Mensuration and calculation: know the exact values of sin θ and cos θ for θ = 0°, 30°, 45°, 60° and 90° know the exact value of tan θ for θ = 0°, 30°, 45°, 60°

Overview

Exact trigonometric values turn familiar angles into precise answers. For 0°, 30°, 45°, 60° and 90°, you must know the sine and cosine values, and for 0°, 30°, 45° and 60°, the tangent values. This is G21, assessed at both Foundation and Higher, despite the specification describing it as Additional foundation content 1. It matters whenever a question asks for an exact length, area, proof-style simplification or multiple-choice answer: a calculator decimal can lose the precision the question demands. You will connect one right-angled-triangle ratio to surds, Pythagoras, algebraic simplification and, at Higher, general-triangle methods. Typical questions ask you to write down a value, work out an exact expression, find a missing side, show that a result is exact, or explain an error. Treat the values as a small, dependable toolkit rather than five disconnected facts. The special triangles explain the table; the table then saves time in the exam.

Key Concepts

Required angles and the exact-value table

Scope. The required angles are in degrees. G21 requires sin and cos at 0°, 30°, 45°, 60°, 90°, and tan at 0°, 30°, 45°, 60°. tan 90° is undefined, so it is neither zero nor infinity and is not a required table value 1. Use degree mode when a calculator question contains °. G21 is shared by both tiers. Here, foundation difficulty means an accessible question style, not Foundation-tier-only content.

The sin row rises from 0 to 1; the cos row falls from 1 to 0. In particular, sin 30° = cos 60° = 1/2 and sin 60° = cos 30° = √3/2. Read the matching column before choosing a value. 1/√3 and √3/3 are equivalent forms of tan 30°; the former matches the table. In an MCQ, decimal options are not automatically wrong: 0.5 is exactly 1/2, as are 0 and 1. A rounded value such as 0.87 is not an exact replacement for √3/2.

Ratios, special triangles and endpoints

For an acute angle θ in a right-angled triangle, name sides relative to θ, not by where they happen to be drawn. The hypotenuse is opposite the right angle and is always longest. SOHCAHTOA chooses the correct ratio: sin uses opposite and hypotenuse, cos uses adjacent and hypotenuse, tan uses opposite and adjacent.

Derive 45°. Start with an isosceles right triangle whose two shorter sides are 1 and 1. Pythagoras gives hypotenuse √2. Therefore sin 45° = cos 45° = 1/√2, written as √2/2 after rationalisation, and tan 45° = 1/1 = 1.

Derive 30° and 60°. Split an equilateral triangle of side 2 down the middle. One half has hypotenuse 2, short side 1 and other side √3 by Pythagoras: it is a 30°-60°-90° triangle. At 30°, sin = 1/2, cos = √3/2 and tan = 1/√3. At 60°, the opposite and adjacent swap, so sin = √3/2, cos = 1/2 and tan = √3. This gives a reasoned rebuild if you forget a table entry.

A normal right triangle cannot represent the endpoints: at 0° or 90° it would collapse. Use the unit-circle endpoint interpretation instead. A point at angle θ has horizontal coordinate cos θ and vertical coordinate sin θ. At 0° it is (1, 0), giving cos 0° = 1 and sin 0° = 0; at 90° it is (0, 1), giving cos 90° = 0 and sin 90° = 1. Since tan θ = sin θ/cos θ, tan 90° would require division by zero, so it is undefined. This is not evidence that tan 90° is a very large number.

Mathematical Relationships and Formula-Sheet Status

Only relationships used here or linked to the stated extensions are listed. The 2027 inserts supply some tools, but not the exact-value table 3 4 5.

Rationalising a surd denominator, for example 1/√3 = √3/3, is algebraic N8 Higher-only content 7. It helps present equivalent forms, but the shared G21 value tan 30° may be recalled as 1/√3. Higher-only sine and cosine rules are supplied on the insert but are not used in G21 calculations 5.

Worked Method and Illustrative Examples

Every original example, practice question and mark allocation below is an illustrative teaching scheme, not an official AQA mark scheme. Questions are AQA-style, not reproduced past-paper questions. The worked examples carry 3, 4 and 6 marks with suggested working times of about 3, 4 and 6 minutes. It shows where clear working could earn credit; candidates should still answer the exact question set.

Example 1: basic exact arithmetic, 3 marks

Question: Work out exactly sin 30° + cos 60°.

Solution. 1. sin 30° = 1/2. [1 mark]
2. cos 60° = 1/2. [1 mark]
3. 1/2 + 1/2 = 1. [1 mark]

Examiner commentary. The values must be from the correct rows and columns. Writing 0.5 + 0.5 reaches the same exact number, but fractions make the exact reasoning visible.

Example 2: a 30° right triangle, 4 marks

Higher-style exact-surd application: G21 recall is shared; calculating exactly with surds extends into N8 Higher content 7.

Question: Calculate the two shorter sides of a right-angled triangle with hypotenuse 12 cm and angle 30°. Give exact answers.

