Small Angle Approximations

When theta is small (measured in radians), the trigonometric functions can be approximated by simpler expressions. These approximations are given in the Edexcel formula booklet and are tested at A-Level (9MA0).

---

The Three Approximations

For small theta (in radians):

sin(theta) ≈ theta

cos(theta) ≈ 1 - theta²/2

tan(theta) ≈ theta

These become more accurate as theta approaches 0.

---

Understanding the Approximations

Function	Approximation	Why it works
sin(theta)	theta	The sine curve is almost linear near the origin
cos(theta)	1 - theta²/2	This is the start of the Taylor series for cosine
tan(theta)	theta	For small angles, tan ≈ sin ≈ theta

Numerical Evidence

theta (radians)	sin(theta)	theta	cos(theta)	1 - theta²/2	tan(theta)
0.01	0.009999833	0.01	0.99995	0.99995	0.01000003
0.1	0.09983	0.1	0.99500	0.99500	0.10033
0.2	0.19867	0.2	0.98007	0.98000	0.20271
0.5	0.47943	0.5	0.87758	0.87500	0.54630

The approximations are very good for theta < 0.1 and reasonable for theta < 0.3.

---

Using Small Angle Approximations

Worked Example 1

Use small angle approximations to estimate sin(0.05).

sin(0.05) ≈ 0.05

Calculator: sin(0.05) = 0.04998... Very close.

Worked Example 2

Use small angle approximations to estimate cos(0.1).

cos(0.1) ≈ 1 - (0.1)²/2 = 1 - 0.005 = 0.995

Calculator: cos(0.1) = 0.99500... Exact to 5 d.p.

Worked Example 3

Show that for small theta, (1 - cos(theta))/theta² ≈ 1/2.

Using cos(theta) ≈ 1 - theta²/2:

(1 - cos(theta))/theta² ≈ (1 - (1 - theta²/2))/theta²

= (theta²/2)/theta²

= 1/2

---

Simplifying Expressions

Worked Example 4

For small theta, simplify (sin(3theta))/(theta).

sin(3theta) ≈ 3theta (since 3theta is also small)

(sin(3theta))/(theta) ≈ 3theta/theta = 3

Worked Example 5

For small x, find an approximate expression for (tan(2x) - sin(x))/(x²).

tan(2x) ≈ 2x

sin(x) ≈ x

(2x - x)/x² = x/x² = 1/x

Hmm, this approaches infinity as x tends to 0, which suggests we need higher-order terms.

Let me redo this more carefully:

tan(2x) ≈ 2x + (2x)³/3 + ... ≈ 2x (to first order) sin(x) ≈ x - x³/6 + ... ≈ x (to first order)

To first order: (2x - x)/x² = x/x² = 1/x, which is not finite.

Actually, let me reconsider. The correct approach:

tan(2x) ≈ 2x (first-order small angle approximation) sin(x) ≈ x

(tan(2x) - sin(x))/x² ≈ (2x - x)/x² = x/x² = 1/x

This expression tends to infinity as x tends to 0, which is correct — the function is not bounded near 0.

Worked Example 6

For small theta, approximate (sin(theta) x tan(theta))/(1 - cos(theta)).

sin(theta) ≈ theta tan(theta) ≈ theta 1 - cos(theta) ≈ theta²/2

(theta x theta)/(theta²/2) = theta²/(theta²/2) = 2

---

Finding Approximate Solutions

Worked Example 7

Find an approximate solution to sin(x) + cos(x) = 1.1 for small positive x.

x + (1 - x²/2) = 1.1

x + 1 - x²/2 = 1.1

x - x²/2 = 0.1

x²/2 - x + 0.1 = 0

x² - 2x + 0.2 = 0

x = (2 ± sqrt(4 - 0.8))/2 = (2 ± sqrt(3.2))/2

x = (2 - 1.789)/2 = 0.106 (taking the smaller root for small x)

x ≈ 0.106 radians

Check: sin(0.106) + cos(0.106) ≈ 0.1058 + 0.9944 = 1.100. Confirmed.

---

Exam Tips

Tip	Detail
Radians only	These approximations only work in radians, not degrees
State the approximation	Always write "for small theta, sin(theta) ≈ theta" etc. before using
Higher order	For cos, the approximation 1 - theta²/2 includes the second-order term
Combine carefully	When simplifying fractions, use the right level of approximation
Verify	Check your approximate answer against the exact value when possible

---

Practice Problems

1. Use small angle approximations to estimate tan(0.08). [Answer: ≈ 0.08]
2. For small theta, simplify (1 - cos(2theta))/theta². [Answer: 2]
3. For small x, approximate (sin(5x))/(3x). [Answer: 5/3]
4. Find an approximate solution to 2sin(x) = x + 0.3 for small x. [Answer: x = 0.3]
5. For small theta, show that sin(theta) x tan(theta) ≈ theta². [Answer: theta x theta = theta²]

---

Summary

• For small theta (in radians): sin(theta) ≈ theta, cos(theta) ≈ 1 - theta²/2, tan(theta) ≈ theta.
• These come from the first terms of the Taylor series.
• Use them to simplify expressions, estimate values, and find approximate solutions.
• Always state the approximation before using it.
• These work well when theta is close to 0 (typically |theta| < 0.3 radians).

