Year 2 Pure · Chapter 8

Radians

Specification: Edexcel Pure 5.1–5.3 — Trigonometry · AQA section E1, E2, E3

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In this chapter

1. Measuring angles in radians
2. Exact values, graphs and equations in radians
3. Arcs, sectors and segments
4. Small-angle approximations
Practice

Key facts — this chapter

One radian is the angle at the centre of a circle when the arc is as long as the radius. $\pi$ radians $= 180^\circ$ .
To convert, multiply degrees by $\frac{\pi}{180}$ , or radians by $\frac{180}{\pi}$ .
In radians, $\sin$ and $\cos$ have period $2\pi$ , and $\tan$ has period $\pi$ . Set your calculator to radians.
For a sector of radius $r$ and angle $\theta$ radians: arc length $s = r\theta$ and area $A = \frac{1}{2}r^2\theta$ .
A segment has area $\frac{1}{2}r^2(\theta - \sin\theta)$ : the sector minus the triangle.
For small $\theta$ in radians: $\sin\theta \approx \theta$ , $\tan\theta \approx \theta$ and $\cos\theta \approx 1 - \frac{1}{2}\theta^2$ .

There is nothing special about $360$ degrees in a full turn: it’s a choice made thousands of years ago. Radians measure an angle by the circle itself, so the formulas for arcs and sectors lose their stray $\frac{\theta}{360}$ factors. From this chapter on, angles are in radians unless a question gives them in degrees.

1. Measuring angles in radians

Key result

An angle of one radian is the angle at the centre of a circle when the arc it cuts off is exactly one radius long.

A full circle has circumference $2\pi r$ , which is $2\pi$ radii, so a full turn is $2\pi$ radians and half a turn is $\pi$ radians:

\begin{aligned} 2\pi \text{ radians} &= 360^\circ \\ \pi \text{ radians} &= 180^\circ \end{aligned}

One radian is $\frac{180^\circ}{\pi} \approx 57.3^\circ$ .

A circle with centre O and radius r, with an arc also of length r marked on the circumference: the angle it makes at O is one radian, just under 60 degrees — When the arc is as long as the radius, the angle is one radian.

To change degrees into radians, multiply by $\frac{\pi}{180}$ . To change radians into degrees, multiply by $\frac{180}{\pi}$ . The common angles are simple fractions of $\pi$ , so learn them both ways.

Angles to know

\begin{array}{c|cccc} \text{degrees} & 30^\circ & 45^\circ & 60^\circ & 90^\circ \\ \hline \text{radians} & \tfrac{\pi}{6} & \tfrac{\pi}{4} & \tfrac{\pi}{3} & \tfrac{\pi}{2} \end{array}

The rest follow from these and $180^\circ = \pi$ : $120^\circ = \tfrac{2\pi}{3}$ , $135^\circ = \tfrac{3\pi}{4}$ , $150^\circ = \tfrac{5\pi}{6}$ , $270^\circ = \tfrac{3\pi}{2}$ and $360^\circ = 2\pi$ .

Worked example 8.1 · Converting

Convert $150^\circ$ to radians, and $\tfrac{7\pi}{4}$ and $2.5$ radians to degrees.

\begin{aligned} 150^\circ &= 150 \times \tfrac{\pi}{180} = \tfrac{5\pi}{6} \\ \tfrac{7\pi}{4} &= \tfrac{7\pi}{4} \times \tfrac{180}{\pi} = 315^\circ \\ 2.5 &= 2.5 \times \tfrac{180}{\pi} = 143.2^\circ \text{ (1 d.p.)} \end{aligned}

An angle in radians is often written with no unit at all: an angle of $2.5$ means $2.5$ radians. Some books write $2.5^c$ or $2.5$ rad.

Common mistake

Check your calculator’s angle mode before every trig calculation. In radian mode, $\sin 30 = -0.988$ , because the calculator reads $30$ as $30$ radians, which is nearly five full turns.