Solution. 1. Opposite side = 12 sin 30°. [1 mark]
2. = 12 × 1/2 = 6 cm. [1 mark]
3. Adjacent side = 12 cos 30°. [1 mark]
4. = 12 × √3/2 = 6√3 cm. [1 mark]

Examiner commentary. Assign opposite and adjacent relative to the labelled 30° angle. Keep cm on lengths. A candidate who swaps sin and cos gets the two answers reversed.

Example 3: challenging linked measures, 6 marks

Higher-style exact-surd application: the right-triangle method is shared, but manipulating exact surd answers links to Higher N8 7.

Question: A right-angled triangle has an angle of 60° and adjacent side 6 cm. Work out its perimeter and area exactly.

Solution.

1. tan 60° = opposite/adjacent, so opposite = 6 tan 60°. [1 mark]
2. Opposite = 6√3 cm. [1 mark]
3. cos 60° = adjacent/hypotenuse, so hypotenuse = 6 ÷ (1/2). [1 mark]
4. Hypotenuse = 12 cm. [1 mark]
5. Perimeter = 6 + 6√3 + 12 = 18 + 6√3 cm. [1 mark]
6. Area = 1/2 × 6 × 6√3 = 18√3 cm². [1 mark]

Examiner commentary. Each mark rewards a different mathematical step, not a duplicate statement. Perimeter has cm; area has cm².

A genuine comparison point is AQA November 2022 Higher Paper 1, 8300/1H, Question 23, worth four marks. The AQA-authored mark scheme gives M1 for a relevant exact ratio, dependent M1 for correct substitution into 4 sin 30 tan 30 cos 30, A1 for obtaining 1 with values seen, and final A1 for 90; tan 30° as 1/√3 or √3/3 was accepted 6. This is authentic assessment information, not a template to copy into every answer.

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Introduction 00:00; core concepts 01:00; exam tips 06:00; recall quiz 08:00; summary 09:00. Answer aloud before the tutor reveals each quiz response.

Practical Applications

A ladder makes a 60° angle with level ground. If the ladder is 4 m long, its horizontal distance from the wall is 4 cos 60° = 2 m. Its vertical height is 4 sin 60° = 2√3 m, an exact-surd Higher application. The ground distance is adjacent to the given angle, not opposite it. In an exam, a correctly labelled sketch avoids selecting the wrong ratio. A diagram need not be drawn to scale, so never measure a printed side to replace calculation.

A roof cross-section with a 45° pitch is an isosceles right triangle: height and horizontal run are equal because tan 45° = 1. With run 3 m, its rise is also 3 m. If asked for the sloping length exactly, Pythagoras gives 3√2 m. This context links angle information to a length without changing the underlying ratios. Real measurements are approximate, but a question can define idealised angles and lengths exactly. Follow the precision the question requests.

Cover and recall: explain why doubling a ladder's length doubles both horizontal distance and height without changing sine or cosine. Then explain why a 60° ladder reaches higher than a 30° ladder of the same length. Check: the ratios depend on angle, not size, and sin 60° is greater than sin 30°.

Exam Discipline, Units and Extensions

AQA papers are each 90 minutes for 80 marks: 67.5 seconds per mark on average 2. In timed practice, plan about one minute per mark, then use the remaining time to check. It is a sensible practice plan, not an AQA per-question time rule. For one mark, write the requested exact value. For two or three marks, state the ratio or substitution, then simplify. For Show that, retain exact working so the stated result follows. For Explain, use a reason such as “because cos 90° = 0, this divides by zero”. Simplify asks for an equivalent clean exact form. Work out or Calculate requires computation and visible working. Write down, State and Give need only the requested fact. In maths, if Evaluate appears, it means compute an expression, not balanced judgement; this guide uses no invented command meanings.

Convert units before calculating: 1 m = 100 cm, so use one length unit throughout. A perimeter is in a length unit, while an area is in a squared unit: 1 m² = 10,000 cm². Trigonometric ratios themselves have no units. There is no required practical for this topic.

A12 Higher-only graph-reading extension, not G21. Full sine, cosine and tangent graphs are A12 Higher-only 8. If reading such a graph, use labelled axes and a sensible scale, then identify intercepts, turning points or asymptotes only where the graph topic requires them. For a focused sketch between 0° and 90°, plot the five known angles on a degree-labelled horizontal axis and the exact ratios on the vertical axis. Join with a smooth curve, not straight segments. Sine increases and cosine decreases, meeting at (45°, √2/2). Their curves are not straight lines: sine is concave down and cosine changes slope through the interval. For tangent, plot 0°, 30°, 45° and 60°; draw an increasing curve approaching, but never touching, the vertical asymptote θ = 90°. Do not plot a finite point for tan 90°. This is a Higher graph extension, not extra G21 recall. The endpoint values can be interpreted on the unit circle, but 0° and 90° do not make ordinary, non-degenerate right triangles. G22/G23 Higher-only extensions use sine rule, cosine rule and 1/2 ab sin C for non-right-angled general triangles; do not substitute them for SOHCAHTOA in a right triangle.

References

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