Skip to main content
SingaporeAdditional MathematicsSyllabus dot point

How do the unit circle and reference angles extend the trigonometric ratios to any angle?

Define sine, cosine and tangent for any angle using the unit circle, determine signs by quadrant, and use reference angles and special angles

A focused answer to the O-Level A-Maths outcome on trigonometric ratios. The unit-circle definitions, the signs of the ratios by quadrant, reference angles, and the exact values of special angles.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

SEAB wants you to define sine, cosine and tangent for any angle, not just acute ones, using a point moving round the unit circle, to know the sign of each ratio in each quadrant, and to use reference angles together with the exact values of the special angles (30∘,45∘,60∘30^\circ, 45^\circ, 60^\circ). This is the foundation for every later trigonometry topic.

The answer

The unit-circle definitions

Take a point PP on the circle of radius 11 centred at the origin, where the radius to PP makes an angle θ\theta measured anticlockwise from the positive xx-axis. Then:

cos⁡θ=x-coordinate of P,sin⁡θ=y-coordinate of P,tan⁡θ=sin⁡θcos⁡θ.\cos\theta = x\text{-coordinate of } P, \qquad \sin\theta = y\text{-coordinate of } P, \qquad \tan\theta = \frac{\sin\theta}{\cos\theta}.

This extends the right-angled-triangle ratios to any angle, including obtuse and reflex angles.

Signs by quadrant

As PP moves round, the signs of xx and yy change, so the ratios change sign. The pattern (often remembered as "All, Sine, Tangent, Cosine") is:

  • First quadrant (0∘0^\circ to 90∘90^\circ): all positive.
  • Second (90∘90^\circ to 180∘180^\circ): only sine positive.
  • Third (180∘180^\circ to 270∘270^\circ): only tangent positive.
  • Fourth (270∘270^\circ to 360∘360^\circ): only cosine positive.

Reference angles

The reference angle is the acute angle between the radius and the horizontal axis. The size of a ratio equals the ratio of its reference angle; the quadrant fixes the sign. So sin⁡150∘=+sin⁡30∘\sin 150^\circ = +\sin 30^\circ and cos⁡210∘=−cos⁡30∘\cos 210^\circ = -\cos 30^\circ.

Special angles

Know these exact values:

sin⁡30∘=12, cos⁡30∘=32, tan⁡30∘=13;\sin 30^\circ = \tfrac{1}{2},\ \cos 30^\circ = \tfrac{\sqrt{3}}{2},\ \tan 30^\circ = \tfrac{1}{\sqrt{3}};

sin⁡45∘=cos⁡45∘=12, tan⁡45∘=1;sin⁡60∘=32, cos⁡60∘=12, tan⁡60∘=3.\sin 45^\circ = \cos 45^\circ = \tfrac{1}{\sqrt{2}},\ \tan 45^\circ = 1;\quad \sin 60^\circ = \tfrac{\sqrt{3}}{2},\ \cos 60^\circ = \tfrac{1}{2},\ \tan 60^\circ = \sqrt{3}.

Examples in context

Example 1. Resolving forces. Splitting a force into horizontal and vertical parts uses Fcos⁡θF\cos\theta and Fsin⁡θF\sin\theta, and when the angle exceeds 90∘90^\circ the unit-circle signs give the correct directions automatically, which is why physics relies on this extension.

Example 2. Modelling tides. A tide height modelled by h=3sin⁡(30t)∘h = 3\sin(30t)^\circ takes the angle through all four quadrants over a cycle, so the signs of sine produce the rise and fall, a direct application of the unit circle.

Try this

Q1. State the sign of sin⁡200∘\sin 200^\circ and cos⁡200∘\cos 200^\circ. [2 marks]

  • Cue. Third quadrant: both negative.

Q2. Find the exact value of sin⁡135∘\sin 135^\circ. [2 marks]

  • Cue. Second quadrant, reference 45∘45^\circ: +sin⁡45∘=12+\sin 45^\circ = \dfrac{1}{\sqrt{2}}.

Q3. Given cos⁡θ=513\cos\theta = \dfrac{5}{13} with θ\theta acute, find sin⁡θ\sin\theta. [2 marks]

  • Cue. sin⁡θ=1−25169=1213\sin\theta = \sqrt{1 - \tfrac{25}{169}} = \dfrac{12}{13}.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksGiven that sin⁡θ=35\sin\theta = \dfrac{3}{5} and that θ\theta is obtuse, find the exact values of cos⁡θ\cos\theta and tan⁡θ\tan\theta.
Show worked answer →

Use sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1: cos⁡2θ=1−925=1625\cos^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}, so cos⁡θ=±45\cos\theta = \pm\dfrac{4}{5}.

An obtuse angle lies in the second quadrant, where cosine is negative, so cos⁡θ=−45\cos\theta = -\dfrac{4}{5}.

Then tan⁡θ=sin⁡θcos⁡θ=3/5−4/5=−34\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{3/5}{-4/5} = -\dfrac{3}{4}.

Markers reward the Pythagorean identity, choosing the correct sign from the quadrant, and the value of tan⁡θ\tan\theta.

Original3 marksWithout a calculator, find the exact value of cos⁡150∘\cos 150^\circ.
Show worked answer →

150∘150^\circ is in the second quadrant, where cosine is negative. Its reference angle (the acute angle to the horizontal axis) is 180∘−150∘=30∘180^\circ - 150^\circ = 30^\circ.

So cos⁡150∘=−cos⁡30∘=−32\cos 150^\circ = -\cos 30^\circ = -\dfrac{\sqrt{3}}{2}.

Markers reward identifying the quadrant and sign, the reference angle 30∘30^\circ, and the exact value −32-\dfrac{\sqrt{3}}{2}.

Related dot points