Radians and small angles

Radians

A radian is a unit of angle measure. One radian is defined as the size of the central angle subtended by an arc of length $l$ equal to the radius $r$ of a circle.

From the figure, an arc of length equal to the circumference of a circle ($C = 2\pi r$) subtends a central angle of $360^\circ$. An arc of length $l$ subtends a central angle $\theta$ (in degrees). Therefore:

$\frac{\text{Length of arc AB}}{\text{Circumference of circle}} = \frac{\text{Measure of central angle}}{\text{Total measurement in a circle}}$

$\frac{l}{2\pi r} = \frac{\theta}{360^\circ}$

$l = \frac{\pi r \theta}{180^\circ}$

Since $s = \frac{l}{r}$ (where $s$ is the angle in radians), we have:

$s = \frac{\pi \theta}{180^\circ}$

Radians are dimensionless.

Converting between degrees and radians

To convert radians to degrees, multiply by $\frac{180^\circ}{\pi}$. To convert degrees to radians, multiply by $\frac{\pi}{180^\circ}$.

Example 1

Convert the following to radians (in multiples of $\pi$): (a) $60^\circ$ (b) $2970^\circ$ (c) $1^\circ$

Solution:

(a) $60^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{3}$ radians

(b) $2970^\circ \times \frac{\pi}{180^\circ} = \frac{33\pi}{2} = 16.5\pi$ radians

(c) $1^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{180}$ radians

Example 2

Convert the following to degrees: (a) $\frac{2\pi}{3}$ radians (b) $\frac{37\pi}{5}$ radians (c) 1 radian

Solution:

(a) $\frac{2\pi}{3} \times \frac{180^\circ}{\pi} = 120^\circ$

(b) $\frac{37\pi}{5} \times \frac{180^\circ}{\pi} = 1332^\circ$

(c) $1 \times \frac{180^\circ}{\pi} \approx 57.296^\circ$

Example 3

Find in radians an interior angle of a regular nonagon.

Solution:

The sum of the exterior angles of any polygon is $360^\circ$. A nonagon has 9 sides, so each exterior angle is $\frac{360^\circ}{9} = 40^\circ$.

Each interior angle is $180^\circ - 40^\circ = 140^\circ$.

Converting to radians: $140^\circ \times \frac{\pi}{180^\circ} = \frac{7\pi}{9}$ radians.

Example 4

Find the angle between the minute-hand and the hour-hand of a clock at 5:45 p.m. Give your answer (a) in degrees (b) in radians.

Solution:

(a) At 5:00, the hour hand is at 5 and the minute hand is at 12. At 5:45, the minute hand is at 9. The hour hand has moved $\frac{3}{4}$ of the way between 5 and 6.

Each number on the clock represents $\frac{360^\circ}{12} = 30^\circ$.

The minute hand is at $9 \times 30^\circ = 270^\circ$.

The hour hand is at $5 \times 30^\circ + \frac{3}{4} \times 30^\circ = 150^\circ + 22.5^\circ = 172.5^\circ$.

The angle between them is $270^\circ - 172.5^\circ = 97.5^\circ$.

(b) Converting to radians: $97.5^\circ \times \frac{\pi}{180^\circ} = \frac{13\pi}{24}$ radians.

Small angle approximations

When an angle $\theta$ is small and measured in radians, we can use the following approximations:

• $\sin\theta \approx \theta$
• $\cos\theta \approx 1 - \frac{\theta^2}{2}$
• $\tan\theta \approx \theta$

The diagram shows a sector of a circle AOB with radius $r$ and angle $\theta$ (in radians). The tangent at A meets OB extended at C. The following inequality holds:

Area of $\triangle AOB$ < Area of sector AOB < Area of $\triangle AOC$

$\frac{1}{2}r^2\sin\theta < \frac{1}{2}r^2\theta < \frac{1}{2}r^2\tan\theta$

Dividing by $\frac{1}{2}r^2$:

$\sin\theta < \theta < \tan\theta$

Dividing by $\sin\theta$ (assuming $\theta$ is positive):

$1 < \frac{\theta}{\sin\theta} < \frac{1}{\cos\theta}$

As $\theta \to 0$, $\cos\theta \to 1$, so $\frac{\theta}{\sin\theta} \to 1$, which means $\frac{\sin\theta}{\theta} \to 1$, and thus $\sin\theta \approx \theta$.

Using the identity $\cos\theta = 1 - 2\sin^2\frac{\theta}{2}$, and since $\sin\frac{\theta}{2} \approx \frac{\theta}{2}$ for small $\theta$:

$\cos\theta \approx 1 - 2\left(\frac{\theta}{2}\right)^2 = 1 - \frac{\theta^2}{2}$

Since $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and as $\theta \to 0$, $\sin\theta \approx \theta$ and $\cos\theta \approx 1$, then $\tan\theta \approx \theta$.

Example 1

Approximate the value of each of the following functions when $\theta$ is small: (a) $\frac{\sin 4\theta - \tan 2\theta}{3\theta}$ (b) $\frac{1 - \cos 2\theta}{\tan 2\theta \sin\theta}$ (c) $\frac{3\tan\theta - \theta}{\sin 2\theta}$

Solution:

(a) $\frac{\sin 4\theta - \tan 2\theta}{3\theta} \approx \frac{4\theta - 2\theta}{3\theta} = \frac{2\theta}{3\theta} = \frac{2}{3}$

(b) $\frac{1 - \cos 2\theta}{\tan 2\theta \sin\theta} \approx \frac{1 - (1 - \frac{(2\theta)^2}{2})}{2\theta \cdot \theta} = \frac{2\theta^2}{2\theta^2} = 1$

(c) $\frac{3\tan\theta - \theta}{\sin 2\theta} \approx \frac{3\theta - \theta}{2\theta} = \frac{2\theta}{2\theta} = 1$

Example 2

Find an approximation of $\frac{\cos(\alpha + \theta) - \cos\alpha}{\theta}$, when $\theta$ is small.

Solution:

Using the factor formula:

$\frac{\cos(\alpha + \theta) - \cos\alpha}{\theta} = \frac{-2\sin\left(\frac{\alpha + \theta + \alpha}{2}\right)\sin\left(\frac{\alpha + \theta - \alpha}{2}\right)}{\theta}$

$= \frac{-2\sin\left(\alpha + \frac{\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)}{\theta}$

When $\theta$ is small, $\sin\frac{\theta}{2} \approx \frac{\theta}{2}$:

$\approx \frac{-2\sin\alpha \cdot \frac{\theta}{2}}{\theta} = -\sin\alpha$

Example 3

Show that if $\theta$ is small, $\sin\left(\frac{\pi}{6} + \theta\right) \approx \frac{1}{2} + \frac{\sqrt{3}}{2}\theta$.

Solution:

$\sin\left(\frac{\pi}{6} + \theta\right) = \sin\frac{\pi}{6}\cos\theta + \cos\frac{\pi}{6}\sin\theta$

$= \frac{1}{2}\cos\theta + \frac{\sqrt{3}}{2}\sin\theta$

When $\theta$ is small, $\cos\theta \approx 1$ and $\sin\theta \approx \theta$:

$\approx \frac{1}{2}(1) + \frac{\sqrt{3}}{2}\theta = \frac{1}{2} + \frac{\sqrt{3}}{2}\theta$