A-Level Deep Dive: Small Angle Approximations

Spec mapping

Edexcel 9MA0-02 specification section 5 — Trigonometry, sub-strand 5.9 covers the standard small angle approximations $\sin\theta \approx \theta$sinθ≈θ, $\cos\theta \approx 1 - \tfrac{\theta^2}{2}$cosθ≈1−2θ2​ and $\tan\theta \approx \theta$tanθ≈θ, where $\theta$θ is measured in radians and is small (refer to the official specification document for exact wording). Although examined as Year 1 Pure content, the material extends through Year 2 calculus (where the limit $\lim_{x \to 0}\sin x / x = 1$limx→0​sinx/x=1 underpins the derivative $\tfrac{d}{dx}\sin x = \cos x$dxd​sinx=cosx) and into the Further Mathematics Maclaurin/Taylor series treatment. Synoptic links to Pure include section 7 (Differentiation, where the small-angle limits provide the first-principles justification of the derivatives of sine and cosine), section 8 (Integration, where small-angle expansions yield approximate values of integrals such as $\int_0^{0.1}\sin x\,dx$∫00.1​sinxdx) and section 5 itself (radian measure, since the approximations are only valid in radians — substituting degrees gives nonsense). Beyond Pure, the result reappears in mechanics for the simple-pendulum derivation $T \approx 2\pi\sqrt{L/g}$T≈2πL/g​, which linearises $\sin\theta\approx\theta$sinθ≈θ in the equation of motion.

---

Worked example with full mark scheme

Question (8 marks):

(a) Use the small-angle approximations for $\sin\theta$sinθ and $\tan\theta$tanθ to show that, for small $x$x, $\dfrac{\sin 3x - \tan 3x}{x^3} \approx -\dfrac{27}{2}.$x3sin3x−tan3x​≈−227​. (5)

(b) Hence, or otherwise, find the value of $\lim_{x \to 0}\dfrac{\sin 3x - \tan 3x}{x^3}.$limx→0​x3sin3x−tan3x​. (3)

Solution with mark scheme:

(a) Step 1 — replace $\sin 3x$sin3x to sufficient order.

The small-angle approximation $\sin\theta \approx \theta - \tfrac{\theta^3}{6}$sinθ≈θ−6θ3​ (the Maclaurin expansion to order three) gives, 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 — using $\sin\theta \approx \theta$sinθ≈θ extended to the next non-zero term $-\theta^3/6$−θ3/6. The Edexcel specification quotes only $\sin\theta \approx \theta$sinθ≈θ, but to evaluate a limit of order $x^3$x3 the next term is required; this is where Year 2/Further Maths techniques bite. Common error: candidates use only $\sin 3x \approx 3x$sin3x≈3x, which leaves a numerator of $3x - 3x = 0$3x−3x=0 and produces the indeterminate form $0/x^3$0/x3 — they then have nothing to work with.

Step 2 — replace $\tan 3x$tan3x to sufficient order.

The expansion $\tan\theta \approx \theta + \tfrac{\theta^3}{3}$tanθ≈θ+3θ3​ (also Maclaurin to order three) gives $\tan 3x \approx 3x + \dfrac{(3x)^3}{3} = 3x + \dfrac{27x^3}{3} = 3x + 9x^3.$tan3x≈3x+3(3x)3​=3x+327x3​=3x+9x3.

M1 — extending $\tan\theta \approx \theta$tanθ≈θ to its third-order term $+\theta^3/3$+θ3/3. Note the plus sign, contrasting with the minus on $\sin\theta$sinθ: confusing the two is one of the most frequent errors at this level.

Step 3 — subtract.

$\sin 3x - \tan 3x \approx \left(3x - \dfrac{9x^3}{2}\right) - \left(3x + 9x^3\right) = -\dfrac{9x^3}{2} - 9x^3 = -\dfrac{9x^3 + 18x^3}{2} = -\dfrac{27x^3}{2}.$sin3x−tan3x≈(3x−29x3​)−(3x+9x3)=−29x3​−9x3=−29x3+18x3​=−227x3​.

A1 — correct simplified numerator $-\tfrac{27x^3}{2}$−227x3​. The leading $3x$3x terms cancel exactly; the surviving terms are both of order $x^3$x3, so the cancellation occurs at the right place to produce a finite limit.

Step 4 — divide.

