Radians and the Unit Circle

In short: A radian is the angle that cuts off an arc as long as the radius, and $π$ radians $= 18 0^{\circ}$ . In radians, arc length is $l = r θ$ and sector area is $A = \frac{1}{2} r^{2} θ$ , and the point at angle $θ$ on the unit circle is $(cos θ, sin θ)$ .

Starting point: Years 9 and 10

Circles

For a circle of radius $r$ :

$Circumference = 2 π r Area = π r^{2}$

Arcs and sectors in degrees

An arc is a fraction of the circumference. A sector is the same fraction of the area. For an angle of $θ^{\circ}$ at the centre:

$arc length = \frac{θ}{360} \times 2 π r sector area = \frac{θ}{360} \times π r^{2}$

One slice is a sector with a 45° angle at the centre. Its crust is the arc: 45/360 of the whole circumference.

Angles bigger than 90 degrees

On a set of axes, measure angles anticlockwise from the positive x-axis. The four quarters are called quadrants, numbered 1 to 4 anticlockwise. Angles in quadrant 2 are between $9 0^{\circ}$ and $18 0^{\circ}$ .

Angles start on the positive x-axis and turn anticlockwise. 150° is between 90° and 180°, so it ends in quadrant 2.

Angles and circles

Degree formulas

Arc: θ/360 × 2πr
Sector: θ/360 × πr²

Radians

π rad = 180°
Arc l = rθ
Sector A = ½r²θ

Unit circle

Radius 1
x = cos θ
y = sin θ

Signs

All, Sin, Tan, Cos
Related angle

Year 11Radians

One radian is the angle that makes an arc equal to the radius. A full turn is $2 π$ radians, so

$π radians = 18 0^{\circ}$

Bend one radius around the circle. The angle it makes at the centre is 1 radian, about 57.3°. About 6.28 radii (2π) fit around the whole circle.

To convert degrees to radians, multiply by $\frac{π}{180}$ . To convert back, multiply by $\frac{180}{π}$ .

Degrees	$3 0^{\circ}$	$4 5^{\circ}$	$6 0^{\circ}$	$9 0^{\circ}$	$18 0^{\circ}$	$36 0^{\circ}$
Radians	$\frac{π}{6}$	$\frac{π}{4}$	$\frac{π}{3}$	$\frac{π}{2}$	$π$	$2 π$

With $θ$ in radians, the formulas lose the 360:

$l = r θ A = \frac{1}{2} r^{2} θ$

Ruby's slice: $θ = \frac{π}{4}$ , so $l = 20 \times \frac{π}{4} = 5 π$ . Same answer, less work.

Segment area. The region between a chord and its arc is a sector minus a triangle:

$A = \frac{1}{2} r^{2} (θ - sin θ)$

The chord cuts the sector into a triangle and a segment. Segment = sector − triangle = ½r²θ − ½r² sin θ.

Perimeter of a sector is two radii plus the arc: $P = 2 r + r θ$ .

Working backwards. Rearrange the formulas: from $A = \frac{1}{2} r^{2} θ$ , a sector of area 12 and radius 4 has $θ = \frac{2 A}{r ^{2}} = 1.5$ radians, and arc length $l = 4 \times 1.5 = 6$ . If you know $l$ and $A$ , divide: $\frac{A}{l} = \frac{r}{2}$ , so $r = \frac{2 A}{l}$ .

Rolling a sector into a cone. The arc becomes the circumference of the cone's base. A sector of radius 9 and angle $\frac{2 π}{3}$ has arc $6 π$ , so the base radius is $\frac{6 π}{2 π} = 3$ (and the slant height is 9).

The arc (6π) wraps round to make the base circle, so 2π × base radius = 6π and the base radius is 3. The radius 9 becomes the slant height.

Straight-line distance across an arc. The chord joining the ends of an arc subtends angle $θ$ at the centre. Find it with the cosine rule: $c^{2} = 2 r^{2} - 2 r^{2} cos θ$ .

