# Radians and the Unit Circle

> Arcs and sectors, measuring angles in radians, and trig ratios for any angle using the unit circle.

From HSC Maths Advanced on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-advanced/radians-and-unit-circle/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The pizza slice
Ruby's family orders a large pizza, 40 cm across, cut into 8 equal slices. Ruby wants to know how much crust she gets on one slice and how big her slice is.

Each slice is a **sector** of a circle, and the crust is an **arc**. To work these out, you need the angle of the slice. Mathematicians found that measuring angles in a new unit, the **radian**, makes these formulas beautifully simple. Radians are also the unit calculus needs.

**In short:** A radian is the angle that cuts off an arc as long as the radius, and $\pi$ radians $= 180^\circ$. In radians, arc length is $l = r\theta$ and sector area is $A = \dfrac{1}{2}r^2\theta$, and the point at angle $\theta$ on the unit circle is $(\cos\theta, \sin\theta)$.

## Starting point: Years 9 and 10

### Circles

For a circle of radius $r$:

$$\text{Circumference} = 2\pi r \qquad \text{Area} = \pi r^2$$

### 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 $\theta^\circ$ at the centre:

$$\text{arc length} = \frac{\theta}{360} \times 2\pi r \qquad \text{sector area} = \frac{\theta}{360} \times \pi r^2$$

*Diagram: A pizza cut into 8 slices. One slice is a sector with angle 45 degrees and radius 20 centimetres, and its crust is an arc*
*One slice is a sector with a 45° angle at the centre. Its crust is the arc: 45/360 of the whole circumference.*

### Worked example: Ruby's slice
Radius 20 cm, angle $\dfrac{360^\circ}{8} = 45^\circ$.

Crust: $\dfrac{45}{360} \times 2\pi(20) = 5\pi \approx 15.7$ cm.

Slice area: $\dfrac{45}{360} \times \pi(20)^2 = 50\pi \approx 157$ cm².

### Calculator tip: Radian mode and the π key
Switch to radian mode (`R` on the screen) when angles are in radians, and back to degrees afterwards. Use the `π` key, not 3.14, so the answer is not rounded too early.

To convert $135^\circ$ to radians, type `135 × π ÷ 180` and press `S⇔D` to see $\frac{3\pi}{4}$.

### 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 $90^\circ$ and $180^\circ$.

*Diagram: Axes split into quadrants 1 to 4, numbered anticlockwise, with an angle of 150 degrees measured anticlockwise from the positive x-axis into quadrant 2*
*Angles start on the positive x-axis and turn anticlockwise. 150° is between 90° and 180°, so it ends in quadrant 2.*

