CHAPTER 04: COMPLETE TRIGONOMETRY – A-LEVEL PURE MATHEMATICS

Mathematics Hip-Hop Academy | A-Level Pure Mathematics

Chapter 04: Trigonometry

Part 1: Trigonometric Functions, Angles and Graphs

Build a strong foundation in trigonometric ratios, angular measurement, the unit circle, special angles, quadrants and trigonometric graphs.

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1. Learning Objectives

By the end of this lesson, you should be able to:

• Define and use the six trigonometric ratios.
• Understand positive and negative angles.
• Identify the four quadrants and the signs of trigonometric ratios.
• Calculate exact values of standard angles.
• Determine reference angles.
• Sketch and interpret sine, cosine and tangent graphs.

2. Introduction to Trigonometry

Trigonometry is the branch of mathematics concerned with relationships between angles and sides of triangles. It also studies circular functions and their graphs.

It is used in geometry, surveying, navigation, engineering, physics and many other fields.

3. The Six Trigonometric Ratios

Consider a right-angled triangle containing an acute angle θ. Relative to θ, its sides are called the opposite, adjacent and hypotenuse.

Important: The hypotenuse is always opposite the right angle. The opposite and adjacent sides are identified relative to the angle being considered.

sin θ = Opposite / Hypotenuse
cos θ = Adjacent / Hypotenuse
tan θ = Opposite / Adjacent

The remaining three ratios are reciprocals:

cosec θ = 1 / sin θ = Hypotenuse / Opposite
sec θ = 1 / cos θ = Hypotenuse / Adjacent
cot θ = 1 / tan θ = Adjacent / Opposite

Worked Example 1

A right-angled triangle has sides 3 cm, 4 cm and 5 cm. Let θ be the angle opposite the side of length 3 cm. Find all six trigonometric ratios.

Solution:

Opposite = 3, adjacent = 4, hypotenuse = 5.

sin θ = 3/5; cos θ = 4/5; tan θ = 3/4

cosec θ = 5/3; sec θ = 5/4; cot θ = 4/3

4. Angles: Degrees and Direction

An angle measures the amount of rotation between two directions.

• Anticlockwise rotation is positive.
• Clockwise rotation is negative.
• One complete revolution is 360°.
• A half revolution is 180°.
• A quarter revolution is 90°.

For example, +60° represents an anticlockwise rotation of 60°, while −60° represents a clockwise rotation of 60°.

Angles differing by 360° have the same terminal arm. For example, 30°, 390° and −330° are coterminal angles.

5. The Unit Circle

The unit circle is a circle of radius 1 centred at the origin of a Cartesian coordinate system.

If a point P on the unit circle corresponds to an angle θ measured from the positive x-axis, its coordinates are:

(x, y) = (cos θ, sin θ)

Therefore, the horizontal coordinate gives the cosine and the vertical coordinate gives the sine.

This relationship allows trigonometric functions to be defined for angles of any size, including negative angles and angles greater than 360°.

6. The Four Quadrants and Signs

Quadrant	Angle range	Positive ratios
I	0° to 90°	All six
II	90° to 180°	Sine and cosecant
III	180° to 270°	Tangent and cotangent
IV	270° to 360°	Cosine and secant

A useful memory aid is ASTC: All, Sine, Tangent, Cosine. Starting in quadrant I and moving anticlockwise, this indicates which ratios are positive.

Worked Example 2

Evaluate sin 150°.

Solution: 150° is in quadrant II, where sine is positive.

Reference angle = 180° − 150° = 30°.

sin 150° = sin 30° = 1/2

Worked Example 3

Evaluate cos 225°.

Solution: 225° is in quadrant III, where cosine is negative.

Reference angle = 225° − 180° = 45°.

cos 225° = −cos 45° = −√2/2

7. Special Angles and Exact Values

The standard angles 0°, 30°, 45°, 60° and 90° occur frequently in examination questions. Their exact values should be memorised.

θ	0°	30°	45°	60°	90°
sin θ	0	1/2	√2/2	√3/2	1
cos θ	1	√3/2	√2/2	1/2	0
tan θ	0	1/√3	1	√3	Undefined

Warning: tan 90° is undefined because tan θ = sin θ / cos θ and cos 90° = 0. Division by zero is not defined.

8. Reference Angles

A reference angle is the acute positive angle formed between the terminal arm of an angle and the x-axis.

• Quadrant I: reference angle = θ.
• Quadrant II: reference angle = 180° − θ.
• Quadrant III: reference angle = θ − 180°.
• Quadrant IV: reference angle = 360° − θ.

Worked Example 4

Find the reference angle of 330° and evaluate tan 330°.

Reference angle = 360° − 330° = 30°.

Tangent is negative in quadrant IV.

tan 330° = −tan 30° = −1/√3

9. Graphs of Trigonometric Functions

9.1 The Sine Graph

The basic sine function is y = sin θ.

• Domain: all real values of θ.
• Range: −1 ≤ y ≤ 1.
• Period: 360°.
• Maximum value: 1.
• Minimum value: −1.

θ	0°	90°	180°	270°	360°
sin θ	0	1	0	−1	0

The sine graph begins at zero, rises to 1, returns to zero, falls to −1 and returns to zero after one complete cycle.

9.2 The Cosine Graph

The basic cosine function is y = cos θ.

• Domain: all real values of θ.
• Range: −1 ≤ y ≤ 1.
• Period: 360°.
• Maximum value: 1.
• Minimum value: −1.

θ	0°	90°	180°	270°	360°
cos θ	1	0	−1	0	1

The cosine graph begins at 1, falls to zero, reaches −1, rises to zero and returns to 1.

9.3 The Tangent Graph

The basic tangent function is y = tan θ.

• Domain: all real θ except θ = 90° + 180°k, where k is an integer.
• Range: all real numbers.
• Period: 180°.
• Vertical asymptotes: θ = 90° + 180°k.

θ	0°	45°	90°	135°	180°
tan θ	0	1	Undefined	−1	0

The tangent graph is not defined at 90° and 270° within one complete revolution. These positions are vertical asymptotes.

10. Examination Practice

Attempt all the questions before opening the answers.

1. Define the six trigonometric ratios using the sides of a right-angled triangle.
2. Find the six trigonometric ratios of an angle in a triangle whose opposite, adjacent and hypotenuse sides are 5, 12 and 13 respectively.
3. State the quadrant in which each angle lies: 120°, 210° and 315°.
4. Evaluate exactly: (a) sin 120°; (b) cos 240°; (c) tan 315°.
5. Find the reference angle of 225°.
6. State the period and range of y = sin θ.
7. State the period of y = tan θ and give its vertical asymptotes.
8. Explain why tan 90° is undefined.

11. Answers and Explanations

Open Answers 1–4

1. Sine = opposite/hypotenuse; cosine = adjacent/hypotenuse; tangent = opposite/adjacent. Cosecant, secant and cotangent are the reciprocals of sine, cosine and tangent respectively.

2.

sin θ = 5/13; cos θ = 12/13; tan θ = 5/12

cosec θ = 13/5; sec θ = 13/12; cot θ = 12/5

3. 120° is in quadrant II; 210° is in quadrant III; 315° is in quadrant IV.

4. (a) √3/2; (b) −1/2; (c) −1.

Open Answers 5–8

5. 225° − 180° = 45°.

6. Period = 360°; range = −1 ≤ y ≤ 1.

7. Period = 180°. Vertical asymptotes occur at θ = 90° + 180°k, where k is an integer.

8. tan θ = sin θ / cos θ. Since cos 90° = 0, tan 90° would require division by zero, which is undefined.

12. Part 1 Summary

• The six trigonometric ratios describe relationships between sides of a right-angled triangle.
• Anticlockwise angles are positive and clockwise angles are negative.
• The unit circle extends trigonometric definitions to angles of any magnitude.
• The quadrant determines the sign of each trigonometric ratio.
• Exact values of standard angles must be known.
• Sine and cosine have period 360°, while tangent has period 180°.

Preparation for Part 2: In the next lesson, we shall study trigonometric identities, including the Pythagorean identities, reciprocal identities and methods of proving identities.