Area between two tangents and an arc. Tangents meet the radius at right angles. The area is the kite made by the two tangents and two radii, minus the sector. With radius 3 and the outside point 5 from the centre, each tangent is 4, the kite has area $2 \times \frac{1}{2} \times 3 \times 4 = 12$ , and the angle at the centre is $2 cos^{- 1} \frac{3}{5}$ .

Each radius meets its tangent at a right angle, so each half of the kite is a 3, 4, 5 triangle. Shaded area = kite − sector, with θ = 2cos⁻¹(3/5) at the centre.

Scaling a sector. Area is proportional to $r^{2} θ$ . If the radius grows by 20%, $r^{2}$ is multiplied by $1.44$ , so to keep the area the same the angle must be divided by $1.44$ , a drop of about 30.6%.

Year 11The unit circle

The unit circle has centre $(0, 0)$ and radius 1. Draw an angle $θ$ from the positive x-axis. The point where it meets the circle is

$P = (cos θ, sin θ)$

This defines sine and cosine for every angle, and $tan θ = \frac{sin θ}{cos θ}$ .

The point on the circle has coordinates (cos θ, sin θ). The letters show which ratios are positive in each quadrant.

ASTC ("All Stations To Central") tells you which ratios are positive:

Quadrant 1: All positive.
Quadrant 2: only Sin positive.
Quadrant 3: only Tan positive.
Quadrant 4: only Cos positive.

Which ratios are positive in each quadrant. On the unit circle cos θ = x, sin θ = y and tan θ = y/x, so the signs of x and y decide each one. Every other ratio in a quadrant is negative.

Finding any exact value. Find the related angle (the acute angle to the x-axis), use its exact value, then fix the sign with ASTC.

$cos 15 0^{\circ}$ : related angle $3 0^{\circ}$ , quadrant 2 where cos is negative, so $cos 15 0^{\circ} = - \frac{3}{2}$ .

150° ends in quadrant 2. The related angle to the x-axis is 30°, so the sizes come from 30°. Cos is the x value, which is negative here: cos 150° = −√3/2.

$sin \frac{4 π}{3}$ : related angle $\frac{π}{3}$ , quadrant 3 where sin is negative, so $- \frac{3}{2}$ .

Reciprocal ratios. $cosec θ = \frac{1}{sin θ}$ , $sec θ = \frac{1}{cos θ}$ and $cot θ = \frac{1}{tan θ}$ . They have the same sign as the ratio they come from. For example $tan \frac{4 π}{3} = 3$ , so $cot \frac{4 π}{3} = \frac{1}{3}$ .

Quick summary

Arc length $= \frac{θ}{360} \times 2 π r$ and sector area $= \frac{θ}{360} \times π r^{2}$ in degrees.
$π$ radians $= 18 0^{\circ}$ . In radians: $l = r θ$ and $A = \frac{1}{2} r^{2} θ$ .
On the unit circle, the point at angle $θ$ is $(cos θ, sin θ)$ .
ASTC gives the sign in each quadrant. Use the related angle for the size.
Calculus always uses radians.

Quiz

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What is the circumference of a circle with radius 5 cm?
1. A $5 π$ cm
2. B $10 π$ cm
3. C $25 π$ cm
4. D $2.5 π$ cm
Show answer
B: $10 π$ cm
$2 π r = 10 π$ .
Find the arc length for an angle of $9 0^{\circ}$ in a circle of radius 8.
1. A $8 π$
2. B $16 π$
3. C $4 π$
4. D $2 π$
Show answer
C: $4 π$
$\frac{90}{360} \times 16 π = 4 π$ .
Find the area of a sector with angle $6 0^{\circ}$ and radius 6.
1. A $6 π$
2. B $12 π$
3. C $36 π$
4. D $2 π$
Show answer
A: $6 π$
$\frac{60}{360} \times 36 π = 6 π$ .