta \approx 2 \theta \end{aligned}

This gives $\dfrac{2\theta - 2\theta}{\theta^3} = \dfrac{0}{\theta^3} = 0$ , but this is not accurate enough — we need to keep more terms.

Using better approximations:

\begin{aligned} \tan 2\theta &\approx 2\theta + \frac{(2\theta)^{3}}{3} = 2\theta + \frac{8\theta^{3}}{3} \\ \sin 2\theta &\approx 2\theta - \frac{(2\theta)^{3}}{6} = 2\theta - \frac{8\theta^{3}}{6} = 2\theta - \frac{4\theta^{3}}{3} \end{aligned}

Therefore:

\begin{aligned} \tan 2\theta - \sin 2\theta &\approx \left(2\theta + \frac{8\theta^{3}}{3}\right) - \left(2\theta - \frac{4\theta^{3}}{3}\right) = \frac{8\theta^{3}}{3} + \frac{4\theta^{3}}{3} = \frac{12\theta^{3}}{3} = 4\theta^{3} \\ \frac{\tan 2\theta - \sin 2\theta}{\theta^{3}} &\approx \frac{4\theta^{3}}{\theta^{3}} = 4 \end{aligned}

Example 4: Given that $\theta$ is small, find an approximate expression for $\dfrac{3\theta + \sin\theta}{2 - \cos\theta}$ as a linear expression in $\theta$ .

\begin{aligned} \text{Numerator: } 3\theta + \sin\theta &\approx 3\theta + \theta = 4\theta \\ \text{Denominator: } 2 - \cos\theta &\approx 2 - \left(1 - \frac{\theta^{2}}{2}\right) = 1 + \frac{\theta^{2}}{2} \end{aligned}

For small $\theta$ , $\theta^2$ is negligible compared to 1, so denominator $\approx 1$ .

\frac{3\theta + \sin\theta}{2 - \cos\theta} \approx \frac{4\theta}{1} = 4\theta

Example 5: Show that for small $\theta$ , $\dfrac{1 - \cos 2\theta}{\sin\theta \tan\theta} \approx 2$ .

\begin{aligned} 1 - \cos 2\theta &\approx 1 - \left(1 - \frac{(2\theta)^{2}}{2}\right) = 1 - 1 + 2\theta^{2} = 2\theta^{2} \\ \sin\theta &\approx \theta \\ \tan\theta &\approx \theta \\ \sin\theta \tan\theta &\approx \theta \times \theta = \theta^{2} \\[4pt] \frac{1 - \cos 2\theta}{\sin\theta \tan\theta} &\approx \frac{2\theta^{2}}{\theta^{2}} = 2 \qquad\blacksquare \end{aligned}

Connection to Limits

The small angle approximations are directly connected to important limits:

\begin{aligned} \lim_{\theta \to 0} \frac{\sin\theta}{\theta} &= 1 \\ \lim_{\theta \to 0} \frac{1 - \cos\theta}{\theta^{2}} &= \frac{1}{2} \\ \lim_{\theta \to 0} \frac{\tan\theta}{\theta} &= 1 \end{aligned}

These limits are fundamental in proving the derivatives of $\sin x$ and $\cos x$ from first principles.

Example 6: Using the small angle approximation, explain why $\tfrac{d}{dx}(\sin x) = \cos x$ .

From first principles:

\begin{aligned} \frac{d}{dx}(\sin x) &= \lim_{h \to 0} \frac{\sin(x + h) - \sin x}{h} \\ &= \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h} \\ &= \lim_{h \to 0} \frac{\sin x(\cos h - 1) + \cos x \sin h}{h} \\ &= \sin x \cdot \lim_{h \to 0}\frac{\cos h - 1}{h} + \cos x \cdot \lim_{h \to 0}\frac{\sin h}{h} \end{aligned}

Using small angle approximations:

\begin{aligned} \cos h - 1 &\approx -\frac{h^{2}}{2}, \ \text{so} \ \frac{\cos h - 1}{h} \approx -\frac{h}{2} \to 0 \\ \sin h &\approx h, \ \text{so} \ \frac{\sin h}{h} \approx 1 \end{aligned}

