Chapter 04: Trigonometry – Part 4: Compound Angles, Double Angles and Multiple Angles

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.

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