MATHEMATICS HIP-HOP
A-Level Pure Mathematics | Chapter 04 — Trigonometry | Part 1
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CHAPTER 04: TRIGONOMETRY

PART 2: TRIGONOMETRIC IDENTITIES AND FORMULAE

Introduction

In Part 1, we studied trigonometric functions, angles, special angles and graphs. We now develop an important algebraic side of trigonometry: identities.

Trigonometric identities are relationships involving trigonometric functions that remain true for every value of the variable for which both sides are defined.

They are useful in simplifying expressions, proving mathematical statements and solving trigonometric equations.

Learning Objectives

By the end of this lesson, you should be able to:

• Explain the meaning of a trigonometric identity.
• State and use the fundamental Pythagorean identities.
• Apply reciprocal and quotient identities.
• Rearrange identities to obtain alternative forms.
• Simplify trigonometric expressions.
• Prove identities using valid algebraic steps.
• Apply identities to examination-style questions.

1. Meaning of a Trigonometric Identity

An identity is an equality that is true for all permissible values of the variable.

Example of an identity
sin²θ + cos²θ = 1

This relationship is true for every angle θ because the squares of the sine and cosine of the same angle always add up to one.

Important distinction

An identity is not the same as an equation that is true only for particular angles.

For example, sin θ = 1 is true only for particular values of θ, whereas sin²θ + cos²θ = 1 is an identity.

2. The Fundamental Pythagorean Identity

The fundamental identity is:

sin²θ + cos²θ = 1

It follows from the Pythagorean theorem applied to a right-angled triangle or from the unit circle.

On a unit circle, a point corresponding to angle θ has coordinates (cos θ, sin θ). Since the radius is one, the circle equation gives:

cos²θ + sin²θ = 1² = 1

Therefore:

sin²θ + cos²θ = 1

2.1 Deriving the Other Pythagorean Identities

Starting with:

sin²θ + cos²θ = 1

Divide every term by cos²θ, provided cos θ ≠ 0.

sin²θ/cos²θ + cos²θ/cos²θ = 1/cos²θ

tan²θ + 1 = sec²θ

1 + tan²θ = sec²θ

Similarly, divide the original identity by sin²θ, provided sin θ ≠ 0.

sin²θ/sin²θ + cos²θ/sin²θ = 1/sin²θ

1 + cot²θ = cosec²θ

1 + cot²θ = cosec²θ

2.2 The Three Fundamental Identities

Identity	Formula
First identity	sin²θ + cos²θ = 1
Second identity	1 + tan²θ = sec²θ
Third identity	1 + cot²θ = cosec²θ

Remember: These identities are valid wherever the expressions involved are defined. Always check that denominators are not zero.

3. Reciprocal Identities

The reciprocal identities connect sine, cosine and tangent with their corresponding reciprocal functions.

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

These can also be rearranged as:

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Worked Example 1

Question: If sin θ = 3/5, find cosec θ.

Step 1: Use the reciprocal identity.
cosec θ = 1/sin θ

Step 2: Substitute sin θ = 3/5.
cosec θ = 1/(3/5)

Step 3: Invert the fraction.
cosec θ = 5/3

Answer: cosec θ = 5/3.

4. Quotient Identities

The quotient identities express tangent and cotangent in terms of sine and cosine.

tan θ = sin θ/cos θ

cot θ = cos θ/sin θ

These follow from the definitions of the trigonometric ratios and are valid where the denominators are non-zero.

Worked Example 2

Question: Given sin θ = 5/13 and cos θ = 12/13, find tan θ.

tan θ = sin θ/cos θ

tan θ = (5/13)/(12/13)

tan θ = (5/13) × (13/12)

Answer: tan θ = 5/12.

5. Rearranging Pythagorean Identities

Identities can be rearranged to make a particular trigonometric function the subject.

From sin²θ + cos²θ = 1:

sin²θ = 1 − cos²θ

cos²θ = 1 − sin²θ

From 1 + tan²θ = sec²θ:

tan²θ = sec²θ − 1

sec²θ = 1 + tan²θ

From 1 + cot²θ = cosec²θ:

cot²θ = cosec²θ − 1

cosec²θ = 1 + cot²θ

Care with square roots:

If sin²θ = 1 − cos²θ, then sin θ = ±√(1 − cos²θ), not automatically the positive root. The correct sign depends on the quadrant of θ.

Worked Example 3

Question: Given cos θ = 4/5 and θ is acute, find sin θ.

Use sin²θ + cos²θ = 1.
sin²θ + (4/5)² = 1

sin²θ = 1 − 16/25
sin²θ = 9/25

sin θ = ±3/5

Since θ is acute, sine is positive.

Answer: sin θ = 3/5.

6. Simplifying Trigonometric Expressions

Simplification involves using identities and algebraic operations to write an expression in a shorter or more convenient form.

Useful techniques include:

• Replacing reciprocal functions with their definitions.
• Replacing tangent and cotangent with quotient forms.
• Using the Pythagorean identities.
• Factorising expressions where possible.
• Combining fractions carefully.

Worked Example 4

Question: Simplify (1 − cos²θ)/sin θ.

Step 1: Use sin²θ + cos²θ = 1.
1 − cos²θ = sin²θ

Step 2: Substitute.
sin²θ/sin θ

Step 3: Cancel the common factor, where sin θ ≠ 0.
= sin θ

Answer: sin θ.

Worked Example 5

Question: Simplify (sec²θ − 1)/tan θ.

Step 1: Recall that sec²θ − 1 = tan²θ.

Step 2: Substitute.
tan²θ/tan θ

Step 3: Cancel the common factor, where tan θ ≠ 0.
= tan θ

Answer: tan θ.

7. Proving Trigonometric Identities

Proving an identity means showing, through valid mathematical steps, that one side of an expression is equal to the other.

A useful approach is to begin with the more complicated side and transform it until it becomes the simpler side.

General Method

1. Write down the side you intend to simplify.
2. Identify an identity that can be applied.
3. Substitute or rearrange carefully.
4. Simplify algebraically.
5. Continue until the required expression is obtained.

Important examination advice

Do not assume that the identity is true and then use the result to justify itself. Show a sequence of valid equalities, preferably starting with one side only.

Worked Example 6

Prove that:
(1 − cos²θ)/cos²θ = tan²θ

Start with the left-hand side:
(1 − cos²θ)/cos²θ

Using sin²θ + cos²θ = 1:
= sin²θ/cos²θ

Using tan θ = sin θ/cos θ:
= tan²θ

Hence proved.

Worked Example 7

Prove that:
sec²θ − tan²θ = 1

Start with the left-hand side:
sec²θ − tan²θ

From 1 + tan²θ = sec²θ:
= (1 + tan²θ) − tan²θ

Cancel the equal terms:
= 1

Hence proved.

Worked Example 8

Prove that:
cosec²θ − cot²θ = 1

Start with the left-hand side:
cosec²θ − cot²θ

From 1 + cot²θ = cosec²θ:
= (1 + cot²θ) − cot²θ

= 1

Hence proved.

8. Combining Quotient and Reciprocal Identities

Sometimes a question contains several trigonometric functions. Replacing them with sine and cosine can make the expression easier to simplify.

Worked Example 9

Question: Simplify (tan θ × cos θ).

tan θ × cos θ

= (sin θ/cos θ) × cos θ

Cancel cos θ, where cos θ ≠ 0.
= sin θ

Answer: sin θ.

Worked Example 10

Question: Simplify (sec θ × cos θ) + (cosec θ × sin θ).

Replace reciprocal functions:
= (1/cos θ × cos θ) + (1/sin θ × sin θ)

= 1 + 1

Answer: 2.

9. Common Errors to Avoid

• Confusing sec θ with 1/sin θ. Remember sec θ = 1/cos θ.
• Forgetting that tan θ = sin θ/cos θ.
• Using sin²θ + cos²θ = 1 incorrectly by omitting the square.
• Taking only the positive square root without considering the quadrant.
• Cancelling terms across addition or subtraction signs.
• Dividing by an expression that may equal zero without checking its domain.
• Changing both sides of an identity in an unclear way during a proof.

