Unit-circle meaning | Year 10

01 / LEARN

Understand the idea.

Coordinates extend the triangle ratios

On the unit circle, turn anticlockwise from $(1, 0)$ through angle $θ$ . The resulting point is $(cos θ, sin θ)$ . This extends the right-triangle definitions to angles beyond the first quadrant.

Cosine is the horizontal coordinate and sine is the vertical coordinate. Determine their signs from the point's quadrant: left means negative cosine, while below the horizontal axis means negative sine.

Equal-scale unit circle with x and y axes and unit ticks. The radius reaches point P in quadrant two at one hundred and fifty degrees anticlockwise from positive x. Fine dashed projections reach both axes; the point coordinates are not labelled.

Use a reference angle and a quadrant

A reference angle is the acute angle between the radius and the horizontal axis. Use a special triangle or inverse ratio to find coordinate magnitudes, then apply the signs determined by the quadrant.

A full turn of $360^{\circ}$ returns to the same point. Also, $tan θ = \frac{sin θ}{cos θ}$ is undefined when the horizontal coordinate is zero; division by zero has no value.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

An obtuse angle

Find the exact unit-circle coordinates at 150°.

Worked solution

1
Find the reference angle.
$180^{\circ} - 150^{\circ} = 30^{\circ}$
The radius lies in quadrant II, $30^{\circ}$ above the negative horizontal axis.
2
Read the coordinate magnitudes.
$∣ x ∣ = \frac{\sqrt{3}}{2}, ∣ y ∣ = \frac{1}{2}$
The $30^{\circ}$ special triangle gives horizontal and vertical magnitudes.
3
Apply the quadrant signs.
$(cos 150^{\circ}, sin 150^{\circ}) = (- \frac{\sqrt{3}}{2}, \frac{1}{2})$
Quadrant II is left of the vertical axis and above the horizontal axis.

Answer

$(- \sqrt{3} / 2, 1 / 2)$

EXAMPLE 02

Recover an angle from coordinates

A unit-circle point is (−3/5,−4/5). Find its quadrant and angle between 0° and 360° to 0.1°.

Worked solution

1
Verify the point is on the unit circle.
${(- \frac{3}{5})}^{2} + {(- \frac{4}{5})}^{2} = \frac{9}{25} + \frac{16}{25} = 1$
Unit-circle coordinates satisfy $x^{2} + y^{2} = 1$ .
2
Identify the quadrant.
$x < 0, y < 0$
A point left of and below the origin is in quadrant III.
3
Find the acute reference angle.
$α = {tan}^{- 1} (\frac{4 / 5}{3 / 5}) = {tan}^{- 1} (\frac{4}{3}) \approx {53.1301}^{\circ}$
The vertical magnitude is opposite and the horizontal magnitude is adjacent.
4
Locate the full angle.
$θ = 180^{\circ} + α \approx {233.1}^{\circ}$
Quadrant III starts at a half-turn and adds the reference angle.

Answer

Quadrant III; 233.1°

03 / YOUR TURN

Try it yourself.

Check your understanding

Find the point at 90°.

Show answer & reasoning

(0, 1)

At $90^{\circ}$ the radius points vertically upwards, so its horizontal coordinate is $0$ and vertical coordinate is $1$ : the point is $(0, 1)$ .

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6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Give the exact unit-circle coordinates at $60^{\circ}$ .
2 marks
Find $sin 210^{\circ}$ exactly.
2 marks
Find $cos 240^{\circ}$ exactly.
2 marks
Find $tan 135^{\circ}$ exactly.
2 marks
Do $45^{\circ}$ and $765^{\circ}$ identify the same unit-circle point? Explain.
2 marks
For a quadrant-II angle, $cos θ = - \frac{3}{5}$ . Find $sin θ$ and $tan θ$ exactly.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Give the exact unit-circle coordinates at $60^{\circ}$ .
2 marks
Coordinates are $(cos 60^{\circ}, sin 60^{\circ})$ .
$(\frac{1}{2}, \frac{\sqrt{3}}{2})$
Find $sin 210^{\circ}$ exactly.
2 marks
$210^{\circ}$ is in quadrant III with reference angle $30^{\circ}$ .
Sine is negative, so $sin 210^{\circ} = - \frac{1}{2}$ .
$- \frac{1}{2}$
Find $cos 240^{\circ}$ exactly.
2 marks
The reference angle is $60^{\circ}$ in quadrant III.
$cos 240^{\circ} = - cos 60^{\circ} = - \frac{1}{2}$ .
$- \frac{1}{2}$
Find $tan 135^{\circ}$ exactly.
2 marks
$tan 135^{\circ} = \frac{\frac{\sqrt{2}}{2}}{- \frac{\sqrt{2}}{2}} = - 1$ .
$- 1$
Do $45^{\circ}$ and $765^{\circ}$ identify the same unit-circle point? Explain.
2 marks
$765^{\circ} - 45^{\circ} = 720^{\circ}$ , which is two complete turns.
Yes.
For a quadrant-II angle, $cos θ = - \frac{3}{5}$ . Find $sin θ$ and $tan θ$ exactly.
3 marks
${sin}^{2} θ = 1 - \frac{9}{25} = \frac{16}{25}$ .
Sine is positive in quadrant II.
$tan θ = \frac{\frac{4}{5}}{- \frac{3}{5}} = - \frac{4}{3}$ .
$sin θ = \frac{4}{5}$ ; $tan θ = - \frac{4}{3}$

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Curriculum reference & lesson scope

Australian Curriculum v9 and South Australian school course scope

Right-triangle applications include AC9M10M03. The approved chapter also develops non-right-triangle methods, exact values and trigonometric functions within one normal Year 10 course.

Official curriculum source ↗

In this lesson

Understand the idea.

Coordinates extend the triangle ratios

Use a reference angle and a quadrant

See the method in action.

An obtuse angle

Find the reference angle.

Read the coordinate magnitudes.

Apply the quadrant signs.

Recover an angle from coordinates

Verify the point is on the unit circle.

Identify the quadrant.

Find the acute reference angle.

Locate the full angle.

Try it yourself.

Check your understanding

Keep practising