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Compound & Double Angle Formulae

CSEC Additional Mathematics Essential Knowledge: Compound angle and double angle formulae are powerful tools that allow us to simplify trigonometric expressions, solve equations, and find exact values for non-standard angles. These formulae connect trigonometric functions of sums and differences of angles to products and powers of trigonometric functions.

Key Concept: Compound angle formulae express trigonometric functions of sums or differences of angles (A ± B) in terms of trigonometric functions of A and B separately. Double angle formulae are special cases where A = B, giving expressions for functions of 2A in terms of functions of A.

Part 1: Compound Angle Formulae

The Six Fundamental Formulae

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What Are Compound Angles?

Compound angles are angles that can be expressed as the sum or difference of two angles: A + B or A – B. For example, 75° = 45° + 30°, or 15° = 45° – 30°.

Sine Formulae

\[ \sin(A \pm B) \]

\[ \sin(A+B) = \sin A \cos B + \cos A \sin B \]

\[ \sin(A-B) = \sin A \cos B – \cos A \sin B \]

Cosine Formulae

\[ \cos(A \pm B) \]

\[ \cos(A+B) = \cos A \cos B – \sin A \sin B \]

\[ \cos(A-B) = \cos A \cos B + \sin A \sin B \]

Tangent Formulae

\[ \tan(A \pm B) \]

\[ \tan(A+B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \]

\[ \tan(A-B) = \frac{\tan A – \tan B}{1 + \tan A \tan B} \]

📝 Example 1: Finding Exact Values

Find the exact value of sin 75° using compound angle formula.

Express as sum: 75° = 45° + 30°

Use formula: sin(45° + 30°) = sin45° cos30° + cos45° sin30°

Substitute known values:
sin45° = \(\frac{\sqrt{2}}{2}\), cos45° = \(\frac{\sqrt{2}}{2}\)
cos30° = \(\frac{\sqrt{3}}{2}\), sin30° = \(\frac{1}{2}\)

Calculate:
\(\frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}\)

Simplify: \(\frac{\sqrt{6} + \sqrt{2}}{4}\)

📝 Example 2: Cosine Difference

Find the exact value of cos 15° using compound angle formula.

Express as difference: 15° = 45° – 30°

Use formula: cos(45° – 30°) = cos45° cos30° + sin45° sin30°

Substitute:
cos45° = \(\frac{\sqrt{2}}{2}\), sin45° = \(\frac{\sqrt{2}}{2}\)
cos30° = \(\frac{\sqrt{3}}{2}\), sin30° = \(\frac{1}{2}\)

Calculate:
\(\frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}\)

Simplify: \(\frac{\sqrt{6} + \sqrt{2}}{4}\)

Memory Aid: Remember “S C, C S” for sine: Sine formula has Sine, Cosine then Cosine, Sine. For cosine: “C C, S S” – Cosine formula has Cosine, Cosine then Sine, Sine with a sign change.

Part 2: Double Angle Formulae

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Special Case: When A = B

Deriving Double Angle Formulae

Double angle formulae are obtained by setting B = A in the compound angle formulae:

Starting point: \(\sin(A+B) = \sin A \cos B + \cos A \sin B\)
Set B = A: \(\sin(A+A) = \sin A \cos A + \cos A \sin A\)

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\(\sin 2A = \sin A \cos A + \sin A \cos A = 2 \sin A \cos A\)

Sine Double Angle

\[ \sin 2A = 2 \sin A \cos A \]

Derivation: From sin(A+B) with B = A

Cosine Double Angle

\[ \cos 2A = \cos^2 A – \sin^2 A \]

\[ \cos 2A = 2\cos^2 A – 1 \]

\[ \cos 2A = 1 – 2\sin^2 A \]

Three equivalent forms

Tangent Double Angle

\[ \tan 2A = \frac{2\tan A}{1 – \tan^2 A} \]

Derivation: From tan(A+B) with B = A

Note: Valid when \(\tan A \neq \pm 1\) and \(\tan A\) exists

📝 Example 3: Using Double Angle Formulae

If \(\sin A = \frac{3}{5}\) and A is acute, find the exact value of sin 2A and cos 2A.

Find cos A: Using \(\sin^2 A + \cos^2 A = 1\)
\(\left(\frac{3}{5}\right)^2 + \cos^2 A = 1 \Rightarrow \frac{9}{25} + \cos^2 A = 1\)
\(\cos^2 A = 1 – \frac{9}{25} = \frac{16}{25} \Rightarrow \cos A = \frac{4}{5}\) (positive since A acute)

sin 2A: \(\sin 2A = 2 \sin A \cos A = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25}\)

cos 2A: \(\cos 2A = \cos^2 A – \sin^2 A = \left(\frac{4}{5}\right)^2 – \left(\frac{3}{5}\right)^2 = \frac{16}{25} – \frac{9}{25} = \frac{7}{25}\)

📝 Example 4: Choosing the Right Form of cos 2A

Express \(\cos 2A\) in terms of \(\sin A\) only.

