Small Angle Approximations

This lesson covers the small angle approximations for sin θ, cos θ, and tan θ when θ 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.

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The Three Approximations

When θ is small and measured in radians:

sin θ ≈ θ
cos θ ≈ 1 − θ²/2
tan θ ≈ θ

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

Crucially, θ must be in radians. The approximations do not work with degrees.

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Why Do These Work?

The sin θ ≈ θ Approximation

Consider the unit circle. For a small angle θ (in radians), the arc length subtended is θ (since s = rθ = 1 × θ), and the opposite side of the triangle (which equals sin θ) 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 θ = θ − θ³/3! + θ⁵/5! − ...

For small θ, the higher-order terms (θ³, θ⁵, ...) are negligibly small, so sin θ ≈ θ.

The cos θ ≈ 1 − θ²/2 Approximation

From the Maclaurin series:

cos θ = 1 − θ²/2! + θ⁴/4! − ...

For small θ, the terms from θ⁴ onwards are negligible, so cos θ ≈ 1 − θ²/2.

Alternatively, using cos θ = 1 − 2sin²(θ/2) and the approximation sin(θ/2) ≈ θ/2:

cos θ ≈ 1 − 2(θ/2)² = 1 − θ²/2

The tan θ ≈ θ Approximation

Since tan θ = sin θ / cos θ, and for small θ:

tan θ ≈ θ / (1 − θ²/2) ≈ θ × (1 + θ²/2 + ...) ≈ θ

The error term is of order θ³, just as for sin θ.

From the Maclaurin series:

tan θ = θ + θ³/3 + 2θ⁵/15 + ...

For small θ, tan θ ≈ θ.

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Numerical Verification

Let us check the approximations for θ = 0.1 radians (about 5.7°):

Function	Exact Value	Approximation	Error
sin(0.1)	0.0998334...	0.1	0.0001666...
cos(0.1)	0.9950042...	1 − 0.005 = 0.995	0.0000042...
tan(0.1)	0.1003347...	0.1	0.0003347...

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

For θ = 0.5 radians (about 28.6°):

Function	Exact Value	Approximation	Error
sin(0.5)	0.4794...	0.5	0.0206...
cos(0.5)	0.8776...	1 − 0.125 = 0.875	0.0026...
tan(0.5)	0.5463...	0.5	0.0463...

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

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Using Small Angle Approximations to Simplify Expressions

Example 1: Use small angle approximations to estimate (sin 3θ)/(2θ) when θ is small.

sin 3θ ≈ 3θ (since 3θ is also small when θ is small)

(sin 3θ)/(2θ) ≈ 3θ/(2θ) = 3/2

Example 2: Simplify (1 − cos θ)/θ² for small θ.

cos θ ≈ 1 − θ²/2
1 − cos θ ≈ θ²/2
(1 − cos θ)/θ² ≈ (θ²/2)/θ² = 1/2

Example 3: Simplify (tan 2θ − sin 2θ)/θ³ for small θ.

tan 2θ ≈ 2θ
sin 2θ ≈ 2θ

This gives (2θ − 2θ)/θ³ = 0/θ³ = 0, but this is not accurate enough — we need to keep more terms.

Using better approximations:

tan 2θ ≈ 2θ + (2θ)³/3 = 2θ + 8θ³/3
sin 2θ ≈ 2θ − (2θ)³/6 = 2θ − 8θ³/6 = 2θ − 4θ³/3

Therefore:

tan 2θ − sin 2θ ≈ (2θ + 8θ³/3) − (2θ − 4θ³/3) = 8θ³/3 + 4θ³/3 = 12θ³/3 = 4θ³
(tan 2θ − sin 2θ)/θ³ ≈ 4θ³/θ³ = 4

Example 4: Given that θ is small, find an approximate expression for (3θ + sin θ)/(2 − cos θ) as a linear expression in θ.

Numerator: 3θ + sin θ ≈ 3θ + θ = 4θ
Denominator: 2 − cos θ ≈ 2 − (1 − θ²/2) = 1 + θ²/2

For small θ, θ² is negligible compared to 1, so denominator ≈ 1.

(3θ + sin θ)/(2 − cos θ) ≈ 4θ/1 = 4θ

Example 5: Show that for small θ, (1 − cos 2θ)/(sin θ tan θ) ≈ 2.