**Mind map**

```text
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 11: Radians
One **radian** is the angle that makes an arc equal to the radius. A full turn is $2\pi$ radians, so

$$\pi \text{ radians} = 180^\circ$$

*Diagram: A circle of radius r. An arc as long as the radius makes an angle of 1 radian at the centre, about 57.3 degrees*
*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 $\dfrac{\pi}{180}$. To convert back, multiply by $\dfrac{180}{\pi}$.

| Degrees | $30^\circ$ | $45^\circ$ | $60^\circ$ | $90^\circ$ | $180^\circ$ | $360^\circ$ |
|---|---|---|---|---|---|---|
| Radians | $\dfrac{\pi}{6}$ | $\dfrac{\pi}{4}$ | $\dfrac{\pi}{3}$ | $\dfrac{\pi}{2}$ | $\pi$ | $2\pi$ |

With $\theta$ in radians, the formulas lose the 360:

$$l = r\theta \qquad A = \frac{1}{2}r^2\theta$$

Ruby's slice: $\theta = \dfrac{\pi}{4}$, so $l = 20 \times \dfrac{\pi}{4} = 5\pi$. Same answer, less work.

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

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

*Diagram: A sector with angle theta split by a chord into a triangle and a segment. The segment is the sector minus the triangle*
*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: $P = 2r + r\theta$.

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

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

*Diagram: A sector of radius 9 and angle 2 pi over 3 rolled into a cone. The arc of 6 pi becomes the base circle of radius 3, and the radius 9 becomes the slant height*
*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 $\theta$ at the centre. Find it with the cosine rule: $c^2 = 2r^2 - 2r^2\cos\theta$.

**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 $2\cos^{-1}\frac{3}{5}$.

*Diagram: Two tangents from a point 5 units from the centre of a circle of radius 3. The tangents are 4 long. The shaded area between the tangents and the arc is the kite minus the sector*
*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 $r^2\theta$. 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 11: The unit circle
The **unit circle** has centre $(0, 0)$ and radius 1. Draw an angle $\theta$ from the positive x-axis. The point where it meets the circle is

$$P = (\cos\theta, \sin\theta)$$

This defines sine and cosine for **every** angle, and $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$.

*Diagram: Unit circle with point P at angle theta, and the quadrant signs All, Sin, Tan, Cos*
*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: **A**ll positive.
- Quadrant 2: only **S**in positive.
- Quadrant 3: only **T**an positive.
- Quadrant 4: only **C**os positive.

*Diagram: The four quadrants with the ratios that are positive in each. Quadrant 1, 0 to 90 degrees: All. Quadrant 2, 90 to 180 degrees: Sin and cosec. Quadrant 3, 180 to 270 degrees: Tan and cot. Quadrant 4, 270 to 360 degrees: Cos and sec*
*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.

$\cos 150^\circ$: related angle $30^\circ$, quadrant 2 where cos is negative, so $\cos 150^\circ = -\dfrac{\sqrt{3}}{2}$.

*Diagram: Unit circle with the angle 150 degrees. The point P is in quadrant 2 at minus root 3 over 2, one half. The related angle to the negative x-axis is 30 degrees*
*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.*

$\sin \dfrac{4\pi}{3}$: related angle $\dfrac{\pi}{3}$, quadrant 3 where sin is negative, so $-\dfrac{\sqrt{3}}{2}$.

**Reciprocal ratios.** $\operatorname{cosec}\theta = \dfrac{1}{\sin\theta}$, $\sec\theta = \dfrac{1}{\cos\theta}$ and $\cot\theta = \dfrac{1}{\tan\theta}$. They have the same sign as the ratio they come from. For example $\tan\dfrac{4\pi}{3} = \sqrt{3}$, so $\cot\dfrac{4\pi}{3} = \dfrac{1}{\sqrt{3}}$.

### Year 12: Why calculus needs radians
For small angles in radians, $\sin x \approx x$. Try $x = 0.01$: $\sin 0.01 = 0.0099998$. This only works in radians, and it is the reason that

$$\frac{d}{dx}\sin x = \cos x$$

is true only when $x$ is in radians. In degrees an extra factor of $\dfrac{\pi}{180}$ would appear everywhere. From now on, assume radians in every calculus question unless told otherwise.

## Quick summary

- Arc length $= \dfrac{\theta}{360} \times 2\pi r$ and sector area $= \dfrac{\theta}{360} \times \pi r^2$ in degrees.
- $\pi$ radians $= 180^\circ$. In radians: $l = r\theta$ and $A = \dfrac{1}{2}r^2\theta$.
- On the unit circle, the point at angle $\theta$ is $(\cos\theta, \sin\theta)$.
- ASTC gives the sign in each quadrant. Use the related angle for the size.
- Calculus always uses radians.

## Quiz

Try these questions from the chapter quiz.

1. What is the circumference of a circle with radius 5 cm?
   - A. $5\pi$ cm
   - B. $10\pi$ cm
   - C. $25\pi$ cm
   - D. $2.5\pi$ cm
   - Answer: B. $10\pi$ cm. $2\pi r = 10\pi$.

2. Find the arc length for an angle of $90^\circ$ in a circle of radius 8.
   - A. $8\pi$
   - B. $16\pi$
   - C. $4\pi$
   - D. $2\pi$
   - Answer: C. $4\pi$. $\dfrac{90}{360} \times 16\pi = 4\pi$.

3. Find the area of a sector with angle $60^\circ$ and radius 6.
   - A. $6\pi$
   - B. $12\pi$
   - C. $36\pi$
   - D. $2\pi$
   - Answer: A. $6\pi$. $\dfrac{60}{360} \times 36\pi = 6\pi$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-advanced/radians-and-unit-circle/#quiz