2. Exact values, graphs and equations in radians

Everything from Year 1 carries over; only the labels change. The exact values are the same numbers at the same angles:

Exact values in radians

\begin{array}{c|ccccc} \theta & 0 & \tfrac{\pi}{6} & \tfrac{\pi}{4} & \tfrac{\pi}{3} & \tfrac{\pi}{2} \\ \hline \sin\theta & 0 & \tfrac12 & \tfrac{\sqrt2}{2} & \tfrac{\sqrt3}{2} & 1 \\[2pt] \cos\theta & 1 & \tfrac{\sqrt3}{2} & \tfrac{\sqrt2}{2} & \tfrac12 & 0 \\[2pt] \tan\theta & 0 & \tfrac{1}{\sqrt3} & 1 & \sqrt3 & \text{—} \end{array}

Graphs of y = sin x and y = cos x from 0 to 2 pi, with the x-axis marked at pi over 2, pi, 3 pi over 2 and 2 pi — sin and cos in radians: one full wave every 2π.

The graphs have the same shapes, with the axis now marked in multiples of $\pi$ . $\sin x$ and $\cos x$ repeat every $2\pi$ . $\tan x$ repeats every $\pi$ , with asymptotes at $x = \frac{\pi}{2}, \frac{3\pi}{2}, \dots$ The symmetries you used in degrees become

\begin{aligned} \sin(\pi - x) &= \sin x \\ \cos(2\pi - x) &= \cos x \\ \tan(\pi + x) &= \tan x \end{aligned}

Worked example 8.2 · Exact values past a right angle

Find the exact values of $\sin\tfrac{4\pi}{3}$ and $\cos\tfrac{7\pi}{4}$ .

$\tfrac{4\pi}{3} = \pi + \tfrac{\pi}{3}$ is in the third quadrant, where sine is negative:
$\sin\tfrac{4\pi}{3} = -\sin\tfrac{\pi}{3} = -\tfrac{\sqrt3}{2}$
$\tfrac{7\pi}{4} = 2\pi - \tfrac{\pi}{4}$ is in the fourth quadrant, where cosine is positive:
$\cos\tfrac{7\pi}{4} = \cos\tfrac{\pi}{4} = \tfrac{\sqrt2}{2}$

Worked example 8.3 · A multiple angle in radians

Solve $\sin 2x = \tfrac{\sqrt3}{2}$ for $0 \le x \le 2\pi$ .

As $x$ runs from $0$ to $2\pi$ , $2x$ runs from $0$ to $4\pi$ . Solve in this wider interval.
The principal value is $2x = \tfrac{\pi}{3}$ , and sine is also positive at $\pi - \tfrac{\pi}{3} = \tfrac{2\pi}{3}$ . Adding $2\pi$ to each gives the rest:
$2x = \tfrac{\pi}{3},\ \tfrac{2\pi}{3},\ \tfrac{7\pi}{3},\ \tfrac{8\pi}{3}$
Halve each one:
$x = \tfrac{\pi}{6},\ \tfrac{\pi}{3},\ \tfrac{7\pi}{6},\ \tfrac{4\pi}{3}$

This is Year 1’s Example 9.3 in radians: the answers are $30^\circ$ , $60^\circ$ , $210^\circ$ and $240^\circ$ again.

Worked example 8.4 · An answer that isn't exact

Solve $3\tan x = 5$ for $0 \le x < 2\pi$ , giving your answers to 3 significant figures.

$\tan x = \tfrac{5}{3}$ , so the principal value is $x = \tan^{-1}\tfrac{5}{3} = 1.0303\ldots$
$\tan$ repeats every $\pi$ , so the other solution in range is $1.0303\ldots + \pi = 4.1719\ldots$ So $x = 1.03$ or $x = 4.17$ (3 s.f.).

