G21: Exact Trigonometric Values
Know the exact values of sin θ and cos θ for θ = 0°, 30°, 45°, 60°, 90°; know exact value of tan θ for θ = 0°, 30°, 45°, 60°
Know the exact values of sin θ and cos θ for θ = 0°, 30°, 45°, 60°, 90°; know exact value of tan θ for θ = 0°, 30°, 45°, 60°
| Angle | sin θ |
|---|---|
| 0° | 0 |
| 30° | ½ |
| 45° | √2/2 |
| 60° | √3/2 |
| 90° | 1 |
| Angle | cos θ |
|---|---|
| 0° | 1 |
| 30° | √3/2 |
| 45° | √2/2 |
| 60° | ½ |
| 90° | 0 |
| Angle | tan θ |
|---|---|
| 0° | 0 |
| 30° | √3/3 or 1/√3 |
| 45° | 1 |
| 60° | √3 |
Find the exact value of 2 sin 30° + cos 60°.
Solution:
sin 30° = ½, cos 60° = ½
2 sin 30° + cos 60° = 2(½) + ½ = 1 + ½ = 3/2
Find the exact value of sin 45° × cos 45°.
Solution:
sin 45° = √2/2, cos 45° = √2/2
sin 45° × cos 45° = (√2/2)² = 2/4 = ½
Find the exact value of tan² 45° + sin 60°.
Solution:
tan 45° = 1, sin 60° = √3/2
tan² 45° + sin 60° = 1² + √3/2 = 1 + √3/2 = (2 + √3)/2
In a right-angled triangle, angle = 30°, hypotenuse = 10 cm. Find the opposite side exactly.
Solution:
sin 30° = opp/hyp
½ = opp/10
opp = 10 × ½ = 5 cm
In a right-angled triangle, angle = 60°, adjacent = 6 cm. Find the hypotenuse exactly.
Solution:
cos 60° = adj/hyp
½ = 6/hyp
hyp = 6 × 2 = 12 cm
Exact trig values are 'write down' marks — but only from the table or the two special triangles. Never reach for a calculator: √3/2 typed as 0.866… scores zero.
Write down the exact value of cos 30°
From the table: cos 30° = √3/2.
Where it comes from: half an equilateral triangle of side 2 gives sides 1, 2, √3 — cos 30° = adjacent/hypotenuse = √3/2.
B1 for √3/2. Writing 0.866 scores 0 — the question says exact.
Write down the exact value of tan 30°
Same half-equilateral triangle: tan 30° = opposite/adjacent = 1/√3.
Rationalised: 1/√3 = √3/3 (multiply top and bottom by √3).
B1 for 1/√3 or √3/3 — both forms are accepted. 0.577 is not.
Exam wording to expect: "Write down the exact value of …" — the word exact bans decimals. Memorise the table for 0°, 30°, 45°, 60°, 90°.
Q1: Write down the exact value of sin 60°.
Q2: Write down the exact value of tan 45°.
Q3: Calculate sin 30° × cos 60°.
Q4: Find sin² 45° + cos² 45°.
Q5: A right-angled triangle has angle 45° and hypotenuse 8 cm. Find the opposite side exactly.
Q6: Write down the exact value of cos 0°
Q7: Write down the exact value of tan 60°
A right-angled triangle has a 30° angle and a hypotenuse of 10 cm. Write down the exact length of the side opposite the 30° angle.
sin 30° = opposite ÷ hypotenuse
sin 30° = 1/2, so opposite = 10 × 1/2 = 5 cm
M1 for using sin 30° = 1/2 (exact form), A1 for 5. Writing 0.5 instead of 1/2 still scores here, but quoting 1/2 keeps 'exact' habits intact.
Every mark in GCSE maths comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. sin 60° = √3/2 (from the table, or the half-equilateral triangle)
Q2. tan 45° = 1 — the isosceles right triangle has equal legs
Q3. sin 30° = 1/2 and cos 60° = 1/2, so 1/2 × 1/2 = 1/4
Q4. (√2/2)² + (√2/2)² = 1/2 + 1/2 = 1 — this is sin²+cos² = 1, worth knowing
Q5. sin 45° = opposite/hypotenuse → √2/2 = opp/8 → opp = 8 × √2/2 = 4√2 cm
Q6. cos 0° = 1
Q7. tan 60° = √3
Memorise the table (angles 0°, 30°, 45°, 60°, 90°):
| 0° | 30° | 45° | 60° | 90° | |
|---|---|---|---|---|---|
| sin | 0 | 1/2 | √2/2 | √3/2 | 1 |
| cos | 1 | √3/2 | √2/2 | 1/2 | 0 |
| tan | 0 | 1/√3 (=√3/3) | 1 | √3 | — |
Memory hooks: sin climbs 0, 1, 2, 3 over √-denominators (√0, √1, √2, √3, √4)/2; cos is sin reversed; tan = sin ÷ cos.
Exact values are "write down" 1-mark questions on Paper 1 (non-calculator). They also gate trig-graph and exact-area questions later, so the table pays off repeatedly.
Links to: trigonometry ratios (R12/G), trig graphs (A12), Pythagoras (G).
Thirty-second checks — cover the answers, work each one, then mark yourself. If any takes longer than a minute, revisit the matching section above.
Spec J560. Three papers of 100 marks each: Paper 1 non-calculator, Papers 2 and 3 calculator, at Foundation/Higher tiers. OCR marks problem-solving explicitly, and its papers tend to front-load multi-step worded contexts — practise extracting the mathematics from prose before calculating.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
Find the exact value of sin(60) x cos(30) + sin(30) x cos(60) without using a calculator.
Solution: sin(60) = root3/2, cos(30) = root3/2, sin(30) = 1/2, cos(60) = 1/2. Result = (root3/2)(root3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1. This equals sin(90) = 1, confirming the addition formula sin(A+B).
1. Wrong: Writing sin(45) = 0.707 instead of 1/root2 (or root2/2) Correct: Exact values must be left in surd form. sin(45) = 1/root2 = root2/2. Decimal approximations lose marks when exact values are requested.
2. Wrong: Confusing sin and cos values: saying sin(60) = 1/2 Correct: sin(60) = root3/2 and sin(30) = 1/2. Cos has the reverse pattern: cos(30) = root3/2 and cos(60) = 1/2.
3. Wrong: Rationalising incorrectly: writing 1/root2 as root2 instead of root2/2 Correct: 1/root2 = root2/2 (multiply top and bottom by root2). root2 is approximately 1.414, while 1/root2 is approximately 0.707.
6 marks: An equilateral triangle has side length 2 cm. (a) Find the exact height using exact trig values. (b) Find the exact area. (c) A regular hexagon is made from 6 equilateral triangles. Find the exact area of the hexagon with side 2 cm.
(a) Height = 2 x sin(60) = 2 x root3/2 = root3 cm.
(b) Area = 1/2 x 2 x root3 = root3 cm squared.
(c) 6 equilateral triangles, each with area root3. Total area = 6 x root3 cm squared.
Mark scheme: M1 sin(60), A1 root3, M1 area formula, A1 root3, M1 hexagon structure, A1 6 x root3
A student calculates cos(30) as 0.866 on their calculator but the exam requires an exact answer.
(a) Write cos(30) as an exact value in surd form.
(b) Show that (root3/2) squared = 3/4.
(c) A question asks for the exact area of a triangle with sides 10 cm, 10 cm and included angle 60 degrees. A student gives 43.3 cm squared. What should the exact answer be?
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