10. Practice Exercises

Exercise A: Basic Knowledge

1. State the three fundamental Pythagorean identities.

2. Write the reciprocal identity for:

• (a) cosec θ
• (b) sec θ
• (c) cot θ

3. Express tan θ and cot θ in terms of sine and cosine.

4. Make sin²θ the subject of sin²θ + cos²θ = 1.

5. State the identity connecting sec²θ and tan²θ.

Exercise B: Calculations

6. If sin θ = 8/17 and θ is acute, find cos θ.

7. If cos θ = 12/13 and θ is acute, find sin θ and tan θ.

8. If tan θ = 3/4 and θ is acute, find sec θ.

9. Given cosec θ = 5/3, find sin θ.

10. If sec θ = 13/5 and θ is acute, find tan θ.

Exercise C: Simplification

11. Simplify (1 − sin²θ)/cos θ.

12. Simplify (sec²θ − 1)/sec²θ.

13. Simplify tan θ × cos θ.

14. Simplify (cosec θ × sin θ) − (sec θ × cos θ).

15. Simplify (1 + cot²θ)/cosec²θ.

Exercise D: Identity Proofs

16. Prove that (1 − sin²θ)/cos²θ = 1.

17. Prove that (sec²θ − 1)/tan²θ = 1.

18. Prove that (cosec²θ − 1)/cot²θ = 1.

19. Prove that sin θ × sec θ = tan θ.

20. Prove that (1 + tan²θ)cos²θ = 1.

11. Answers and Worked Solutions

Answers to Exercise A

1.

• sin²θ + cos²θ = 1
• 1 + tan²θ = sec²θ
• 1 + cot²θ = cosec²θ

2.

• (a) cosec θ = 1/sin θ
• (b) sec θ = 1/cos θ
• (c) cot θ = 1/tan θ

3. tan θ = sin θ/cos θ; cot θ = cos θ/sin θ.

4. sin²θ = 1 − cos²θ.

5. sec²θ = 1 + tan²θ.

Worked solutions to Exercise B

6.

cos²θ = 1 − sin²θ

= 1 − 64/289 = 225/289.

Since θ is acute, cos θ = 15/17.

7.

sin²θ = 1 − 144/169 = 25/169.

Since θ is acute, sin θ = 5/13.

tan θ = (5/13)/(12/13) = 5/12.

8.

sec²θ = 1 + tan²θ = 1 + 9/16 = 25/16.

Since θ is acute, sec θ = 5/4.

9.

sin θ = 1/cosec θ = 3/5.

10.

tan²θ = sec²θ − 1 = 169/25 − 1 = 144/25.

Since θ is acute, tan θ = 12/5.

Worked solutions to Exercise C

11.

(1 − sin²θ)/cos θ = cos²θ/cos θ = cos θ.

12.

(sec²θ − 1)/sec²θ = tan²θ/sec²θ.

Using tan θ = sin θ/cos θ and sec θ = 1/cos θ, the result is sin²θ.

13.

tan θ × cos θ = (sin θ/cos θ) × cos θ = sin θ.

14.

(cosec θ × sin θ) − (sec θ × cos θ) = 1 − 1 = 0.

15.

(1 + cot²θ)/cosec²θ = cosec²θ/cosec²θ = 1.

Worked solutions to Exercise D

16. Start with the left-hand side:

(1 − sin²θ)/cos²θ = cos²θ/cos²θ = 1.

Hence proved.

17.

(sec²θ − 1)/tan²θ = tan²θ/tan²θ = 1.

Hence proved.

18.

(cosec²θ − 1)/cot²θ = cot²θ/cot²θ = 1.

Hence proved.

19.

sin θ × sec θ = sin θ × (1/cos θ) = tan θ.

Hence proved.

20.

(1 + tan²θ)cos²θ = sec²θ × cos²θ.

= (1/cos²θ) × cos²θ = 1.

Hence proved.

12. Chapter Summary

In this lesson, we have studied the algebraic relationships connecting trigonometric functions.

Fundamental identities
sin²θ + cos²θ = 1
1 + tan²θ = sec²θ
1 + cot²θ = cosec²θ

Reciprocal identities
cosec θ = 1/sin θ
sec θ = 1/cos θ
cot θ = 1/tan θ

Quotient identities
tan θ = sin θ/cos θ
cot θ = cos θ/sin θ

These identities form the foundation for more advanced trigonometric work, particularly solving equations and proving more complicated relationships.

Looking Ahead: Part 3

In the next lesson, we shall study Trigonometric Equations and General Solutions. We will learn how to solve equations involving sine, cosine and tangent, determine solutions within specified intervals and express general solutions.

MATHEMATICS HIP-HOP
Making Mathematics Easier, One Concept at a Time.
A-Level Pure Mathematics — Chapter 04, Part 2

CHAPTER 04: TRIGONOMETRY

PART 3: TRIGONOMETRIC EQUATIONS AND GENERAL SOLUTIONS

Introduction

In Part 1, we studied trigonometric functions, angles and graphs. In Part 2, we examined identities and formulae.

We now use this knowledge to solve trigonometric equations. Unlike ordinary algebraic equations, trigonometric equations can have several solutions because sine, cosine and tangent are periodic functions.

Our task is to find every solution in a specified interval and, where required, express the complete set of solutions using general solution formulae.

Learning Objectives

By the end of this lesson, you should be able to:

• Explain what a trigonometric equation is.
• Solve simple sine, cosine and tangent equations.
• Determine all solutions within a given interval.
• Write general solutions using integer multiples of the period.
• Solve trigonometric equations involving squared functions.
• Factorise quadratic trigonometric equations.
• Use identities to transform equations before solving.
• Check solutions and avoid losing valid answers.

1. Meaning of a Trigonometric Equation

A trigonometric equation is an equation containing one or more trigonometric functions of an unknown angle.

Examples include:

sin θ = 1/2

cos θ = 0

tan θ = √3

2sin²θ − 3sin θ + 1 = 0

These equations are solved by finding the angle or angles that make the equation true.

Important:

Always read the interval carefully. An equation may have infinitely many solutions, but a specified interval contains only a limited number of them.

2. Understanding Periodicity

Trigonometric functions repeat their values at regular intervals. This repeating property is called periodicity.

Function	Period in degrees	Period in radians
sin θ	360°	2π
cos θ	360°	2π
tan θ	180°	π

For example:

sin 30° = sin 390° = sin 750°

Each angle differs from the preceding one by 360°.

Similarly:

tan 45° = tan 225° = tan 405°

Each angle differs by 180°.

Remember:

Sine and cosine repeat after 360°, whereas tangent repeats after 180°. This difference is essential when writing general solutions.

3. Solving Sine Equations

Consider the equation:

sin θ = 1/2

From the special angles, the reference angle is 30°.

Sine is positive in the first and second quadrants. Therefore, between 0° and 360°, the solutions are:

θ = 30° or θ = 150°

Because sine has a period of 360°, the general solutions are:

θ = 30° + 360°n
or
θ = 150° + 360°n

where n is any integer.

General Rule for Sine

If sin θ = sin α, the general solutions in degrees are:

θ = α + 360°n
or
θ = 180° − α + 360°n

For radians:

θ = α + 2πn
or
θ = π − α + 2πn

These forms give the two families of angles having the same sine value.

Worked Example 1

Solve sin θ = 1/2 for 0° ≤ θ ≤ 360°.

The reference angle is 30° because sin 30° = 1/2.

Sine is positive in Quadrants I and II.

Quadrant I: θ = 30°.
Quadrant II: θ = 180° − 30° = 150°.

Answer: θ = 30°, 150°.

Worked Example 2

Solve sin θ = −√3/2 for 0° ≤ θ ≤ 360°.

The reference angle is 60°.

Sine is negative in Quadrants III and IV.

Quadrant III: θ = 180° + 60° = 240°.
Quadrant IV: θ = 360° − 60° = 300°.

Answer: θ = 240°, 300°.

4. Solving Cosine Equations

Consider:

cos θ = 1/2

The reference angle is 60°.

Cosine is positive in Quadrants I and IV.

θ = 60° or θ = 300°

Since cosine has a period of 360°, the general solution is:

θ = α + 360°n
or
θ = 360° − α + 360°n

Equivalently:

θ = ±α + 360°n

For radians:

θ = ±α + 2πn

Worked Example 3

Solve cos θ = −1/2 for 0° ≤ θ ≤ 360°.