3. Arcs, sectors and segments

A sector with angle $\theta$ is the fraction $\frac{\theta}{2\pi}$ of the whole circle. So its arc is that fraction of the circumference, and its area is that fraction of the circle’s area.

Key result

For a sector of radius $r$ and angle $\theta$ in radians,

\begin{aligned} s &= \frac{\theta}{2\pi} \times 2\pi r = r\theta \\ A &= \frac{\theta}{2\pi} \times \pi r^2 = \frac{1}{2}r^2\theta \end{aligned}

Two copies of a circle with centre O and radius r. On the left a sector with angle theta is shaded, with its arc of length r theta marked. On the right only the segment between the chord and the arc is shaded — A sector (left), and the segment cut off by its chord (right).

Worked example 8.5 · A sector

A sector has radius $6$ cm and angle $1.2$ radians. Find its arc length, area and perimeter.

\begin{aligned} s &= 6 \times 1.2 = 7.2 \text{ cm} \\ A &= \tfrac{1}{2} \times 6^2 \times 1.2 = 21.6 \text{ cm}^2 \end{aligned}

The perimeter is the arc plus two radii: $7.2 + 6 + 6 = 19.2$ cm.

Worked example 8.6 · Working backwards

A sector has area $15$ cm² and perimeter $16$ cm. Find the possible values of its radius $r$ and angle $\theta$ .

The two facts give two equations:
$\tfrac{1}{2}r^2\theta = 15 \quad\text{and}\quad 2r + r\theta = 16$
From the second, $r\theta = 16 - 2r$ . Substitute into the first, written as $\tfrac{1}{2}r(r\theta) = 15$ :
$\begin{aligned} \tfrac{1}{2}r(16 - 2r) &= 15 \\ r^2 - 8r + 15 &= 0 \\ (r - 3)(r - 5) &= 0 \end{aligned}$
If $r = 3$ , then $3\theta = 10$ and $\theta = \tfrac{10}{3}$ . If $r = 5$ , then $5\theta = 6$ and $\theta = 1.2$ . Both angles are less than $2\pi$ , so both sectors exist.

Change the angle and the radius and watch the arc and the area. In radians the angle is simply how many radii long the arc is. Switch to small angles to see sin θ, θ and tan θ close up.

A chord cuts a circle into two segments. The minor segment is what’s left of the sector when you take away the triangle made by the two radii and the chord. The triangle has area $\frac{1}{2}ab\sin C = \frac{1}{2}r^2\sin\theta$ .

Key result

The segment cut off by a chord that makes angle $\theta$ at the centre has area

\frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta = \frac{1}{2}r^2(\theta - \sin\theta)

Worked example 8.7 · A segment

A chord of a circle of radius $8$ cm makes an angle of $\tfrac{\pi}{3}$ at the centre. Find the exact area of the minor segment, and give it to 3 significant figures.

\begin{aligned} A &= \tfrac{1}{2} \times 8^2\left(\tfrac{\pi}{3} - \sin\tfrac{\pi}{3}\right) \\ &= 32\left(\tfrac{\pi}{3} - \tfrac{\sqrt3}{2}\right) \\ &= \tfrac{32\pi}{3} - 16\sqrt3 \\ &= 5.80 \text{ cm}^2 \text{ (3 s.f.)} \end{aligned}

Common mistake

The formulas $s = r\theta$ and $A = \frac{1}{2}r^2\theta$ only work with $\theta$ in radians. If a question gives the angle in degrees, convert it first.

4. Small-angle approximations

When $\theta$ is small and in radians, $\sin\theta$ , $\tan\theta$ and $\theta$ itself are almost equal, and $\cos\theta$ is almost $1$ .

Key result

For small $\theta$ , measured in radians,

\begin{aligned} \sin\theta &\approx \theta \\ \tan\theta &\approx \theta \\ \cos\theta &\approx 1 - \tfrac{1}{2}\theta^2 \end{aligned}

A narrow sector of a circle of radius 1 with a small angle theta. Three nearly equal vertical lengths are marked: the height sin theta inside the sector, the arc theta, and the tangent length tan theta just outside — On a circle of radius 1: sin θ < θ < tan θ. As θ shrinks, the three lengths close up.