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

M1 — dividing through by $x^3$x3 (valid since we are working with the leading-order expansion and $x \neq 0$x=0 in the limit process).

A1 — final stated result $-\tfrac{27}{2}$−227​, matching the printed form.

(b) Step 1 — interpret part (a) as a limit statement.

Because the higher-order remainders are $O(x^5)$O(x5) for $\sin 3x$sin3x and $O(x^5)$O(x5) for $\tan 3x$tan3x, the error in the approximation $-\tfrac{27}{2}$−227​ is itself of order $x^2$x2 (after dividing by $x^3$x3). As $x \to 0$x→0 this error vanishes.

M1 — recognising that the approximation becomes exact in the limit because the discarded terms are higher-order.

Step 2 — state the limit.

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

M1 — writing the limit explicitly using the approximation result.

A1 — exact value $-\tfrac{27}{2}$−227​, equivalent to $-13.5$−13.5 but presented as a fraction since "exact" is implied.

Total: 8 marks (M5 A3, split as shown).

---

Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): Given that $\theta$θ is small (measured in radians):

(a) Show, using small-angle approximations, that $\dfrac{\cos 2\theta - 1}{\theta\sin\theta} \approx -2.$θsinθcos2θ−1​≈−2. (4)

(b) Hence, or otherwise, evaluate $\lim_{\theta \to 0}\dfrac{\cos 2\theta - 1}{\theta\sin\theta}$limθ→0​θsinθcos2θ−1​. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — applying $\cos 2\theta \approx 1 - \tfrac{(2\theta)^2}{2} = 1 - 2\theta^2$cos2θ≈1−2(2θ)2​=1−2θ2, so $\cos 2\theta - 1 \approx -2\theta^2$cos2θ−1≈−2θ2.
• M1 (AO1.1a) — applying $\sin\theta \approx \theta$sinθ≈θ, so $\theta\sin\theta \approx \theta \cdot \theta = \theta^2$θsinθ≈θ⋅θ=θ2.
• M1 (AO1.1b) — forming the quotient: $\dfrac{-2\theta^2}{\theta^2}$θ2−2θ2​.
• A1 (AO2.5) — simplifying to $-2$−2 and stating the printed form.

(b)

• M1 (AO2.4) — recognising that the discarded higher-order terms are $O(\theta^4)$O(θ4) in the numerator and $O(\theta^4)$O(θ4) in the denominator, so they vanish under the limit.
• A1 (AO1.1b) — $\lim_{\theta \to 0}\dfrac{\cos 2\theta - 1}{\theta\sin\theta} = -2$limθ→0​θsinθcos2θ−1​=−2.

Total: 6 marks split AO1 = 4, AO2 = 2. This question is AO1-led (procedural application of the standard approximations) but the final A mark in (a) and the M1 in (b) reward AO2 reasoning — specifically, justifying that the approximation captures the exact limiting value rather than merely an "approximate" one. Edexcel small-angle questions repeatedly carry this hidden AO2 weight: the routine substitution earns most marks, but the limit-justification mark separates A* from A.

---

Synoptic links

Connects to:

1. Year 2 Pure / Further Maths — Maclaurin and Taylor series: the standard small-angle approximations are the first non-zero terms of the Maclaurin series $\sin\theta = \theta - \tfrac{\theta^3}{6} + \tfrac{\theta^5}{120} - \cdots$sinθ=θ−6θ3​+120θ5​−⋯ and $\cos\theta = 1 - \tfrac{\theta^2}{2} + \tfrac{\theta^4}{24} - \cdots$cosθ=1−2θ2​+24θ4​−⋯. Recognising the approximations as truncated series clarifies why the cosine approximation needs the $\theta^2/2$θ2/2 correction (the $\theta$θ coefficient is zero) while the sine approximation can drop straight to $\theta$θ.

2. Section 7 — limit definition of the derivative: the derivative $\tfrac{d}{dx}\sin x = \cos x$dxd​sinx=cosx is proved from first principles by writing $\tfrac{\sin(x+h) - \sin x}{h} = \sin x \cdot \tfrac{\cos h - 1}{h} + \cos x \cdot \tfrac{\sin h}{h}$hsin(x+h)−sinx​=sinx⋅hcosh−1​+cosx⋅hsinh​ and invoking $\lim_{h\to 0}\sin h / h = 1$limh→0​sinh/h=1 together with $\lim_{h\to 0}(\cos h - 1)/h = 0$limh→0​(cosh−1)/h=0 — both of which follow directly from the small-angle approximations.

3. Mechanics — pendulum and SHM: the equation of motion for a simple pendulum is $\ddot\theta = -(g/L)\sin\theta$θ¨=−(g/L)sinθ. This is not simple harmonic motion until the small-angle approximation $\sin\theta \approx \theta$sinθ≈θ linearises it to $\ddot\theta = -(g/L)\theta$θ¨=−(g/L)θ, giving the standard period $T \approx 2\pi\sqrt{L/g}$T≈2πL/g​. The whole "pendulum clock" calculation depends on this single approximation.