The reference angle is 60°.

Cosine is negative in Quadrants II and III.

Quadrant II: θ = 180° − 60° = 120°.
Quadrant III: θ = 180° + 60° = 240°.

Answer: θ = 120°, 240°.

5. Solving Tangent Equations

Consider:

tan θ = 1

The reference angle is 45°.

Tangent is positive in Quadrants I and III.

θ = 45° or θ = 225°

Since tangent has a period of 180°, its general solution is:

θ = α + 180°n

For radians:

θ = α + πn

Worked Example 4

Solve tan θ = −√3 for 0° ≤ θ ≤ 360°.

The reference angle is 60°.

Tangent is negative in Quadrants II and IV.

Quadrant II: θ = 180° − 60° = 120°.
Quadrant IV: θ = 360° − 60° = 300°.

Answer: θ = 120°, 300°.

6. Finding All Solutions in a Given Interval

When solving an equation over an interval larger than one complete revolution, more solutions may occur.

For example, consider:

Solve sin θ = 1/2 for 0° ≤ θ ≤ 720°.

The two families of solutions are:

θ = 30° + 360°n

θ = 150° + 360°n

For the first family:

• n = 0 gives 30°.
• n = 1 gives 390°.

For the second family:

• n = 0 gives 150°.
• n = 1 gives 510°.

Other integer values give angles outside the interval.

Answer:

θ = 30°, 150°, 390°, 510°

Worked Example 5

Solve cos θ = √2/2 for −360° ≤ θ ≤ 360°.

The reference angle is 45°.

Cosine is positive in Quadrants I and IV.

General solutions:
θ = 45° + 360°n
or θ = −45° + 360°n

First family:

• n = 0 gives 45°.
• n = −1 gives −315°.

Second family:

• n = 0 gives −45°.
• n = 1 gives 315°.

Answer:

θ = −315°, −45°, 45°, 315°

7. Equations Involving Squared Trigonometric Functions

Some trigonometric equations contain sin²θ, cos²θ or tan²θ. These can often be solved using algebraic methods.

For example:

2sin²θ − 1 = 0

Rearrange:

2sin²θ = 1

sin²θ = 1/2

Taking square roots:

sin θ = ±√2/2

We must solve both the positive and negative cases.

Worked Example 6

Solve 2sin²θ − 1 = 0 for 0° ≤ θ ≤ 360°.

2sin²θ − 1 = 0
2sin²θ = 1
sin²θ = 1/2

sin θ = ±√2/2

For sin θ = √2/2:

θ = 45°, 135°.

For sin θ = −√2/2:

θ = 225°, 315°.

Answer:

θ = 45°, 135°, 225°, 315°

8. Quadratic Trigonometric Equations

A quadratic trigonometric equation can be formed when a trigonometric function appears as the variable in a quadratic expression.

For example:

2sin²θ − 3sin θ + 1 = 0

We can solve this using factorisation in the same way as an ordinary quadratic equation.

Worked Example 7

Solve 2sin²θ − 3sin θ + 1 = 0 for 0° ≤ θ ≤ 360°.

Factorise:
(2sin θ − 1)(sin θ − 1) = 0

Therefore:
2sin θ − 1 = 0
or
sin θ − 1 = 0

First equation:
sin θ = 1/2
θ = 30°, 150°.

Second equation:
sin θ = 1
θ = 90°.

Answer:

θ = 30°, 90°, 150°

Important examination warning

Never cancel a trigonometric factor before considering whether it can equal zero.

For example, if:

cos θ(sin θ + 1) = 0

the correct method is to set each factor equal to zero:

cos θ = 0

or sin θ + 1 = 0

Cancelling cos θ would lose the solutions obtained from cos θ = 0.

Worked Example 8

Solve cos θ(sin θ + 1) = 0 for 0° ≤ θ ≤ 360°.

Set each factor equal to zero.
cos θ = 0
or sin θ + 1 = 0.

From cos θ = 0:
θ = 90°, 270°.

From sin θ = −1:
θ = 270°.

Combining the distinct solutions:

Answer: θ = 90°, 270°.

9. Using Pythagorean Identities in Equations

Sometimes an equation contains more than one trigonometric function. An identity can help convert it into a simpler equation.

Recall:

sin²θ + cos²θ = 1

Therefore:

cos²θ = 1 − sin²θ

We can use this substitution when an equation contains both sine and cosine.

Worked Example 9

Solve 2cos²θ = 1 for 0° ≤ θ ≤ 360°.

2cos²θ = 1
cos²θ = 1/2

cos θ = ±√2/2

For cos θ = √2/2:
θ = 45°, 315°.

For cos θ = −√2/2:
θ = 135°, 225°.

Answer:

θ = 45°, 135°, 225°, 315°

10. Equations in Radians

Trigonometric equations may also be expressed in radians. The same methods apply, but the periods are written in terms of π.

Function	Period
sin θ	2π
cos θ	2π
tan θ	π

Worked Example 10

Solve sin θ = 1/2 for 0 ≤ θ ≤ 2π.

The reference angle is π/6.

Sine is positive in Quadrants I and II.

Quadrant I:
θ = π/6.

Quadrant II:
θ = π − π/6 = 5π/6.

Answer:

θ = π/6, 5π/6

11. Common Errors to Avoid

• Giving only one solution when the equation has more than one solution in the interval.
• Using 360° as the period of tangent instead of 180°.
• Forgetting the negative square-root case when solving sin²θ or cos²θ equations.
• Ignoring the interval stated in the question.
• Using the wrong quadrants for positive and negative trigonometric values.
• Cancelling a trigonometric factor that could be zero.
• Mixing degrees and radians in the same calculation.
• Repeating a solution or including an angle outside the specified interval.

12. Practice Exercises

Exercise A: Basic Equations

1. Solve sin θ = 1/2 for 0° ≤ θ ≤ 360°.

2. Solve cos θ = −1/2 for 0° ≤ θ ≤ 360°.

3. Solve tan θ = √3 for 0° ≤ θ ≤ 360°.

4. Solve sin θ = −1/2 for 0° ≤ θ ≤ 360°.

5. Solve cos θ = √3/2 for 0° ≤ θ ≤ 360°.

6. Solve tan θ = −1 for 0° ≤ θ ≤ 360°.

Exercise B: Extended Intervals

7. Solve sin θ = 1/2 for 0° ≤ θ ≤ 720°.

8. Solve cos θ = 1/2 for −360° ≤ θ ≤ 360°.

9. Solve tan θ = 1 for −180° ≤ θ ≤ 540°.

10. Solve sin θ = −√3/2 for 0° ≤ θ ≤ 720°.

Exercise C: Quadratic Equations

11. Solve 2sin²θ − 1 = 0 for 0° ≤ θ ≤ 360°.

12. Solve 2cos²θ − 1 = 0 for 0° ≤ θ ≤ 360°.

13. Solve 2sin²θ − 3sin θ + 1 = 0 for 0° ≤ θ ≤ 360°.

14. Solve 2cos²θ + cos θ − 1 = 0 for 0° ≤ θ ≤ 360°.

15. Solve sin θ(2sin θ − 1) = 0 for 0° ≤ θ ≤ 360°.

Exercise D: Mixed Equations

16. Solve cos θ(sin θ + 1) = 0 for 0° ≤ θ ≤ 360°.

17. Solve sin²θ = cos²θ for 0° ≤ θ ≤ 360°.

18. Solve 1 + tan²θ = 4 for 0° ≤ θ ≤ 360°.

19. Solve sin θ = 1/2 for 0 ≤ θ ≤ 2π.

20. Solve cos θ = −1/2 for 0 ≤ θ ≤ 2π.

13. Answers and Worked Solutions

Answers to Exercise A

1. Reference angle = 30°. Sine is positive in Quadrants I and II.

θ = 30°, 150°.

2. Reference angle = 60°. Cosine is negative in Quadrants II and III.

θ = 120°, 240°.

3. Reference angle = 60°. Tangent is positive in Quadrants I and III.

θ = 60°, 240°.

4. Reference angle = 30°. Sine is negative in Quadrants III and IV.

θ = 210°, 330°.

5. Reference angle = 30°. Cosine is positive in Quadrants I and IV.

θ = 30°, 330°.