Where the approximations come from

On a circle of radius $1$ , the arc has length $\theta$ . The height of its end above the first radius is $\sin\theta$ , which is a little shorter. The tangent at the start, cut off by the second radius, has length $\tan\theta$ , which is a little longer. So $\sin\theta < \theta < \tan\theta$ , and as $\theta$ shrinks the three lengths close up.

For cosine, use $\cos^2\theta = 1 - \sin^2\theta \approx 1 - \theta^2$ . Then the binomial series from Chapter 7 gives $\cos\theta \approx (1 - \theta^2)^{\frac{1}{2}} \approx 1 - \frac{1}{2}\theta^2$ .

Worked example 8.8 · Simplifying an expression

Show that, when $\theta$ is small, $\dfrac{\sin 4\theta + \cos 2\theta - 1}{3\tan\theta} \approx \dfrac{4}{3} - \dfrac{2}{3}\theta$ .

Replace each function, keeping the whole angle together: $\sin 4\theta \approx 4\theta$ , $\tan\theta \approx \theta$ and $\cos 2\theta \approx 1 - \tfrac{1}{2}(2\theta)^2 = 1 - 2\theta^2$ .
Substitute and simplify:
$\begin{aligned} &\frac{\sin 4\theta + \cos 2\theta - 1}{3\tan\theta} \\ &\approx \frac{4\theta + (1 - 2\theta^2) - 1}{3\theta} \\ &= \frac{4\theta - 2\theta^2}{3\theta} = \frac{4}{3} - \frac{2}{3}\theta \end{aligned}$

Common mistake

Square the whole angle in the cosine approximation: $\cos 2\theta \approx 1 - \frac{1}{2}(2\theta)^2 = 1 - 2\theta^2$ , not $1 - \theta^2$ .

Worked example 8.9 · How good are they?

Compare each approximation with the true value when $\theta = 0.1$ , and find the percentage errors.

	Value	Estimate	Error
$\sin 0.1$	$0.099833$	$0.1$	$0.17\%$
$\tan 0.1$	$0.100335$	$0.1$	$0.33\%$
$\cos 0.1$	$0.995004$	$0.995$	$0.0004\%$

Each error is the difference as a percentage of the true value, for example $\frac{0.1 - 0.099833}{0.099833} \times 100 = 0.17\%$ . All three are good, and the cosine one is far better, because its first missing term is a $\theta^4$ term.

Practice

Twelve questions in three tiers. Try each one properly before opening the solution — the solutions show full method, not just answers.

Warm-up

1 [3 marks]

Convert $240^\circ$ to radians, as an exact multiple of $\pi$ . [1]
Convert $\tfrac{7\pi}{12}$ radians to degrees. [1]
Convert $0.8$ radians to degrees, to 1 decimal place. [1]

Show solution

(a) $240 \times \tfrac{\pi}{180} = \tfrac{4\pi}{3}$ .

(b) $\tfrac{7\pi}{12} \times \tfrac{180}{\pi} = 105^\circ$ .

(c) $0.8 \times \tfrac{180}{\pi} = 45.8^\circ$ (1 d.p.).

2 [4 marks]

Find the exact values of $\sin\tfrac{5\pi}{4}$ , $\cos\tfrac{2\pi}{3}$ and $\tan\tfrac{5\pi}{3}$ . [3]
State the period of $y = \cos 3x$ , in radians. [1]

Show solution

(a) Each is the related acute angle’s value, with the sign of its quadrant:

\begin{aligned} \sin\tfrac{5\pi}{4} &= -\sin\tfrac{\pi}{4} = -\tfrac{\sqrt2}{2} \\ \cos\tfrac{2\pi}{3} &= -\cos\tfrac{\pi}{3} = -\tfrac{1}{2} \\ \tan\tfrac{5\pi}{3} &= -\tan\tfrac{\pi}{3} = -\sqrt3 \end{aligned}

(b) $\cos x$ has period $2\pi$ , so $\cos 3x$ has period $\tfrac{2\pi}{3}$ .

3 [3 marks]

A sector has radius $5$ cm and angle $0.9$ radians. Find its arc length, its area and its perimeter.

Show solution

\begin{aligned} s &= 5 \times 0.9 = 4.5 \text{ cm} \\ A &= \tfrac{1}{2} \times 5^2 \times 0.9 = 11.25 \text{ cm}^2 \end{aligned}

The perimeter is $4.5 + 5 + 5 = 14.5$ cm.

4 [3 marks]

Find the approximate value of $\dfrac{\theta\sin\theta}{1 - \cos\theta}$ when $\theta$ is small.

Show solution

$\sin\theta \approx \theta$ and $1 - \cos\theta \approx \tfrac{1}{2}\theta^2$ , so

\frac{\theta\sin\theta}{1 - \cos\theta} \approx \frac{\theta^2}{\frac{1}{2}\theta^2} = 2

Standard

5 [3 marks]

Solve $\cos x = -0.4$ for $0 \le x \le 2\pi$ , giving your answers to 3 significant figures.

Show solution

The principal value is $x = \cos^{-1}(-0.4) = 1.9823\ldots$ The other solution comes from the symmetry $\cos(2\pi - x) = \cos x$ : $2\pi - 1.9823\ldots = 4.3008\ldots$

So $x = 1.98$ or $x = 4.30$ (3 s.f.).

6 [4 marks]

Solve $2\sin^2 x = \sin x$ for $0 \le x \le 2\pi$ , giving exact answers.

Show solution

Don’t divide by $\sin x$ : that would lose the solutions where $\sin x = 0$ . Factorise instead:

\sin x\,(2\sin x - 1) = 0

$\sin x = 0$ gives $x = 0,\ \pi,\ 2\pi$ .
$\sin x = \tfrac{1}{2}$ gives $x = \tfrac{\pi}{6},\ \tfrac{5\pi}{6}$ .

So $x = 0,\ \tfrac{\pi}{6},\ \tfrac{5\pi}{6},\ \pi,\ 2\pi$ .

7 [3 marks]

A sector of a circle of radius $8$ cm has area $24$ cm². Find the angle of the sector and its perimeter.

Show solution

$\tfrac{1}{2} \times 8^2 \times \theta = 24$ , so $32\theta = 24$ and $\theta = 0.75$ radians.

The arc is $8 \times 0.75 = 6$ cm, so the perimeter is $6 + 8 + 8 = 22$ cm.

8 [5 marks]

A sector of a circle has radius $r$ cm and perimeter $20$ cm.

Show that the area of the sector is $A = 10r - r^2$ . [3]
Find the largest possible area, and the angle of the sector when the area is largest. [2]

Show solution

(a) The perimeter is $2r + r\theta = 20$ , so $r\theta = 20 - 2r$ . Then

\begin{aligned} A = \tfrac{1}{2}r^2\theta &= \tfrac{1}{2}r(r\theta) \\ &= \tfrac{1}{2}r(20 - 2r) \\ &= 10r - r^2 \end{aligned}

(b) Complete the square: $A = 25 - (r - 5)^2$ . So the largest area is $25$ cm², when $r = 5$ . Then $5\theta = 20 - 10$ , so $\theta = 2$ radians.

