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Small angle approximationsEdexcel A-Level Maths: Revision notes

Section 1

Why small angles are special

For a small angle θ\theta measured in radians, the arc, the opposite side and the tangent are almost the same length, so sin⁡θ\sin\theta, tan⁡θ\tan\theta and θ\theta are almost equal; cos⁡θ\cos\theta is close to 11 but slightly smaller. The approximations below are only valid in radians. For example sin⁡0.1=0.0998\sin0.1=0.0998 is close to 0.10.1, but sin⁡10\sin10 is not close to 1010.

Key termssmall angle
Common mistake

Using the approximations with a degree value such as sin⁡10∘≈10\sin10^\circ\approx10. Convert to radians first.

Section 2

The three standard approximations

For small θ\theta in radians: sin⁡θ≈θ,cos⁡θ≈1−θ22,tan⁡θ≈θ.\sin\theta\approx\theta,\qquad\cos\theta\approx1-\frac{\theta^2}{2},\qquad\tan\theta\approx\theta. These replace the trig function by a simple polynomial. They give 1−cos⁡θ≈θ221-\cos\theta\approx\frac{\theta^2}{2} directly. Multiples of the angle follow by substitution: sin⁡3θ≈3θ\sin3\theta\approx3\theta, tan⁡2θ≈2θ\tan2\theta\approx2\theta and cos⁡4x≈1−(4x)22=1−8x2\cos4x\approx1-\frac{(4x)^2}{2}=1-8x^2, so 1−cos⁡4x≈8x21-\cos4x\approx8x^2.

Key termsapproximation
Common mistake

Forgetting to square the coefficient: cos⁡3x≈1−9x22\cos3x\approx1-\frac{9x^2}{2}, not 1−3x221-\frac{3x^2}{2}.

Section 3

How accurate are they?

The approximations improve as θ\theta gets smaller. Example: cos⁡0.2≈1−0.042=0.98\cos0.2\approx1-\frac{0.04}{2}=0.98, while the calculator gives 0.98007…0.98007\ldots The percentage error is ∣approx−exact∣exact×100=0.0068%\frac{|\text{approx}-\text{exact}|}{\text{exact}}\times100=0.0068\%. At θ=0.5\theta=0.5, sin⁡θ≈θ\sin\theta\approx\theta is out by about 4.3%4.3\%, so the approximation is poorer. Compare an approximation with the calculator value by working in radian mode.

Key termspercentage error
Exam tip

The larger θ\theta is, the larger the error. Say that the approximation is only reliable for small θ\theta.

Section 4

Approximating expressions with several terms

Replace each trigonometric function and simplify, keeping terms up to the order the question needs. Example: 4cos⁡θ+3sin⁡θ≈4(1−θ22)+3θ=4+3θ−2θ24\cos\theta+3\sin\theta\approx4\left(1-\frac{\theta^2}{2}\right)+3\theta=4+3\theta-2\theta^2. Setting this equal to 4.24.2 gives 2θ2−3θ+0.2=02\theta^2-3\theta+0.2=0 with roots 0.06990.0699 and 1.431.43. Only the small root is valid, because 1.431.43 is not small and the approximation does not apply there; check: 4cos⁡0.0699+3sin⁡0.0699=4.19984\cos0.0699+3\sin0.0699=4.1998.

Exam tip

Reject roots that are not small: the approximation cannot be trusted there.

Section 5

Quotients and limits

For a fraction, approximate numerator and denominator separately and cancel powers of xx. Example (specification type): cos⁡3x−1xsin⁡4x≈−9x22x(4x)=−98\frac{\cos3x-1}{x\sin4x}\approx\frac{-\frac{9x^2}{2}}{x(4x)}=-\frac98. Similarly 1−cos⁡4xxtan⁡2x≈8x22x2=4\frac{1-\cos4x}{x\tan2x}\approx\frac{8x^2}{2x^2}=4 and cos⁡5x−1xsin⁡2x≈−254\frac{\cos5x-1}{x\sin2x}\approx-\frac{25}{4}; with x=0.01x=0.01 the calculator gives −6.249-6.249, close to −6.25-6.25.

Key termsnumeratordenominator
Common mistake

Using only cos⁡x≈1\cos x\approx1 for the numerator: then cos⁡3x−1≈0\cos3x-1\approx0 and the answer is lost. Keep the −θ22-\frac{\theta^2}{2} term.

Section 6

Using the approximations in context

In applied problems with small angles, such as a pendulum of length LL with h=L(1−cos⁡θ)h=L(1-\cos\theta) and d=Lsin⁡θd=L\sin\theta, the approximations give h≈Lθ22h\approx\frac{L\theta^2}{2} and d≈Lθd\approx L\theta. For L=1.5L=1.5 and θ=0.1\theta=0.1: h≈0.0075h\approx0.0075 m with exact value 0.00749370.0074937 m (percentage error 0.083%0.083\%), and d≈0.15d\approx0.15 m. State clearly that the model assumes θ\theta is small.

Exam tip

Quote units and say the approximation is valid because the angle is small.

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Carry on to the next subtopic.

Exam questions on Small angle approximations

  1. In this question θ\theta is small and measured in radians, so the standard small angle approximations may be used.
    Calculate the percentage error in the estimate cos⁡0.2≈0.98\cos0.2\approx0.98, using the calculator value of cos⁡0.2\cos0.2. Give your answer to 2 significant figures.2 marks
  2. For small xx in radians, consider the expression E=1−cos⁡4xxtan⁡2xE=\frac{1-\cos4x}{x\tan2x}.
    Hence find the approximate value of EE when xx is small.2 marks
  3. A pendulum of length 1.51.5 m swings through a small angle θ\theta radians from the vertical. The horizontal displacement of the bob from the vertical is d=1.5sin⁡θd=1.5\sin\theta m and its height above its lowest point is h=1.5(1−cos⁡θ)h=1.5(1-\cos\theta) m.
    Use a small angle approximation to show that, for small θ\theta, h≈0.75θ2h\approx0.75\theta^2.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).