4. Numerical analysis / computer graphics: real-time graphics engines often replace $\sin\theta$sinθ and $\cos\theta$cosθ with their truncated Taylor expansions for small rotation increments because polynomial evaluation is much cheaper than calling a transcendental function. The error bound $O(\theta^3)$O(θ3) for sine and $O(\theta^4)$O(θ4) for cosine is what makes this engineering trade-off acceptable.

5. Optics — paraxial lens equation: Snell's law $n_1\sin\theta_1 = n_2\sin\theta_2$n1​sinθ1​=n2​sinθ2​ is intractable in general, but for paraxial (near-axis) rays the approximation $\sin\theta \approx \theta$sinθ≈θ linearises Snell to $n_1\theta_1 = n_2\theta_2$n1​θ1​=n2​θ2​, from which the entire thin-lens equation $\tfrac{1}{u} + \tfrac{1}{v} = \tfrac{1}{f}$u1​+v1​=f1​ is derived. School-level optics is, mathematically, the small-angle approximation.

---

Mark-scheme literacy

Small-angle questions on 9MA0-02 split AO marks more evenly than surd questions, because the topic genuinely tests reasoning as well as procedure:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Stating and substituting the standard approximations correctly; expanding to higher orders when required
AO2 (reasoning / interpretation)	30–40%	Justifying why a given approximation is sufficient; identifying that radians (not degrees) are required; interpreting an approximation as a limit
AO3 (problem-solving)	0–15%	Applying approximations in unfamiliar contexts (limits, integrals, mechanics) where the technique must be selected, not signalled

Examiner-rewarded phrasing: "for $\theta$θ in radians and small, applying the small-angle approximation $\sin\theta \approx \theta$sinθ≈θ"; "since the discarded terms are of order $\theta^3$θ3, the relative error is $O(\theta^2)$O(θ2)"; "in the limit $\theta \to 0$θ→0 the higher-order terms vanish, hence the approximation gives the exact value of the limit". Phrases that lose marks: "for small $\theta$θ, $\sin\theta = \theta$sinθ=θ" (using equals instead of approximately equals); "applying $\sin\theta \approx \theta$sinθ≈θ" without the radians qualifier on a question where degrees might be implied; or — most damagingly — using $\cos\theta \approx 1$cosθ≈1 when the question requires $\theta^2/2$θ2/2 accuracy.

A specific Edexcel pattern to watch: questions of the form "find $\lim_{x\to 0} f(x)/g(x)$limx→0​f(x)/g(x)" where both numerator and denominator vanish to leading order. The mark scheme typically requires expansion to one order beyond the obvious — the leading-order cancellation is the question, not the answer. Reading the limit-form clue and matching the order of expansion to the order of cancellation is the AO2 skill being assessed.

---

Grade-band model answers

3-mark question

Question: Use the small-angle approximation $\cos\theta \approx 1 - \tfrac{\theta^2}{2}$cosθ≈1−2θ2​ to estimate $\cos 0.1$cos0.1 to 4 decimal places, where $0.1$0.1 is in radians. Compare your answer with the calculator value.

Grade C response (~210 words):

Using $\cos\theta \approx 1 - \tfrac{\theta^2}{2}$cosθ≈1−2θ2​ with $\theta = 0.1$θ=0.1:

$\cos 0.1 \approx 1 - \tfrac{(0.1)^2}{2} = 1 - \tfrac{0.01}{2} = 1 - 0.005 = 0.9950$cos0.1≈1−2(0.1)2​=1−20.01​=1−0.005=0.9950.

The calculator gives $\cos 0.1 = 0.99500\ldots$cos0.1=0.99500…, so the approximation matches to 4 decimal places.

Examiner commentary: Full marks (3/3). The candidate substitutes correctly, simplifies cleanly, and reports the answer to the requested precision. The comparison with the calculator value confirms the result and demonstrates self-checking. The radian implication of the question is honoured implicitly (the candidate does not convert to degrees). This is the standard Grade C answer for a procedural application question — the maths is right, the working is shown, and the precision matches the question's demand. A common error at this level is to compute $\theta^2/2$θ2/2 as $0.05$0.05 instead of $0.005$0.005, dropping a power of ten; another is to write $0.995$0.995 instead of $0.9950$0.9950 and lose the accuracy mark for not matching 4 d.p. Neither slip occurs here.

Grade A response (~290 words):*

Apply the small-angle approximation $\cos\theta \approx 1 - \tfrac{\theta^2}{2}$cosθ≈1−2θ2​, valid for $\theta$θ in radians and small. With $\theta = 0.1$θ=0.1 rad: