Derive exact special-angle ratios and handle 0° and 90°
Derive the 30°,45°,60° ratios · State endpoint values using the unit circle · Keep exact radicals and recognise undefined tangent
Section Paper 4 · Historical Extended · calculator
Exam scope and assessment notes
Extended content; not a Core requirement. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.
Before you begin
- Right-triangle ratios
- Pythagoras and a unit circle
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should you establish before applying the method in this lesson?
What you will learn
- Derive the 30°,45°,60° ratios
- State endpoint values using the unit circle
- Keep exact radicals and recognise undefined tangent
Two triangles produce the exact values
A 45°–45°–90° triangle has sides 1,1,√2, so sine and cosine at 45° are √2/2 and tangent is 1. Bisect an equilateral triangle of side 2: the right triangle has sides 1,√3,2. Thus sin30°=1/2, cos30°=√3/2, tan30°=√3/3; sin60°=√3/2, cos60°=1/2, tan60°=√3.
Endpoints use the circle definition
On the unit circle, the point at θ is (cosθ,sinθ). At 0° it is (1,0); at 90° it is (0,1). Therefore sin0°=0, cos0°=1, tan0°=0; sin90°=1 and cos90°=0. Tangent is sine divided by cosine, so tan90° is undefined. These endpoint values do not require a non-degenerate triangle with an acute angle of 0° or 90°.
Sources for this lesson
Pause and explain
Is tan90° equal to zero?
Worked example
State the exact sine, cosine and tangent values for 60°.
Show the worked solution
- The bisected equilateral triangle gives opposite √3, adjacent 1 and hypotenuse 2.
- Form the three ratios without decimal rounding.
Answer sin60°=√3/2; cos60°=1/2; tan60°=√3.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
State exact sin30°, cos30° and tan30°.
Give me a hint
Use sides 1,√3,2.
Compare my reasoning
- sin30°=1/2 and cos30°=√3/2.
- tan30°=1/√3=√3/3.
1/2; √3/2; √3/3.
Look for these in your work
- Use the side roles
- Retain exact radicals
State exact sin45°, cos45°, tan45°, sin60°, cos60° and tan60°.
Give me a hint
Use the two special right triangles.
Compare my reasoning
- At 45°: √2/2, √2/2, 1.
- At 60°: √3/2, 1/2, √3.
45°: √2/2, √2/2, 1; 60°: √3/2, 1/2, √3.
Look for these in your work
- Distinguish the special triangles
- Keep the angle order
Compare all three ratios at 0° and 90°. Explain the exceptional tangent value.
Give me a hint
Use the unit-circle coordinates, then the quotient.
Compare my reasoning
- At 0°: sine 0, cosine 1, tangent 0.
- At 90°: sine 1, cosine 0, tangent undefined because its denominator is zero.
0°: 0,1,0; 90°: 1,0,undefined.
Look for these in your work
- Use the endpoint definition
- Check division by zero
Common mistakes
- Undefined tangent is not zero.
- Use exact radicals when an exact value is requested.
What can you explain now?
Is tan90° equal to zero?
Compare with the explanation
No; it is undefined because cos90°=0.
The quotient sin90°/cos90° would require division by zero.
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Make it stick
Is tan90° equal to zero?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.