6. Reference angle = 45°. Tangent is negative in Quadrants II and IV.

θ = 135°, 315°.

Worked solutions to Exercise B

7.

sin θ = 1/2.

General solutions:

θ = 30° + 360°n or θ = 150° + 360°n.

Within the interval:

θ = 30°, 150°, 390°, 510°.

8.

cos θ = 1/2.

General solutions:

θ = 60° + 360°n or θ = −60° + 360°n.

Within the interval:

θ = −300°, −60°, 60°, 300°.

9.

tan θ = 1.

General solution:

θ = 45° + 180°n.

Within the interval:

θ = −135°, 45°, 225°, 405°.

10.

Reference angle = 60°.

General solutions:

θ = 240° + 360°n or θ = 300° + 360°n.

Within the interval:

θ = 240°, 300°, 600°, 660°.

Worked solutions to Exercise C

11.

2sin²θ − 1 = 0

sin²θ = 1/2

sin θ = ±√2/2.

θ = 45°, 135°, 225°, 315°.

12.

2cos²θ − 1 = 0

cos²θ = 1/2

cos θ = ±√2/2.

θ = 45°, 135°, 225°, 315°.

13.

2sin²θ − 3sin θ + 1 = 0

(2sin θ − 1)(sin θ − 1) = 0.

sin θ = 1/2 or sin θ = 1.

θ = 30°, 90°, 150°.

14.

2cos²θ + cos θ − 1 = 0

(2cos θ − 1)(cos θ + 1) = 0.

cos θ = 1/2 or cos θ = −1.

θ = 60°, 180°, 300°.

15.

sin θ(2sin θ − 1) = 0.

sin θ = 0 or sin θ = 1/2.

From sin θ = 0: θ = 0°, 180°, 360°.

From sin θ = 1/2: θ = 30°, 150°.

θ = 0°, 30°, 150°, 180°, 360°.

Worked solutions to Exercise D

16.

cos θ(sin θ + 1) = 0.

cos θ = 0 or sin θ = −1.

From cos θ = 0: θ = 90°, 270°.

From sin θ = −1: θ = 270°.

θ = 90°, 270°.

17.

sin²θ = cos²θ.

Dividing by cos²θ, where cos θ ≠ 0:

tan²θ = 1.

tan θ = ±1.

Checking the original equation gives:

θ = 45°, 135°, 225°, 315°.

18.

1 + tan²θ = 4.

tan²θ = 3.

tan θ = ±√3.

θ = 60°, 120°, 240°, 300°.

19.

sin θ = 1/2.

In the interval 0 to 2π:

θ = π/6, 5π/6.

20.

cos θ = −1/2.

In the interval 0 to 2π:

θ = 2π/3, 4π/3.

14. Final Revision Summary

Sine general solution
θ = α + 360°n
or θ = 180° − α + 360°n

Cosine general solution
θ = ±α + 360°n

Tangent general solution
θ = α + 180°n

For radians, replace 360° with 2π and 180° with π.

Here, α is a reference angle appropriate to the equation, and n is any integer. Always select the values that fall within the required interval.

Final examination checklist

• Identify the trigonometric function.
• Find the reference angle.
• Determine the correct quadrants.
• Write all possible solutions.
• Apply the interval limits.
• Check that no valid solutions have been lost.

Looking Ahead: Part 4

In the next lesson, we shall study Compound Angles, Double Angles and Multiple Angles. We will develop the formulae for sin(A ± B), cos(A ± B) and tan(A ± B), and learn how to apply them in mathematical problems.

MATHEMATICS HIP-HOP
Making Mathematics Easier, One Concept at a Time.
A-Level Pure Mathematics — Chapter 04, Part 3

CHAPTER 04: TRIGONOMETRY

PART 4: COMPOUND ANGLES, DOUBLE ANGLES AND MULTIPLE ANGLES

Mathematics HipHop Academy

In this lesson, we extend our knowledge of trigonometric identities by studying angles formed through addition and subtraction. We shall learn how to calculate exact trigonometric values, derive double-angle formulae, establish multiple-angle identities and simplify expressions.

Learning objectives

• Understand compound angles.
• Apply sine, cosine and tangent addition and subtraction formulae.
• Calculate exact values using compound-angle identities.
• Derive and use double-angle formulae.
• Establish selected triple-angle identities.
• Apply these identities in simplification and proofs.

1. Understanding Compound Angles

A compound angle is an angle expressed as the sum or difference of two angles.

A + B    and    A − B

For example:

• 75° = 45° + 30°
• 15° = 45° − 30°
• 105° = 60° + 45°

Instead of calculating the trigonometric ratio of the final angle directly, compound-angle formulae allow us to use the known ratios of the individual angles.

Important warning

Trigonometric functions do not distribute over addition or subtraction.

sin(A + B) ≠ sin A + sin B

cos(A − B) ≠ cos A − cos B

2. The Compound-Angle Formulae

These identities are fundamental tools in advanced trigonometry.

2.1 Sine of the sum of two angles

sin(A + B) = sin A cos B + cos A sin B

2.2 Sine of the difference of two angles

sin(A − B) = sin A cos B − cos A sin B

2.3 Cosine of the sum of two angles

cos(A + B) = cos A cos B − sin A sin B

2.4 Cosine of the difference of two angles

cos(A − B) = cos A cos B + sin A sin B

2.5 Tangent of the sum of two angles

tan(A + B) = (tan A + tan B) / (1 − tan A tan B)

2.6 Tangent of the difference of two angles

tan(A − B) = (tan A − tan B) / (1 + tan A tan B)

Memory guide

• For sine, the sign inside the bracket is retained in the middle of the formula.
• For cosine, the sign inside the bracket changes in the middle of the formula.
• For tangent, use the corresponding numerator sign and the opposite sign in the denominator.

Always write the complete formula before substituting values.

3. Applying Compound-Angle Formulae

Worked Example 1: Find sin 75° exactly.

Solution

First, express 75° as the sum of two familiar angles:

75° = 45° + 30°

Use the sine addition formula:

sin(A + B) = sin A cos B + cos A sin B

Substitute A = 45° and B = 30°:

sin 75° = sin 45° cos 30° + cos 45° sin 30°

Use the exact values:

= (√2/2)(√3/2) + (√2/2)(1/2)

Simplify:

= √6/4 + √2/4

sin 75° = (√6 + √2)/4

Worked Example 2: Find cos 15° exactly.

Solution

Write 15° as a difference:

15° = 45° − 30°

Use the cosine difference formula:

cos(A − B) = cos A cos B + sin A sin B

Substitute:

cos 15° = cos 45° cos 30° + sin 45° sin 30°

= (√2/2)(√3/2) + (√2/2)(1/2)

cos 15° = (√6 + √2)/4

Worked Example 3: Find tan 15° exactly.

Solution

Express 15° as 45° − 30°.

Apply the tangent difference formula:

tan 15° = (tan 45° − tan 30°)/(1 + tan 45° tan 30°)

= (1 − 1/√3)/(1 + 1/√3)

Multiply the numerator and denominator by √3:

= (√3 − 1)/(√3 + 1)

Rationalise the denominator:

= (√3 − 1)² / (3 − 1)

tan 15° = 2 − √3

4. Deriving the Double-Angle Formulae

A double angle occurs when the same angle is added to itself.

A + A = 2A

We obtain the double-angle identities by putting B = A in the compound-angle formulae.

4.1 Double-angle formula for sine

Start with:

sin(A + B) = sin A cos B + cos A sin B

Let B = A:

sin(A + A) = sin A cos A + cos A sin A

sin 2A = 2 sin A cos A

4.2 Double-angle formula for cosine

Start with:

cos(A + B) = cos A cos B − sin A sin B

Let B = A:

cos 2A = cos² A − sin² A

Using the fundamental identity sin² A + cos² A = 1, we can obtain two further forms.

First alternative:

cos² A = 1 − sin² A

cos 2A = 1 − 2 sin² A

Second alternative:

sin² A = 1 − cos² A

cos 2A = 2 cos² A − 1

Remember all three forms of cos 2A:

cos 2A = cos² A − sin² A

cos 2A = 1 − 2 sin² A

cos 2A = 2 cos² A − 1

4.3 Double-angle formula for tangent

Start with the tangent addition formula:

tan(A + B) = (tan A + tan B)/(1 − tan A tan B)

Put B = A:

tan 2A = (tan A + tan A)/(1 − tan² A)

tan 2A = 2 tan A/(1 − tan² A)

This form is valid wherever the expressions involved are defined.

