Unit-circle definitions and exact values | Year 11

01 / LEARN

Understand the idea.

Coordinates define the ratios

Begin at $(1, 0)$ and turn anticlockwise for a positive angle. The point reached on the unit circle has horizontal coordinate $cos θ$ and vertical coordinate $sin θ$ . Negative angles turn clockwise, so the definitions work beyond acute angles.

The circle equation gives ${cos}^{2} θ + {sin}^{2} θ = 1$ . Tangent is the vertical coordinate divided by the horizontal coordinate, so $tan θ = \frac{sin θ}{cos θ}$ is undefined when the horizontal coordinate is zero.

An equal-scale circle of radius one has a point in the first quadrant labelled cosine theta, sine theta. The radius, angle theta, horizontal cosine and vertical sine projection are shown.

Reference angles and signs

Find the acute reference angle to the horizontal axis, then use an exact-value triangle for the coordinate magnitudes. Attach signs from the actual quadrant: left means negative cosine, and below means negative sine. Determine tangent from their quotient.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Find sine, cosine and tangent of $\frac{5 π}{6}$ .

Worked solution

1
Locate the angle and its reference angle.
$\frac{5 π}{6} = π - \frac{π}{6}$
The point is in quadrant II, with reference angle one sixth of a half-turn.
2
Assign the coordinate signs.
$sin \frac{5 π}{6} = \frac{1}{2}, cos \frac{5 π}{6} = - \frac{\sqrt{3}}{2}$
Quadrant II is above the horizontal axis and left of the vertical axis.
3
Divide to find tangent.
$tan \frac{5 π}{6} = \frac{\frac{1}{2}}{- \frac{\sqrt{3}}{2}} = - \frac{1}{\sqrt{3}} = - \frac{\sqrt{3}}{3}$
Tangent is sine divided by the nonzero cosine; rationalising preserves its exact value.

Answer

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

EXAMPLE 02

Worked example 2

Find $sin (- \frac{π}{4})$ and $cos (- \frac{π}{4})$ .

Worked solution

1
Interpret the negative turn.
$- \frac{π}{4}$
A clockwise half of a right angle reaches quadrant IV.
2
Use the exact reference magnitudes.
$| sin (- \frac{π}{4}) | = | cos (- \frac{π}{4}) | = \frac{\sqrt{2}}{2}$
The reference triangle is an isosceles right triangle on the unit circle.
3
Apply the signs of the coordinates.
$sin (- \frac{π}{4}) = - \frac{\sqrt{2}}{2}, cos (- \frac{π}{4}) = \frac{\sqrt{2}}{2}$
The quadrant IV point has negative vertical coordinate and positive horizontal coordinate.

Answer

$- \sqrt{2} / 2$ and $\sqrt{2} / 2$

03 / YOUR TURN

Try it yourself.

Check your understanding

What unit-circle point corresponds to θ=π?

Show answer & reasoning

$(- 1, 0)$

A turn of $π$ radians is a half-turn from $(1, 0)$ to $(- 1, 0)$ . Thus $cos π = - 1$ and $sin π = 0$ .

Keep practising

6 worksheet questions · 18 marks

Open worksheet Worked solutions

Read the questions online

Find $sin (\frac{7 π}{6})$ exactly.
3 marks
Find $cos (\frac{3 π}{4})$ exactly.
3 marks
Find $tan (\frac{5 π}{3})$ exactly.
3 marks
Find $sin (\frac{11 π}{6})$ and $cos (\frac{11 π}{6})$ .
3 marks
If $cos θ = - \frac{3}{5}$ in quadrant III, find sine and tangent.
3 marks
Explain why $tan (\frac{π}{2})$ is undefined.
3 marks

Read the worked solutions

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

Find $sin (\frac{7 π}{6})$ exactly.
3 marks
Reference angle $\frac{π}{6}$ ; quadrant III has negative sine.
$- \frac{1}{2}$
Find $cos (\frac{3 π}{4})$ exactly.
3 marks
Reference angle $\frac{π}{4}$ ; the horizontal coordinate is negative.
$- \frac{\sqrt{2}}{2}$
Find $tan (\frac{5 π}{3})$ exactly.
3 marks
Reference angle $\frac{π}{3}$ in quadrant IV; divide negative sine by positive cosine.
$- \sqrt{3}$
Find $sin (\frac{11 π}{6})$ and $cos (\frac{11 π}{6})$ .
3 marks
Reference angle $\frac{π}{6}$ in quadrant IV.
$- \frac{1}{2}, \frac{\sqrt{3}}{2}$
If $cos θ = - \frac{3}{5}$ in quadrant III, find sine and tangent.
3 marks
${sin}^{2} θ = 1 - \frac{9}{25} = \frac{16}{25}$ .
Select the negative sine; divide to obtain tangent.
$sin θ = - \frac{4}{5}, tan θ = \frac{4}{3}$
Explain why $tan (\frac{π}{2})$ is undefined.
3 marks
$cos (\frac{π}{2}) = 0$ ; tangent is sine divided by cosine.
It would require division by zero.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: triangle rules, circular measure and trigonometric functions; school sequencing may vary.

Official curriculum source ↗

In this lesson

Understand the idea.

Coordinates define the ratios

Reference angles and signs

See the method in action.

Worked example 1

Locate the angle and its reference angle.

Assign the coordinate signs.

Divide to find tangent.

Worked example 2

Interpret the negative turn.

Use the exact reference magnitudes.

Apply the signs of the coordinates.

Try it yourself.

Check your understanding

Keep practising