Therefore:

\frac{d}{dx}(\sin x) = \sin x \cdot 0 + \cos x \cdot 1 = \cos x

Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

Exam-Style Problems

Problem 1: When $\theta$ is small, find approximate expressions for:

(a) $\dfrac{\sin 5\theta}{3\theta}$

\approx 5 \theta /(3 \theta ) = 5/3

(b) $\dfrac{1 - \cos 4\theta}{8\theta^2}$

\begin{aligned} \cos 4 \theta \approx 1 - (4 \theta )^{2}/2 = 1 - 8 \theta ^{2} \\ 1 - \cos 4 \theta \approx 8 \theta ^{2} \\ (1 - \cos 4 \theta )/(8 \theta ^{2}) \approx 8 \theta ^{2}/(8 \theta ^{2}) = 1 \end{aligned}

\approx \frac{\theta + \theta}{2\theta} = \frac{2\theta}{2\theta} = 1

Problem 2: The expression $\dfrac{2 - 2\cos\theta + \sin^2\theta}{4\theta^2}$ is approximately equal to $k$ when $\theta$ is small. Find the value of $k$ .

\begin{aligned} 2 - 2\cos\theta &\approx 2 - 2\left(1 - \frac{\theta^{2}}{2}\right) = 2 - 2 + \theta^{2} = \theta^{2} \\ \sin^{2}\theta &\approx \theta^{2} \\[4pt] \frac{\theta^{2} + \theta^{2}}{4\theta^{2}} &= \frac{2\theta^{2}}{4\theta^{2}} = \frac{1}{2} \end{aligned}

So $k = \tfrac{1}{2}$ .

Synoptic Links

This topic connects to:

Numerical methods (aqa-alevel-maths-pure-1 / numerical-methods) — small-angle truncation is the same logic as truncating a Maclaurin series for fast computation.
Differential equations modelling (aqa-alevel-maths-calculus-applications / differential-equations-modelling) — the pendulum $\theta'' + \tfrac{g}{L}\sin\theta = 0$ linearises to SHM via $\sin\theta \approx \theta$ .
Differentiation from first principles (aqa-alevel-maths-pure-1 / differentiation) — the derivative $\tfrac{d}{dx}(\sin x) = \cos x$ depends on the limit $\sin h / h \to 1$ .

Summary

For small $\theta$ in radians: $\sin\theta \approx \theta$ , $\cos\theta \approx 1 - \theta^2/2$ , $\tan\theta \approx \theta$ .
These come from truncating the Maclaurin series.
$\theta$ must be in radians — these do not work with degrees.
Used to simplify expressions, evaluate limits, and approximate calculations.
When first-order terms cancel, you may need to include higher-order terms ( $\theta^3$ terms).
The key limits: $\sin\theta/\theta \to 1$ and $(1 - \cos\theta)/\theta^2 \to \tfrac{1}{2}$ as $\theta \to 0$ .

Exam Tip: Small angle approximations are quick to apply but you must state clearly that you are using them (e.g., "Using the small angle approximation $\sin\theta \approx \theta$ ..."). In the exam, you may be asked to find the value of an expression correct to a certain order — make sure you include enough terms of the approximation. Always work in radians. If a question says " $\theta$ is small", this is your cue to use small angle approximations.

A-Level Deep Dive: Small-Angle Approximations

Spec mapping

AQA 7357 specification, Paper 2 — Pure Mathematics, Section E (Trigonometry), Year 2 sub-strand E5 covers the standard small-angle approximations of sine, cosine and tangent: $\sin\theta \approx \theta$ , $\cos\theta \approx 1 - \tfrac{1}{2}\theta^2$ , $\tan\theta \approx \theta$ (where $\theta$ is in radians) (refer to the official specification document for exact wording). This is examined synoptically across 7357/2 (Pure Paper 2), often inside limits, differentiation-from-first-principles questions, and modelling contexts. The AQA formula booklet does not list these approximations — they must be memorised, along with the radian-only restriction. Examiners exploit the radian condition heavily: a candidate who slips into degrees mid-calculation cannot recover.