5. Using Double-Angle Formulae

Worked Example 4: If sin A = 3/5 and A is acute, find cos 2A.

Solution

We are given sin A, so choose:

cos 2A = 1 − 2 sin² A

Substitute:

cos 2A = 1 − 2(3/5)²

= 1 − 18/25

cos 2A = 7/25

Worked Example 5: If cos A = 4/5 and A is acute, find sin 2A.

Solution

First, find sin A using the identity:

sin² A + cos² A = 1

sin² A = 1 − 16/25 = 9/25

Since A is acute, sin A is positive.

sin A = 3/5

Now use the double-angle formula:

sin 2A = 2 sin A cos A

= 2 × 3/5 × 4/5

sin 2A = 24/25

Worked Example 6: Given tan A = 1/2, find tan 2A.

Solution

tan 2A = 2 tan A/(1 − tan² A)

= 2(1/2)/(1 − 1/4)

= 1/(3/4)

tan 2A = 4/3

6. Multiple-Angle Formulae: Triple Angles

A triple angle is an angle expressed as three times a given angle.

3A = 2A + A

We can derive triple-angle identities by applying compound-angle formulae to 2A + A.

6.1 Deriving sin 3A

sin 3A = sin(2A + A)

= sin 2A cos A + cos 2A sin A

Substitute the double-angle identities:

= 2 sin A cos² A + (cos² A − sin² A)sin A

= 3 sin A cos² A − sin³ A

Since cos² A = 1 − sin² A:

sin 3A = 3 sin A − 4 sin³ A

6.2 Deriving cos 3A

cos 3A = cos(2A + A)

= cos 2A cos A − sin 2A sin A

= (cos² A − sin² A)cos A − 2 sin² A cos A

= cos³ A − 3 sin² A cos A

Using sin² A = 1 − cos² A:

cos 3A = 4 cos³ A − 3 cos A

6.3 Triple-angle formulae summary

sin 3A = 3 sin A − 4 sin³ A

cos 3A = 4 cos³ A − 3 cos A

7. Simplification Using Double-Angle Identities

Double-angle formulae can help us replace expressions involving 2A with expressions involving A, or the reverse.

Worked Example 7: Simplify (1 − cos 2A)/sin 2A.

Solution

Use:

1 − cos 2A = 2 sin² A

sin 2A = 2 sin A cos A

Substitute:

(1 − cos 2A)/sin 2A = 2 sin² A/(2 sin A cos A)

= sin A/cos A

= tan A

Answer: tan A, wherever the original expression is defined.

Worked Example 8: Simplify (sin 2A)/(1 + cos 2A).

Solution

sin 2A = 2 sin A cos A

1 + cos 2A = 1 + 2 cos² A − 1 = 2 cos² A

(sin 2A)/(1 + cos 2A) = (2 sin A cos A)/(2 cos² A)

= tan A

The expression is defined when cos A is non-zero.

8. Proving Trigonometric Identities

Compound-angle and double-angle formulae are frequently used in identity proofs.

Good examination practice

• Begin with one side of the identity, usually the more complicated side.
• Replace double-angle expressions with suitable identities.
• Use the fundamental identity where necessary.
• Simplify step by step until the other side is obtained.
• Do not assume the identity is true by moving terms across the equality sign as if solving an equation.

Worked Example 9: Prove that

(1 − cos 2A)/sin 2A = tan A

Proof

Start with the left-hand side:

LHS = (1 − cos 2A)/sin 2A

Apply the double-angle formulae:

= [1 − (1 − 2 sin² A)]/[2 sin A cos A]

= 2 sin² A/(2 sin A cos A)

= sin A/cos A

= tan A = RHS

Therefore, the identity is proved wherever the original expression is defined.

9. Choosing the Correct Formula

Information given	Useful formula
sin A and cos A	sin 2A = 2 sin A cos A
sin A only	cos 2A = 1 − 2 sin² A
cos A only	cos 2A = 2 cos² A − 1
tan A only	tan 2A = 2 tan A/(1 − tan² A)
Two different angles	Compound-angle formulae

Examination tip

Do not use a formula simply because you remember it first. Examine the information given and select the identity that requires the fewest extra calculations.

10. Common Errors to Avoid

1. Writing sin(A + B) = sin A + sin B.
2. Forgetting that the middle sign in the cosine formula changes.
3. Confusing sin 2A with 2 sin A.
4. Using the wrong form of cos 2A when simplifying.
5. Forgetting to establish the missing trigonometric ratio when it is needed.
6. Ignoring restrictions where a denominator is zero.
7. Giving decimal answers when an exact surd is required.

11. Practice Exercises

Instructions: Attempt all questions before opening the answers. Give exact values where possible and show all working.

Exercise A: Compound Angles

1. Express 105° as the sum of two standard angles.
2. Find the exact value of sin 75°.
3. Find the exact value of cos 15°.
4. Find the exact value of tan 15°.
5. Given sin A = 3/5, cos A = 4/5, sin B = 5/13 and cos B = 12/13, find sin(A + B).

Exercise B: Double Angles

6. Derive the formula sin 2A = 2 sin A cos A from the compound-angle formula.
7. Write the three equivalent forms of cos 2A.
8. If sin A = 5/13 and A is acute, find cos 2A.
9. If cos A = 12/13 and A is acute, find sin 2A.
10. If tan A = 1/3, find tan 2A.

Exercise C: Simplification and Proof

11. Simplify (1 − cos 2A)/sin 2A.
12. Simplify sin 2A/(1 + cos 2A).
13. Prove that (1 + cos 2A)/sin 2A = cot A, wherever defined.
14. Prove that sin 2A/(1 − cos 2A) = cot A, wherever defined.

Exercise D: Multiple Angles

15. Show that sin 3A = 3 sin A − 4 sin³ A.
16. Show that cos 3A = 4 cos³ A − 3 cos A.
17. Find sin 3A when sin A = 1/2 and A = 30°.

12. Answers and Selected Solutions

Open each section after attempting the questions.

Exercise A: Answers

1. 105° = 60° + 45°.
2. sin 75° = (√6 + √2)/4.
3. cos 15° = (√6 + √2)/4.
4. tan 15° = 2 − √3.
5. sin(A + B) = sin A cos B + cos A sin B = (3/5)(12/13) + (4/5)(5/13) = 36/65 + 20/65 = 56/65.

Exercise B: Answers

6. sin(A + A) = sin A cos A + cos A sin A = 2 sin A cos A.
7. cos 2A = cos² A − sin² A = 1 − 2 sin² A = 2 cos² A − 1.
8. cos 2A = 1 − 2(5/13)² = 1 − 50/169 = 119/169.
9. First, sin A = 5/13 because A is acute. Therefore sin 2A = 2(5/13)(12/13) = 120/169.
10. tan 2A = [2(1/3)]/[1 − (1/3)²] = (2/3)/(8/9) = 3/4.

Exercise C: Answers

11. Using 1 − cos 2A = 2 sin² A and sin 2A = 2 sin A cos A, the result is tan A.
12. Using 1 + cos 2A = 2 cos² A and sin 2A = 2 sin A cos A, the result is tan A.
13. LHS = (1 + cos 2A)/sin 2A = 2 cos² A/(2 sin A cos A) = cos A/sin A = cot A.
14. LHS = sin 2A/(1 − cos 2A) = 2 sin A cos A/(2 sin² A) = cos A/sin A = cot A.

Exercise D: Answers

15. Use sin 3A = sin(2A + A) = sin 2A cos A + cos 2A sin A. Substituting the double-angle identities and simplifying gives 3 sin A − 4 sin³ A.
16. Use cos 3A = cos(2A + A) = cos 2A cos A − sin 2A sin A. Substituting and simplifying gives 4 cos³ A − 3 cos A.
17. Since A = 30°, sin 3A = sin 90° = 1. Alternatively, 3(1/2) − 4(1/2)³ = 3/2 − 1/2 = 1.

