Double Angle Formulae

Learn to apply double angle identities for sin 2A, cos 2A, and tan 2A

SPEC 5.6 • TRIGONOMETRY

Learning Objectives

By the end of this lesson you will be able to

Know and apply double angle formulae for sin 2A, cos 2A, and tan 2A

Choose the appropriate form of cos 2A based on given information

Recognise and simplify expressions using double angle patterns

Apply double angle formulae to find exact values and prove identities

Formula Booklet

All double angle formulae are given in the formula booklet — focus on knowing when and how to apply them.

The sin 2A Formula

Formula

sin 2A = 2 sin A cos A

Pattern Recognition

Whenever you see 2 sin A cos A, replace it with sin 2A (and vice versa).

ExpressionSimplifies To

2 sin 30° cos 30°sin 60° = √3/2

2 sin x cos xsin 2x

Remember

You need BOTH sin A AND cos A to use this formula.

The cos 2A Formulae

There are three equivalent forms of cos 2A. Choose based on what information you have:

Form 1cos 2A = cos²A − sin²A

Use when you have BOTH sin A and cos A

Form 2cos 2A = 2cos²A − 1

Use when you ONLY have cos A

Form 3cos 2A = 1 − 2sin²A

Use when you ONLY have sin A

Choosing the Right Form

Look at what information you're given. If you only have sin A, use Form 3. If you only have cos A, use Form 2. This avoids unnecessary calculation!

Rearranged Forms (for integration)

cos²A = ½(1 + cos 2A) | sin²A = ½(1 − cos 2A)

The tan 2A Formula

Formula

tan 2 A = \frac{2 tan A}{1 - tan ^{2} A}

When to Use

Use this formula when you're given tan A (or can easily find it).

Caution

This formula is undefined when tan²A = 1 (i.e., when A = 45°, 135°, etc.) because the denominator becomes zero.

Formulae Summary

sin 2A

sin 2A = 2 sin A cos A

cos 2A = cos²A − sin²Acos 2A = 2cos²A − 1cos 2A = 1 − 2sin²A

tan 2A

tan 2 A = \frac{2 tan A}{1 - tan ^{2} A}

Key Patterns to Spot

2 sin x cos x → sin 2x | cos²x − sin²x → cos 2x | 2cos²x − 1 → cos 2x | 1 − 2sin²x → cos 2x

Worked Example01

Given that sin A = 3/5 where A is acute, find the exact value of sin 2A.

1

Use sin 2A = 2 sin A cos A — need to find cos A first

2

Since A is acute and sin A = 3/5: cos²A = 1 − sin²A = 1 − 9/25 = 16/25

3

cos A = 4/5 (positive since A is acute)

4

sin 2A = 2 × (3/5) × (4/5) = 2 × 12/25 = 24/25

Answersin 2A = 24/25

Worked Example02

Given that cos A = 4/5 where A is acute, find the exact value of cos 2A.

1

Choose Form 2: cos 2A = 2cos²A − 1 (we only have cos A)

2

cos 2A = 2 × (4/5)² − 1

3

cos 2A = 2 × 16/25 − 1 = 32/25 − 25/25

Answercos 2A = 7/25

Worked Example03

Find the exact value of 2 sin 15° cos 15°.

1

Recognise the pattern: 2 sin A cos A = sin 2A

2

Here A = 15°

3

So 2 sin 15° cos 15° = sin(2 × 15°) = sin 30°

Answer1/2

Worked Example04

Find the exact value of cos²15° − sin²15°.

1

Recognise this is Form 1: cos²A − sin²A = cos 2A

2

Here A = 15°

3

So cos²15° − sin²15° = cos(2 × 15°) = cos 30°

Answer√3/2

Worked Example05

Find the exact value of 1 − 2sin²22.5°.

1

Recognise this is Form 3: 1 − 2sin²A = cos 2A

2

Here A = 22.5°

3

So 1 − 2sin²22.5° = cos(2 × 22.5°) = cos 45°

Answer√2/2

Worked Example06

Prove that $\frac{1 - cos 2 A}{sin 2 A} = tan A$

1

LHS = (1 − cos 2A) / sin 2A

2

Use cos 2A = 1 − 2sin²A and sin 2A = 2 sin A cos A

3

= (1 − (1 − 2sin²A)) / (2 sin A cos A)

4

= 2sin²A / (2 sin A cos A) = sin A / cos A

5

= tan A = RHS ✓

AnswerQED

Practice

Question: Given that tan A = 2 where A is acute, find the exact value of sin 2A.

Hint: First find sin A and cos A using a right triangle.

1

If tan A = 2 = opposite/adjacent, use right triangle with opp = 2, adj = 1

2

Hypotenuse = √(4 + 1) = √5

3

sin A = 2/√5, cos A = 1/√5

4

sin 2A = 2 sin A cos A = 2 × (2/√5) × (1/√5) = 4/5

Answersin 2A = 4/5

Practice

Question: Find the exact value of 2cos²15° − 1.

Hint: Which form of cos 2A is this?

1

Recognise: 2cos²A − 1 = cos 2A (Form 2)

2

Here A = 15°

3

So 2cos²15° − 1 = cos(2 × 15°) = cos 30°

Answer√3/2

Summary

Key Formulae

FormulaWhen to Use

sin 2A = 2 sin A cos AWhen you have both sin A and cos A

cos 2A = cos²A − sin²AWhen you have both sin A and cos A

cos 2A = 2cos²A − 1When you only have cos A

cos 2A = 1 − 2sin²AWhen you only have sin A

tan 2A = 2tan A / (1 − tan²A)When you have tan A

Coming Next

R-formula (harmonic form), solving trigonometric equations, and proving more complex identities.