Worked example with full mark scheme

Question (8 marks):

(a) Using the small-angle approximations for $\sin x$ and $\tan x$ , evaluate

$\lim_{x \to 0} \dfrac{\sin 3x - \tan 3x}{x^3}$

stating the order of accuracy required at each stage. (6)

(b) Hence explain why the approximation $\tan\theta \approx \theta$ is insufficient when computing this limit, and state which extra term in the Taylor expansion of $\tan\theta$ is required. (2)

Solution with mark scheme:

(a) Step 1 — recognise the order of accuracy required.

The denominator is $x^3$ , so any approximation truncated at lower order than $x^3$ will lose information. We need expansions of $\sin 3x$ and $\tan 3x$ accurate to at least order $x^3$ .

M1 (AO3.1a) — identifying that the cubic denominator forces a cubic-order numerator expansion.

Step 2 — expand $\sin 3x$ to cubic order.

Using $\sin\theta \approx \theta - \tfrac{1}{6}\theta^3$ (the standard approximation extended one further term, from the Maclaurin series), with $\theta = 3x$ :

$\sin 3x \approx 3x - \dfrac{(3x)^3}{6} = 3x - \dfrac{27x^3}{6} = 3x - \dfrac{9x^3}{2}$

M1 — correct expansion to cubic order with $\theta = 3x$ substituted before cubing.

Step 3 — expand $\tan 3x$ to cubic order.

Using $\tan\theta \approx \theta + \tfrac{1}{3}\theta^3$ (Maclaurin extension), with $\theta = 3x$ :

$\tan 3x \approx 3x + \dfrac{(3x)^3}{3} = 3x + \dfrac{27x^3}{3} = 3x + 9x^3$

M1 — correct cubic-order expansion of $\tan 3x$ . Common error: candidates use the bare $\tan\theta \approx \theta$ from the formula sheet and obtain numerator $0$ , giving the indeterminate form $0/0$ they cannot resolve.

Step 4 — subtract.

$\sin 3x - \tan 3x \approx \left(3x - \dfrac{9x^3}{2}\right) - \left(3x + 9x^3\right) = -\dfrac{9x^3}{2} - 9x^3 = -\dfrac{27x^3}{2}$

A1 — the linear $3x$ terms cancel exactly, leaving a clean cubic-order numerator. The cancellation is the entire point of needing higher-order terms.

Step 5 — divide and take the limit.

$\dfrac{\sin 3x - \tan 3x}{x^3} \approx \dfrac{-27x^3/2}{x^3} = -\dfrac{27}{2}$

As $x \to 0$ the higher-order error terms (of order $x^5$ and beyond) vanish, so:

$\lim_{x \to 0} \dfrac{\sin 3x - \tan 3x}{x^3} = -\dfrac{27}{2}$

A1 — final exact answer.

B1 — explicit comment that the omitted terms are $O(x^5)$ and contribute $0$ in the limit.

(b) Step 1 — diagnose the failure mode.

If only $\tan\theta \approx \theta$ is used, then $\sin 3x - \tan 3x \approx -\tfrac{9x^3}{2} - 0 = -\tfrac{9x^3}{2}$ , which would give the (incorrect) limit $-\tfrac{9}{2}$ . The cubic term $+\tfrac{1}{3}\theta^3$ in the Maclaurin expansion of $\tan\theta$ is the missing ingredient — it is the same order as the $-\tfrac{1}{6}\theta^3$ term from $\sin\theta$ , so dropping it changes the numerator by a factor that is not negligible.

B1 — identifying $\tfrac{1}{3}\theta^3$ as the required extra term.

B1 — correct explanation that this term is the same order of magnitude as the cubic term from $\sin\theta$ , so cannot be ignored.

Total: 8 marks (M3 A2 B3, split as shown).