13. Chapter Summary

In this part, we have studied the relationships between compound angles, double angles and multiple angles.

sin(A ± B) = sin A cos B ± cos A sin B

cos(A ± B) = cos A cos B ∓ sin A sin B

tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B)

sin 2A = 2 sin A cos A

cos 2A = cos² A − sin² A

tan 2A = 2 tan A/(1 − tan² A)

sin 3A = 3 sin A − 4 sin³ A

cos 3A = 4 cos³ A − 3 cos A

Final Message

Compound-angle and double-angle identities are important tools in advanced mathematics. Mastering them will strengthen your ability to calculate exact values, simplify expressions, prove identities and solve more demanding trigonometric problems.

Mathematics HipHop Academy — Making Mathematics Easier, One Concept at a Time.

CHAPTER 04: TRIGONOMETRY

PART 5: TRIGONOMETRIC APPLICATIONS AND EXAMINATION PRACTICE

Mathematics HipHop Academy

In this final part of Trigonometry, we apply the ideas studied in Parts 1–4 to practical and examination problems. We shall investigate heights and distances, angles of elevation and depression, bearings, non-right-angled triangles, three-dimensional applications and mixed examination questions.

Learning objectives

• Solve practical problems involving angles of elevation and depression.
• Interpret and calculate three-figure bearings.
• Apply the sine rule and cosine rule.
• Calculate the area of a triangle using trigonometry.
• Solve selected three-dimensional trigonometry problems.
• Choose appropriate methods for mixed examination questions.
• Present accurate mathematical working and answers with suitable units.

1. Trigonometry in Real-Life Applications

Trigonometry is useful whenever measurements involve angles, distances and heights. It is applied in surveying, construction, navigation, architecture, engineering and astronomy.

Some common applications include:

• Finding the height of a building without measuring it directly.
• Calculating the distance across a river.
• Determining the direction of a ship or aircraft.
• Finding inaccessible distances using measured angles.
• Calculating unknown sides and angles in triangular structures.

Important principle

Most application questions become easier when the information is converted into a clear mathematical diagram. Draw the triangle, label the known measurements and identify the unknown quantity before choosing a formula.

2. Angles of Elevation and Depression

2.1 Angle of elevation

The angle of elevation is the angle between the horizontal line through an observer and the line of sight to an object above the observer.

For example, when a person standing on level ground looks upwards towards the top of a tower, the angle between the horizontal and the line of sight is the angle of elevation.

2.2 Angle of depression

The angle of depression is the angle between the horizontal line through an observer and the line of sight to an object below the observer.

Relationship between elevation and depression

When two horizontal lines are parallel, the angle of depression from the higher point is equal to the corresponding angle of elevation from the lower point. This follows from the properties of alternate angles.

2.3 Choosing the correct trigonometric ratio

For a right-angled triangle, remember SOH–CAH–TOA:

sin θ = Opposite / Hypotenuse

cos θ = Adjacent / Hypotenuse

tan θ = Opposite / Adjacent

Worked Example 1: Height of a tower

A student stands 40 m from the foot of a vertical tower. The angle of elevation of the top of the tower is 32°. Find the height of the tower, correct to one decimal place.

Solution

Let the height of the tower be h.

The opposite side is h and the adjacent side is 40 m. Therefore, use tangent.

tan 32° = h / 40

Make h the subject:

h = 40 tan 32°

h = 24.995... m

Answer: 25.0 m

Worked Example 2: Distance from a cliff

A vertical cliff is 60 m high. The angle of depression from the top of the cliff to a boat is 12°. Find the horizontal distance of the boat from the foot of the cliff, to the nearest metre.

Solution

The angle of elevation from the boat is also 12°.

Let the horizontal distance be d.

tan 12° = 60 / d

d = 60 / tan 12°

d = 282.3... m

Answer: 282 m

3. Bearings and Navigation

A bearing describes the direction of one point from another.

True bearings are measured clockwise from the north direction and are written using three figures.

Direction	Bearing
North	000° or 360°
East	090°
South	180°
West	270°

Examples of three-figure bearings include 025°, 090°, 145° and 315°.

3.1 Back bearings

A back bearing gives the direction of the starting point from the destination.

If the bearing is less than 180°:
Back bearing = Bearing + 180°

If the bearing is 180° or greater:
Back bearing = Bearing − 180°

Worked Example 3: Finding a back bearing

The bearing of B from A is 072°. Find the bearing of A from B.

Solution

Back bearing = 072° + 180°

= 252°

Answer: 252°

Worked Example 4: Distance travelled on a bearing

A person walks 200 m on a bearing of 060° from point A to point B. Calculate how far east and how far north B is from A.

Solution

Resolve the 200 m displacement into eastward and northward components.

Eastward distance = 200 sin 60°

= 173.2 m

Northward distance = 200 cos 60°

= 100 m

Answer: 173.2 m east and 100 m north.

Examination reminder

• Always draw a north arrow at the relevant observation point.
• Measure bearings clockwise from north.
• Write three figures, for example 045° rather than 45°.
• Check whether the question asks for a bearing from A to B or from B to A.

4. The Sine Rule

The sine rule is used to solve non-right-angled triangles when a known side and its opposite angle form a pair, together with another suitable measurement.

For triangle ABC, let side a be opposite angle A, side b opposite angle B and side c opposite angle C.

a / sin A = b / sin B = c / sin C

An equivalent form is:

sin A / a = sin B / b = sin C / c

4.1 When to use the sine rule

• When two angles and one side are known.
• When two sides and a non-included opposite angle pair are known.

Worked Example 5: Finding an unknown side

In triangle ABC, angle A = 40°, angle B = 65° and side a = 12 cm. Find side b.

Solution

a / sin A = b / sin B

12 / sin 40° = b / sin 65°

b = 12 sin 65° / sin 40°

b = 16.9 cm (approximately)

Answer: 16.9 cm

Worked Example 6: Finding an angle

In a triangle, side a = 8 cm, side b = 10 cm and angle A = 30°. Find angle B, taking the acute solution.

Solution

a / sin A = b / sin B

8 / sin 30° = 10 / sin B

sin B = 10 sin 30° / 8

sin B = 5/8

B = sin⁻¹(5/8)

B = 38.7°

Answer: 38.7° for the acute solution.

Important: The ambiguous case

When using the sine rule to find an angle, a sine value between 0 and 1 may correspond to two angles between 0° and 180°: θ and 180° − θ. Check the triangle's other angles and the information in the question before selecting the answer.

5. The Cosine Rule

The cosine rule is used in non-right-angled triangles when two sides and their included angle are known, or when all three sides are known.

a² = b² + c² − 2bc cos A

b² = a² + c² − 2ac cos B

c² = a² + b² − 2ab cos C

To calculate an angle, rearrange the formula:

cos A = (b² + c² − a²) / 2bc

5.1 When to use the cosine rule

• When two sides and the included angle are known.
• When all three sides are known and an angle is required.

Worked Example 7: Finding a side

Two sides of a triangle are 5 cm and 8 cm. The included angle is 60°. Find the third side.

Solution

Let the unknown side be c.

c² = 5² + 8² − 2(5)(8)cos 60°

c² = 25 + 64 − 80(0.5)

c² = 49

c = 7 cm

Answer: 7 cm

Worked Example 8: Finding an angle

A triangle has sides 7 cm, 8 cm and 9 cm. Find the angle opposite the side of length 9 cm.

Solution

cos C = (7² + 8² − 9²)/(2 × 7 × 8)

= (49 + 64 − 81)/112

= 32/112 = 2/7

C = cos⁻¹(2/7)

C = 73.4° approximately

Answer: 73.4°

6. Area of a Triangle Using Trigonometry

When two sides and the included angle are known, the area of a triangle can be calculated without finding its perpendicular height.

Area = ½ab sin C

Similarly:

Area = ½bc sin A

Area = ½ac sin B

Worked Example 9: Area of a triangle

Find the area of a triangle whose two sides are 12 cm and 9 cm and whose included angle is 40°.

Solution

Area = ½ab sin C

= ½ × 12 × 9 × sin 40°

= 54 × 0.6428

= 34.7 cm² approximately

Answer: 34.7 cm²

7. Three-Dimensional Trigonometry

Some problems involve objects in three dimensions, such as buildings, rectangular rooms, cuboids and pyramids.

