5.5 Small-angle approximations — Practise with help

Before each estimate, check that the angle is in radians and that it is small. Then choose the right approximation for the function in the question.

Warm-up: remember

Change $0.1$ radians into degrees, to one decimal place.
Work out $0.2^2$ and $0.3^2$ .
Work out $1 - \dfrac{0.2^2}{2}$ .

Answers

$0.1 \times \dfrac{180}{\pi} \approx 5.7^\circ$ .
$0.2^2 = 0.04$ and $0.3^2 = 0.09$ .
$1 - \dfrac{0.04}{2} = 1 - 0.02 = 0.98$ .

If question 3 was hard, revise substituting into a formula before you continue.

Practise with help

1. Estimate $\sin 0.15$ using a small-angle approximation.

Hint

For small angles in radians, sine is almost the angle itself.

Next step

Write $\sin 0.15 \approx 0.15$ directly.

Full working

\sin 0.15 \approx 0.15.

A calculator gives $0.14944$ , so the estimate is good to about three decimal places.

2. Estimate $\cos 0.3$ using a small-angle approximation.

Hint

Use the quadratic approximation for cosine, not $1$ on its own.

Next step

Substitute $\theta = 0.3$ into $1 - \dfrac{\theta^2}{2}$ .

Full working

\cos 0.3 \approx 1 - \frac{0.3^2}{2} = 1 - \frac{0.09}{2} = 1 - 0.045 = 0.955.

A calculator gives $0.95534$ , so the estimate is accurate to three decimal places.

3. Estimate $\tan 0.08$ using a small-angle approximation.

Hint

Tangent behaves like sine near zero.

Next step

Write $\tan 0.08 \approx 0.08$ .

Full working

\tan 0.08 \approx 0.08.

A calculator gives $0.08017$ , so the error is about $0.00017$ .

4. A pendulum of length $0.6$ m swings through a small angle $\theta = 0.12$ radians. Estimate the horizontal displacement of the bob.

Hint

The displacement is $L\sin\theta$ . Replace $\sin\theta$ with $\theta$ .

Next step

Work out $0.6 \times 0.12$ .

Full working

x = L\sin\theta \approx L\theta = 0.6 \times 0.12 = 0.072 \text{ m}.

The displacement is about $7.2$ cm.

5. The error in the approximation $\sin\theta \approx \theta$ is about $\dfrac{\theta^3}{6}$ . Estimate the error at $\theta = 0.2$ and compare it with the table in the Learn page.

Hint

Substitute $\theta = 0.2$ into $\dfrac{\theta^3}{6}$ .

Next step

$\theta^3 = 0.2^3 = 0.008$ . Divide by $6$ .

Full working

\frac{0.2^3}{6} = \frac{0.008}{6} \approx 0.00133.

This matches the table entry for $\theta = 0.2$ , where the error in $\sin\theta$ is $0.00133$ .

6. A ray of light meets a flat surface at a small angle $i = 0.04$ radians. Snell's law gives $\sin i = 1.5\sin r$ , where $r$ is the angle of the refracted ray. Estimate $r$ using small-angle approximations.

Hint

Replace both sines with their angles: $i \approx 1.5r$ .

Next step

Divide both sides by $1.5$ : $r \approx \dfrac{0.04}{1.5}$ .

Full working

0.04 \approx 1.5r, \qquad r \approx \frac{0.04}{1.5} \approx 0.0267 \text{ radians}.

The refracted ray is about $0.027$ radians from the normal.