1 − cos 2θ ≈ 1 − (1 − (2θ)²/2) = 1 − 1 + 2θ² = 2θ²
sin θ ≈ θ
tan θ ≈ θ
sin θ tan θ ≈ θ × θ = θ²

(1 − cos 2θ)/(sin θ tan θ) ≈ 2θ²/θ² = 2 ∎

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Connection to Limits

The small angle approximations are directly connected to important limits:

lim(θ→0) sin θ / θ = 1
lim(θ→0) (1 − cos θ) / θ² = 1/2
lim(θ→0) tan θ / θ = 1

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 d/dx(sin x) = cos x.

From first principles:

d/dx(sin x) = lim(h→0) [sin(x + h) − sin x] / h
            = lim(h→0) [sin x cos h + cos x sin h − sin x] / h
            = lim(h→0) [sin x(cos h − 1) + cos x sin h] / h
            = sin x × lim(h→0)(cos h − 1)/h + cos x × lim(h→0)(sin h)/h

Using small angle approximations:

cos h − 1 ≈ −h²/2, so (cos h − 1)/h ≈ −h/2 → 0
sin h ≈ h, so sin h / h ≈ 1

Therefore:

d/dx(sin x) = sin x × 0 + cos x × 1 = cos x

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Exam-Style Problems

Problem 1: When θ is small, find approximate expressions for:

(a) (sin 5θ)/(3θ)

≈ 5θ/(3θ) = 5/3

(b) (1 − cos 4θ)/(8θ²)

cos 4θ ≈ 1 − (4θ)²/2 = 1 − 8θ²
1 − cos 4θ ≈ 8θ²
(1 − cos 4θ)/(8θ²) ≈ 8θ²/(8θ²) = 1

(c) (θ + tan θ)/(sin 2θ)

≈ (θ + θ)/(2θ) = 2θ/(2θ) = 1

Problem 2: The expression (2 − 2cos θ + sin²θ)/(4θ²) is approximately equal to k when θ is small. Find the value of k.

2 − 2cos θ ≈ 2 − 2(1 − θ²/2) = 2 − 2 + θ² = θ²
sin²θ ≈ θ²

(θ² + θ²)/(4θ²) = 2θ²/(4θ²) = 1/2

So k = 1/2.

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Summary

• For small θ in radians: sin θ ≈ θ, cos θ ≈ 1 − θ²/2, tan θ ≈ θ.
• These come from truncating the Maclaurin series.
• θ 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 (θ³ terms).
• The key limits: sin θ/θ → 1 and (1 − cos θ)/θ² → 1/2 as θ → 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 θ ≈ θ..."). 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 "θ 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: $\sin\theta \approx \theta$sinθ≈θ, $\cos\theta \approx 1 - \tfrac{1}{2}\theta^2$cosθ≈1−21​θ2, $\tan\theta \approx \theta$tanθ≈θ (where $\theta$θ 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.

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Worked example with full mark scheme

Question (8 marks):

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

$\lim_{x \to 0} \dfrac{\sin 3x - \tan 3x}{x^3}$limx→0​x3sin3x−tan3x​

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

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

Solution with mark scheme:

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

The denominator is $x^3$x3, so any approximation truncated at lower order than $x^3$x3 will lose information. We need expansions of $\sin 3x$sin3x and $\tan 3x$tan3x accurate to at least order $x^3$x3.

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

Step 2 — expand $\sin 3x$sin3x to cubic order.

Using $\sin\theta \approx \theta - \tfrac{1}{6}\theta^3$sinθ≈θ−61​θ3 (the standard approximation extended one further term, from the Maclaurin series), with $\theta = 3x$θ=3x:

$\sin 3x \approx 3x - \dfrac{(3x)^3}{6} = 3x - \dfrac{27x^3}{6} = 3x - \dfrac{9x^3}{2}$sin3x≈3x−6(3x)3​=3x−627x3​=3x−29x3​

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

Step 3 — expand $\tan 3x$tan3x to cubic order.

Using $\tan\theta \approx \theta + \tfrac{1}{3}\theta^3$tanθ≈θ+31​θ3 (Maclaurin extension), with $\theta = 3x$θ=3x:

$\tan 3x \approx 3x + \dfrac{(3x)^3}{3} = 3x + \dfrac{27x^3}{3} = 3x + 9x^3$tan3x≈3x+3(3x)3​=3x+327x3​=3x+9x3

M1 — correct cubic-order expansion of $\tan 3x$tan3x. Common error: candidates use the bare $\tan\theta \approx \theta$tanθ≈θ from the formula sheet and obtain numerator $0$0, giving the indeterminate form $0/0$0/0 they cannot resolve.

Step 4 — subtract.

$\sin 3x - \tan 3x \approx \left(3x - \dfrac{9x^3}{2}\right) - \left(3x + 9x^3\right) = -\dfrac{9x^3}{2} - 9x^3 = -\dfrac{27x^3}{2}$sin3x−tan3x≈(3x−29x3​)−(3x+9x3)=−29x3​−9x3=−227x3​

A1 — the linear $3x$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.