The main idea is to identify a suitable right-angled triangle within the three-dimensional figure. Often, a horizontal diagonal is found first, followed by a second triangle involving the vertical height.

Method for three-dimensional problems

1. Draw or study the three-dimensional object carefully.
2. Identify the horizontal plane and any vertical heights.
3. Find a suitable diagonal or horizontal distance using Pythagoras' theorem if necessary.
4. Identify the right-angled triangle containing the required angle or length.
5. Apply the appropriate trigonometric ratio.
6. State the answer with the correct unit.

Worked Example 10: Angle of elevation in a rectangular field

A rectangular field measures 30 m by 40 m. A vertical pole of height 20 m stands at one corner. Find the angle of elevation from the opposite corner of the field to the top of the pole.

Solution

First, find the diagonal of the rectangular field.

d² = 30² + 40²

d² = 900 + 1600 = 2500

d = 50 m

Now use the right-angled triangle formed by the diagonal and the pole.

tan θ = 20 / 50

θ = tan⁻¹(0.4)

θ = 21.8° approximately

Answer: 21.8°

8. Choosing the Correct Method

Given information	Suitable method
Right-angled triangle	SOH–CAH–TOA
Two angles and one side	Sine rule
Two sides and a non-included opposite angle	Sine rule
Two sides and included angle	Cosine rule
Three sides of a triangle	Cosine rule
Two sides and included angle for area	Area = ½ab sin C
Height and horizontal distance	Tangent ratio
Direction and distance	Bearings and trigonometric components

9. Examination Techniques and Common Mistakes

9.1 Read the question carefully

Identify what is given and what must be found. Underline important measurements and units.

9.2 Draw a diagram

Even when a diagram is not provided, sketching one can help you understand the situation and identify the correct angle.

9.3 Choose the correct formula

Do not substitute numbers until you have identified the appropriate formula.

9.4 Use the calculator correctly

Check whether the question uses degrees or radians. Set your calculator to the appropriate mode before calculating.

9.5 Avoid premature rounding

Keep several decimal places during intermediate calculations. Round only the final answer to the requested accuracy.

9.6 Include units

Lengths should have units such as cm, m or km. Areas should have square units such as cm² or m². Bearings should be written as three-figure angles.

Common mistakes

• Measuring an angle of elevation from the vertical instead of the horizontal.
• Measuring bearings anticlockwise instead of clockwise from north.
• Using the sine rule when the cosine rule is required.
• Forgetting to use the included angle in the cosine rule.
• Confusing a length with an area.
• Rounding too early.
• Giving an angle without checking whether it is geometrically possible.

10. Mixed Examination Practice

Instructions: Attempt all questions. Show complete working, use suitable units and give answers to the stated accuracy.

Section A: Right-Angled Applications

1. A ladder 10 m long rests against a vertical wall. The foot of the ladder is 6 m from the wall. Calculate the height reached by the ladder.
2. A tree casts a shadow 15 m long when the angle of elevation of the sun is 38°. Find the height of the tree, correct to one decimal place.
3. From a point 75 m away from the base of a building, the angle of elevation to its top is 28°. Find the height of the building.
4. An aircraft is observed at an angle of elevation of 25° from a point on level ground. If its vertical height is 2,000 m, find its horizontal distance from the observer.

Section B: Bearings

5. The bearing of B from A is 125°. Find the bearing of A from B.
6. A ship travels 10 km due east and then 10 km due north. Find its distance from its starting point and the bearing of its final position from the starting point.
7. A person walks 8 km on a bearing of 030°. Find the northward and eastward components of the displacement.

Section C: Sine Rule and Cosine Rule

8. In triangle ABC, A = 45°, B = 60° and side a = 12 cm. Find side b.
9. Two sides of a triangle are 9 cm and 12 cm, and their included angle is 60°. Find the third side.
10. A triangle has sides 5 cm, 6 cm and 7 cm. Find the angle opposite the longest side.
11. In a triangle, two sides are 10 cm and 14 cm and the included angle is 35°. Find its area.

Section D: Mixed Applications

12. A rectangular garden measures 24 m by 32 m. Find its diagonal.
13. A pole stands vertically at one corner of a rectangular field measuring 20 m by 21 m. If the pole is 15 m high, find the angle of elevation of its top from the opposite corner.
14. Prove that (1 − cos 2A)/sin 2A = tan A, wherever defined.
15. Solve 2 sin x = 1 for 0° ≤ x ≤ 360°.

11. Answers and Selected Solutions

Attempt the questions before opening the answer sections.

Section A: Right-Angled Applications

1. Height² = 10² − 6² = 100 − 36 = 64. Height = 8 m.
2. tan 38° = h/15. Therefore h = 15 tan 38° = 11.7 m approximately.
3. tan 28° = h/75. Therefore h = 75 tan 28° = 39.9 m approximately.
4. tan 25° = 2000/d. Therefore d = 2000/tan 25° = 4,289 m approximately.

Section B: Bearings

5. Back bearing = 125° + 180° = 305°.
6. Distance = √(10² + 10²) = √200 = 14.14 km approximately. Bearing = 045°.
7. Northward component = 8 cos 30° = 6.93 km. Eastward component = 8 sin 30° = 4 km.

Section C: Sine Rule and Cosine Rule

8. b/sin 60° = 12/sin 45°. Therefore b = 12 sin 60°/sin 45° = 14.7 cm approximately.
9. c² = 9² + 12² − 2(9)(12)cos 60° = 81 + 144 − 108 = 117. Therefore c = √117 = 10.8 cm approximately.
10. Let C be opposite the 7 cm side. cos C = (5² + 6² − 7²)/(2 × 5 × 6) = 12/60 = 0.2. Therefore C = cos⁻¹(0.2) = 78.5° approximately.
11. Area = ½ × 10 × 14 × sin 35° = 40.2 cm² approximately.

Section D: Mixed Applications

12. Diagonal² = 24² + 32² = 576 + 1024 = 1600. Diagonal = 40 m.
13. Horizontal diagonal = √(20² + 21²) = √841 = 29 m. tan θ = 15/29. Therefore θ = tan⁻¹(15/29) = 27.3° approximately.
14. Starting from the left-hand side: (1 − cos 2A)/sin 2A = [1 − (1 − 2 sin² A)]/[2 sin A cos A] = 2 sin² A/(2 sin A cos A) = sin A/cos A = tan A. Hence proved.
15. sin x = 1/2. In the interval 0° to 360°, the solutions are x = 30° and x = 150°.

12. Final Chapter Revision: Trigonometry Formula Sheet

The following formulae summarise the main ideas covered throughout Chapter 04.

Basic trigonometric ratios

sin θ = Opposite / Hypotenuse

cos θ = Adjacent / Hypotenuse

tan θ = Opposite / Adjacent

Fundamental identity

sin² θ + cos² θ = 1

Compound-angle identities

sin(A ± B) = sin A cos B ± cos A sin B

cos(A ± B) = cos A cos B ∓ sin A sin B

tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B)

Double-angle identities

sin 2A = 2 sin A cos A

cos 2A = cos² A − sin² A

cos 2A = 1 − 2 sin² A

cos 2A = 2 cos² A − 1

tan 2A = 2 tan A/(1 − tan² A)

Triangle formulae

a/sin A = b/sin B = c/sin C

a² = b² + c² − 2bc cos A

Area = ½ab sin C

Bearings

Back bearing = Bearing ± 180°

13. Chapter 04: Final Summary

Congratulations on completing Chapter 04: Trigonometry.

Across the five parts, we have studied:

1. Part 1: Trigonometric Functions, Angles and Graphs.
2. Part 2: Trigonometric Identities and Formulae.
3. Part 3: Trigonometric Equations and General Solutions.
4. Part 4: Compound Angles, Double Angles and Multiple Angles.
5. Part 5: Trigonometric Applications and Examination Practice.

Final Message

Trigonometry is not simply about memorising formulae. It is about recognising relationships, interpreting diagrams, choosing appropriate methods and applying mathematical reasoning accurately.

Practise regularly, show every important step and always check whether your final answer makes sense in the context of the question.

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