A-level Mathematics

5.5 Small-angle approximations — Practise

Work through these on your own. Check that each angle is in radians and small before you replace a function with its approximation. Where a question asks for an estimate, compare it with a calculator value afterwards.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

  1. Use a small-angle approximation to estimate sin⁡0.1\sin 0.1.
  2. Use a small-angle approximation to estimate cos⁡0.2\cos 0.2.
  3. Use a small-angle approximation to estimate tan⁡0.05\tan 0.05.
  4. A pendulum of length 0.60.6 m swings through a small angle θ=0.12\theta = 0.12 radians. Estimate the horizontal displacement of the bob, in metres.
  5. The error in sin⁡θ≈θ\sin\theta \approx \theta is about θ36\dfrac{\theta^3}{6}. Estimate the error at θ=0.2\theta = 0.2, to three significant figures.
  6. Light meets a surface at a small angle i=0.04i = 0.04 radians, and sin⁡i=1.5sin⁡r\sin i = 1.5\sin r. Estimate the angle rr in radians, to three significant figures.
  7. Use a small-angle approximation to estimate cos⁡0.1\cos 0.1.
  8. Use a small-angle approximation to estimate sin⁡0.02\sin 0.02.
  9. Use small-angle approximations to estimate sin⁡0.1+tan⁡0.1\sin 0.1 + \tan 0.1.
  10. Work out the error in the approximation sin⁡θ≈θ\sin\theta \approx \theta at θ=0.5\theta = 0.5, to three significant figures.
Answers and concise solutions

1. The estimate

Answer: For a small angle in radians, sin⁡θ≈θ\sin\theta \approx \theta. So sin⁡0.1≈0.1\sin 0.1 \approx 0.1. The exact value is 0.099830.09983, so the estimate is within 0.00020.0002.

2. The estimate

Answer: cos⁡θ≈1−θ22\cos\theta \approx 1 - \dfrac{\theta^2}{2}. So cos⁡0.2≈1−0.042=0.98\cos 0.2 \approx 1 - \dfrac{0.04}{2} = 0.98. The exact value is 0.980070.98007, so the estimate is within 0.00010.0001.

3. The estimate

Answer: For a small angle in radians, tan⁡θ≈θ\tan\theta \approx \theta. So tan⁡0.05≈0.05\tan 0.05 \approx 0.05. The exact value is 0.050040.05004, so the estimate is within 0.00010.0001.

4. The displacement

Answer: The displacement is x=Lsin⁡θ≈Lθ=0.6×0.12=0.072x = L\sin\theta \approx L\theta = 0.6 \times 0.12 = 0.072 m, which is about 7.27.2 cm.

5. The error

Answer: The error is about 0.236=0.0086≈0.00133\dfrac{0.2^3}{6} = \dfrac{0.008}{6} \approx 0.00133. The exact difference is 0.2−sin⁡0.2=0.001330.2 - \sin 0.2 = 0.00133, so the estimate matches.

6. The angle rr

Answer: Replace both sines with their angles: i≈1.5ri \approx 1.5r. Then r≈0.041.5≈0.0267r \approx \dfrac{0.04}{1.5} \approx 0.0267 radians.

7. The estimate

Answer: cos⁡0.1≈1−0.012=0.995\cos 0.1 \approx 1 - \dfrac{0.01}{2} = 0.995. The exact value is 0.995000.99500, so the estimate is accurate to five decimal places.

8. The estimate

Answer: sin⁡0.02≈0.02\sin 0.02 \approx 0.02. The exact value is 0.0199990.019999, so the error is about 1.3×10−61.3 \times 10^{-6}.

9. The estimate

Answer: Both sin⁡θ\sin\theta and tan⁡θ\tan\theta are approximately θ\theta, so the sum is approximately 0.1+0.1=0.20.1 + 0.1 = 0.2. The exact value is 0.199990.19999.

10. The error at 0.50.5

Answer: The error is 0.5−sin⁡0.5=0.5−0.47943≈0.02060.5 - \sin 0.5 = 0.5 - 0.47943 \approx 0.0206. At 0.20.2 radians the error was only 0.001330.00133, so the error grows quickly with the angle.

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