Specimen question modelled on the AQA 7357/2 format

Question (6 marks): A pendulum of length $L$ swings under gravity. Its equation of motion for small displacement angle $\theta$ (in radians) is

$\ddot\theta = -\dfrac{g}{L}\sin\theta$

(a) Using a small-angle approximation, show that for small $\theta$ this reduces to

$\ddot\theta \approx -\dfrac{g}{L}\theta$

stating clearly the approximation used and the constraint on $\theta$ . (2)

(b) Given that this linearised equation has solution $\theta(t) = A\cos(\omega t + \phi)$ for some constants, deduce $\omega$ and hence the period $T$ . (3)

(c) Estimate the maximum amplitude (in radians) for which the linearised model is accurate to within $1\%$ , using the next term of the expansion. (1)

Mark scheme decomposition by AO:

(a)

B1 (AO1.1a) — stating $\sin\theta \approx \theta$ for small $\theta$ measured in radians.
A1 (AO2.1) — substituting and writing the linearised equation cleanly.

(b)

M1 (AO1.1b) — comparing $\ddot\theta = -\omega^2 \theta$ with the linearised equation to identify $\omega^2 = g/L$ .
A1 (AO1.1b) — $\omega = \sqrt{g/L}$ .
A1 (AO2.5) — $T = 2\pi/\omega = 2\pi\sqrt{L/g}$ .

(c)

B1 (AO3.4) — using $\sin\theta \approx \theta - \tfrac{1}{6}\theta^3$ , the relative error is $\tfrac{1}{6}\theta^2$ . Setting $\tfrac{1}{6}\theta^2 < 0.01$ gives $\theta < \sqrt{0.06} \approx 0.245$ radians (about $14°$ ).

Total: 6 marks split AO1 = 3, AO2 = 2, AO3 = 1. This is the textbook synoptic mechanics-pure crossover that AQA favours in Year 2: small-angle approximation is the bridge between the (insoluble) nonlinear pendulum and the (tractable) simple harmonic motion model.

Synoptic links

Small-angle approximations sit at a junction of several A-Level threads:

Section E — Differentiation of $\sin x$ from first principles. The proof that $\frac{d}{dx}\sin x = \cos x$ uses $\lim_{h \to 0}\frac{\sin h}{h} = 1$ (equivalent to $\sin h \approx h$ ) and $\lim_{h \to 0}\frac{\cos h - 1}{h} = 0$ (equivalent to $\cos h \approx 1 - \tfrac{1}{2}h^2$ ). Without small-angle results, the derivative cannot be derived; with them, it is two lines.
Section E — Maclaurin / Taylor series (Further Maths context). The small-angle approximations are the first non-zero terms of the Maclaurin expansions $\sin\theta = \theta - \tfrac{\theta^3}{6} + \tfrac{\theta^5}{120} - \ldots$ and $\cos\theta = 1 - \tfrac{\theta^2}{2} + \tfrac{\theta^4}{24} - \ldots$ . AS-Level treats them as memorised facts; Further Maths derives them. Recognising the connection lifts a procedural skill into a structural one.

Small Angle Approximations

Small Angle Approximations

The Three Approximations

Why Do These Work?

The sin⁡θ≈θ\sin\theta \approx \thetasinθ≈θ Approximation

The cos⁡θ≈1−θ2/2\cos\theta \approx 1 - \theta^2/2cosθ≈1−θ2/2 Approximation

The tan⁡θ≈θ\tan\theta \approx \thetatanθ≈θ Approximation

Numerical Verification

Using Small Angle Approximations to Simplify Expressions

Connection to Limits

Specimen Question (Modelled on the AQA Paper Format)

Exam-Style Problems

Synoptic Links

Summary

A-Level Deep Dive: Small-Angle Approximations

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357/2 format

Synoptic links

More in Mathematics

The $\sin\theta \approx \theta$ Approximation

The $\cos\theta \approx 1 - \theta^2/2$ Approximation

The $\tan\theta \approx \theta$ Approximation