$\dfrac{\sin 3x - \tan 3x}{x^3} \approx \dfrac{-27x^3/2}{x^3} = -\dfrac{27}{2}$x3sin3x−tan3x​≈x3−27x3/2​=−227​

As $x \to 0$x→0 the higher-order error terms (of order $x^5$x5 and beyond) vanish, so:

$\lim_{x \to 0} \dfrac{\sin 3x - \tan 3x}{x^3} = -\dfrac{27}{2}$limx→0​x3sin3x−tan3x​=−227​

A1 — final exact answer.

B1 — explicit comment that the omitted terms are $O(x^5)$O(x5) and contribute $0$0 in the limit.

(b) Step 1 — diagnose the failure mode.

If only $\tan\theta \approx \theta$tanθ≈θ is used, then $\sin 3x - \tan 3x \approx -\tfrac{9x^3}{2} - 0 = -\tfrac{9x^3}{2}$sin3x−tan3x≈−29x3​−0=−29x3​, which would give the (incorrect) limit $-\tfrac{9}{2}$−29​. The cubic term $+\tfrac{1}{3}\theta^3$+31​θ3 in the Maclaurin expansion of $\tan\theta$tanθ is the missing ingredient — it is the same order as the $-\tfrac{1}{6}\theta^3$−61​θ3 term from $\sin\theta$sinθ, so dropping it changes the numerator by a factor that is not negligible.

B1 — identifying $\tfrac{1}{3}\theta^3$31​θ3 as the required extra term.

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

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

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Specimen question modelled on the AQA 7357/2 format

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

$\ddot\theta = -\dfrac{g}{L}\sin\theta$θ¨=−Lg​sinθ

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

$\ddot\theta \approx -\dfrac{g}{L}\theta$θ¨≈−Lg​θ

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

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

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

Mark scheme decomposition by AO:

(a)

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

(b)

• M1 (AO1.1b) — comparing $\ddot\theta = -\omega^2 \theta$θ¨=−ω2θ with the linearised equation to identify $\omega^2 = g/L$ω2=g/L.
• A1 (AO1.1b) — $\omega = \sqrt{g/L}$ω=g/L​.
• A1 (AO2.5) — $T = 2\pi/\omega = 2\pi\sqrt{L/g}$T=2π/ω=2πL/g​.

(c)

• B1 (AO3.4) — using $\sin\theta \approx \theta - \tfrac{1}{6}\theta^3$sinθ≈θ−61​θ3, the relative error is $\tfrac{1}{6}\theta^2$61​θ2. Setting $\tfrac{1}{6}\theta^2 < 0.01$61​θ2<0.01 gives $\theta < \sqrt{0.06} \approx 0.245$θ<0.06​≈0.245 radians (about $14°$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.

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Synoptic links

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

1. Section E — Differentiation of $\sin x$sinx from first principles. The proof that $\frac{d}{dx}\sin x = \cos x$dxd​sinx=cosx uses $\lim_{h \to 0}\frac{\sin h}{h} = 1$limh→0​hsinh​=1 (equivalent to $\sin h \approx h$sinh≈h) and $\lim_{h \to 0}\frac{\cos h - 1}{h} = 0$limh→0​hcosh−1​=0 (equivalent to $\cos h \approx 1 - \tfrac{1}{2}h^2$cosh≈1−21​h2). Without small-angle results, the derivative cannot be derived; with them, it is two lines.

2. Section E — Maclaurin / Taylor series (Further Maths context). The small-angle approximations are the first non-zero terms of the Maclaurin expansions $\sin\theta = \theta - \tfrac{\theta^3}{6} + \tfrac{\theta^5}{120} - \ldots$sinθ=θ−6θ3​+120θ5​−… and $\cos\theta = 1 - \tfrac{\theta^2}{2} + \tfrac{\theta^4}{24} - \ldots$cosθ=1−2θ2​+24θ4​−…. AS-Level treats them as memorised facts; Further Maths derives them. Recognising the connection lifts a procedural skill into a structural one.

3. Mechanics (Paper 3) — pendulum and SHM. As shown in the specimen, small-angle linearisation converts the nonlinear pendulum equation into the SHM equation $\ddot\theta = -\omega^2 \theta$θ¨=−ω2θ with period $T \approx 2\pi\sqrt{L/g}$T≈2πL/g​. Every "estimate the period of a pendulum" question rests silently on $\sin\theta \approx \theta$sinθ≈θ.