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G21: Exact Trigonometric Values

Foundation Higher AQAEdexcelOCREduqasCCEA

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°

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📋 Key Concepts

Exact values are the precise values of trigonometric functions for special angles, written as fractions or surds rather than decimals.
Why learn these? They appear frequently in exams and simplify calculations without a calculator.

📝 Sin Values

Anglesin θ
0
30°½
45°√2/2
60°√3/2
90°1
Pattern: sin goes from 0 to 1 in a predictable way: 0, ½, √2/2, √3/2, 1

📝 Cos Values

Anglecos θ
1
30°√3/2
45°√2/2
60°½
90°0
Pattern: cos is sin "backwards": 1, √3/2, √2/2, ½, 0
Note: sin 30° = cos 60°, sin 45° = cos 45°, sin 60° = cos 30°

📝 Tan Values

Angletan θ
0
30°√3/3 or 1/√3
45°1
60°√3
Remember: tan θ = sin θ / cos θ
Note: tan 90° is undefined (division by zero).

📝 Memory Tips

Sin table: Write as fractions with denominator 2:
  • sin 0° = √0/2 = 0
  • sin 30° = √1/2 = ½
  • sin 45° = √2/2
  • sin 60° = √3/2
  • sin 90° = √4/2 = 1
Cos is sin backwards: Just reverse the order!

📝 Using Exact Values in Calculations

Example 1

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

Example 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

Example 3

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

📝 Finding Sides with Exact Values

Example 4

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

Example 5

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

❓ Practice Questions

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.

✅ Answers

  1. √3/2
  2. 1
  3. ½ × ½ = ¼
  4. (√2/2)² + (√2/2)² = ½ + ½ = 1
  5. sin 45° = opp/8, √2/2 = opp/8, opp = 8 × √2/2 = 4√2 cm

🎯 Exam Tips

🧠 Problem-Solving Strategies

Problem-Solving

Memorise exact values: sin 0=0, sin 30=1/2, sin 45=1/root2, sin 60=root3/2, sin 90=1. Cos follows the same values in reverse order. Tan 0=0, tan 30=1/root3, tan 45=1, tan 60=root3. Use the special triangles: 45-45-90 (sides 1:1:root2) and 30-60-90 (sides 1:root3:2).
Multi-Step Problem

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).

⚠️ Common Errors

Watch Out!

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-Mark Exam Question

Extended Answer

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

📊 AO3: Reason & Interpret

Reasoning and Interpretation

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?

Answers: (a) cos(30) = root3/2. (b) (root3/2) squared = 3/4. The root3 squared gives 3, and 2 squared gives 4. (c) Area = 1/2 x 10 x 10 x sin(60) = 50 x root3/2 = 25 x root3 cm squared. The decimal 43.3 is an approximation; the exact answer must be in surd form.

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