9 [5 marks]

Show that, when $\theta$ is small, $\dfrac{\cos 3\theta - 1}{\theta\sin 2\theta} \approx -\dfrac{9}{4}$ . [3]
Find the value of the expression when $\theta = 0.05$ , to 4 decimal places, and the percentage error in the approximation. [2]

Show solution

(a) $\cos 3\theta \approx 1 - \tfrac{1}{2}(3\theta)^2 = 1 - \tfrac{9}{2}\theta^2$ and $\sin 2\theta \approx 2\theta$ , so

\frac{\cos 3\theta - 1}{\theta\sin 2\theta} \approx \frac{-\frac{9}{2}\theta^2}{2\theta^2} = -\frac{9}{4}

(b) In radian mode, the expression is $-2.2495$ (4 d.p.) when $\theta = 0.05$ . The percentage error is

\frac{2.25 - 2.2495\ldots}{2.2495\ldots} \times 100 = 0.02\%

to 1 significant figure.

Exam style

10 [7 marks]

From a point on level ground $400$ m from the foot of a tall building, the angle of elevation of the top of the building is $0.05$ radians.

Use a small-angle approximation to estimate the height of the building. [2]
Find the height to 5 significant figures, and the percentage error in your estimate. [3]
A second building, also $400$ m away, has an angle of elevation of $0.5$ radians. Explain why the same method would be much less reliable for it. [2]

Show solution

(a) The height is $h = 400\tan 0.05$ . For small angles $\tan\theta \approx \theta$ , so $h \approx 400 \times 0.05 = 20$ m.

(b) $h = 400\tan 0.05 = 20.017$ m (5 s.f.). The percentage error is

\frac{20.017 - 20}{20.017} \times 100 = 0.08\%

to 1 significant figure.

(c) The approximation $\tan\theta \approx \theta$ only works when $\theta$ is small. At $0.5$ radians, $\tan 0.5 = 0.546$ , so the estimate $400 \times 0.5 = 200$ m is about $8\%$ below the true height of $219$ m.

11 [5 marks]

Solve $2\cos^2\theta + 3\sin\theta = 3$ for $0 \le \theta < 2\pi$ , giving exact answers.

Show solution

Use $\cos^2\theta = 1 - \sin^2\theta$ to get an equation in $\sin\theta$ only:

\begin{aligned} 2(1 - \sin^2\theta) + 3\sin\theta &= 3 \\ 2\sin^2\theta - 3\sin\theta + 1 &= 0 \\ (2\sin\theta - 1)(\sin\theta - 1) &= 0 \end{aligned}

$\sin\theta = \tfrac{1}{2}$ gives $\tfrac{\pi}{6}$ and $\tfrac{5\pi}{6}$ , and $\sin\theta = 1$ gives $\tfrac{\pi}{2}$ . So $\theta = \tfrac{\pi}{6},\ \tfrac{\pi}{2},\ \tfrac{5\pi}{6}$ .

12 [9 marks]

A sector $OAB$ of a circle with centre $O$ has area $12$ cm² and arc length $6$ cm.

Find the radius of the circle and the angle $AOB$ . [3]
Find the area of the segment between the chord $AB$ and the arc, to 3 significant figures. [3]
Find the perimeter of the segment, to 3 significant figures. [3]

Show solution

(a) $r\theta = 6$ and $\tfrac{1}{2}r^2\theta = 12$ . Writing the area as $\tfrac{1}{2}r(r\theta) = \tfrac{1}{2}r \times 6 = 3r$ gives $3r = 12$ , so $r = 4$ cm and $\theta = \tfrac{6}{4} = 1.5$ radians.

(b)

\begin{aligned} A &= \tfrac{1}{2} \times 4^2 \times (1.5 - \sin 1.5) \\ &= 8(1.5 - 0.9974\ldots) \\ &= 4.02 \text{ cm}^2 \text{ (3 s.f.)} \end{aligned}

(c) Find the chord with the cosine rule:

\begin{aligned} AB^2 &= 4^2 + 4^2 - 2 \times 4 \times 4\cos 1.5 \\ &= 29.736\ldots \\ AB &= 5.4531\ldots \end{aligned}

The perimeter is the arc plus the chord: $6 + 5.4531\ldots = 11.5$ cm (3 s.f.).