From the three forms:
1. \(\cos 2A = \cos^2 A – \sin^2 A\)
2. \(\cos 2A = 2\cos^2 A – 1\)
3. \(\cos 2A = 1 – 2\sin^2 A\)

We want in terms of sin A only: Use \(\cos 2A = 1 – 2\sin^2 A\)

Alternatively: If we had \(\cos 2A\) in terms of cos A only, we would use \(\cos 2A = 2\cos^2 A – 1\)

Part 3: Applications and Problem Solving

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CSEC-Style Problems

Finding Exact Values

Express angles like 75°, 105°, 15° as sums/differences of standard angles (30°, 45°, 60°) to find exact values.

Simplifying Expressions

Use formulae to simplify expressions like sin 3A (as sin(2A+A)) or cos 4A (as cos(2×2A)).

Solving Equations

Solve trigonometric equations like sin 2x = sin x or cos 2x + sin x = 0.

Proving Identities

Prove identities like \(\frac{\sin 2A}{1+\cos 2A} = \tan A\) using double angle formulae.

📝 Example 5: Solving Trigonometric Equations

Solve the equation \(\sin 2x = \sin x\) for \(0 \leq x \leq 2\pi\).

Use double angle: \(\sin 2x = 2 \sin x \cos x\)

Rewrite equation: \(2 \sin x \cos x = \sin x\)

Rearrange: \(2 \sin x \cos x – \sin x = 0\)

Factor: \(\sin x (2 \cos x – 1) = 0\)

Solve:
Case 1: \(\sin x = 0 \Rightarrow x = 0, \pi, 2\pi\)
Case 2: \(2 \cos x – 1 = 0 \Rightarrow \cos x = \frac{1}{2} \Rightarrow x = \frac{\pi}{3}, \frac{5\pi}{3}\)

Solutions: \(x = 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}, 2\pi\)

📝 Example 6: Proving an Identity (Past Paper Style)

Prove that \(\frac{\sin 2A}{1+\cos 2A} = \tan A\).

Left side: \(\frac{\sin 2A}{1+\cos 2A}\)

Apply double angle formulae:
\(\sin 2A = 2 \sin A \cos A\)
\(\cos 2A = 2\cos^2 A – 1\)

Substitute: \(\frac{2 \sin A \cos A}{1 + (2\cos^2 A – 1)}\)

Simplify denominator: \(1 + 2\cos^2 A – 1 = 2\cos^2 A\)

Simplify fraction: \(\frac{2 \sin A \cos A}{2\cos^2 A} = \frac{\sin A}{\cos A} = \tan A\)

Conclusion: LHS = RHS, identity proven.

Real-World Applications:

Physics: Wave interference, superposition of waves, alternating current circuits

Engineering: Signal processing, vibration analysis, electrical engineering

Computer Graphics: 3D rotations, animation transformations

Architecture: Structural design, angle calculations in construction

Navigation: Calculating bearings and distances using trigonometry

Part 4: Half-Angle Formulae and Extensions

Related Formulae

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Deriving Half-Angle Formulae

Half-angle formulae can be derived from double angle formulae by substituting \(A = \frac{\theta}{2}\), so \(2A = \theta\):

From \(\cos 2A = 1 – 2\sin^2 A\):
\(\cos \theta = 1 – 2\sin^2\left(\frac{\theta}{2}\right)\)
Rearranging: \(\sin\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1-\cos\theta}{2}}\)

From \(\cos 2A = 2\cos^2 A – 1\):
\(\cos \theta = 2\cos^2\left(\frac{\theta}{2}\right) – 1\)
Rearranging: \(\cos\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1+\cos\theta}{2}}\)

For tangent: \(\tan\left(\frac{\theta}{2}\right) = \frac{\sin\theta}{1+\cos\theta} = \frac{1-\cos\theta}{\sin\theta}\)

📝 Example 7: Triple Angle Formulae

Express sin 3A in terms of sin A only.

Write as compound angle: sin 3A = sin(2A + A)

Apply sin(A+B) formula: sin(2A + A) = sin 2A cos A + cos 2A sin A

Use double angle formulae:
sin 2A = 2 sin A cos A
cos 2A = 1 – 2 sin² A

Substitute: (2 sin A cos A) cos A + (1 – 2 sin² A) sin A

Simplify: 2 sin A cos² A + sin A – 2 sin³ A

Use identity: cos² A = 1 – sin² A
= 2 sin A (1 – sin² A) + sin A – 2 sin³ A

Final: = 2 sin A – 2 sin³ A + sin A – 2 sin³ A = 3 sin A – 4 sin³ A

Memory Aids for Compound Angle Formulae

For sine: “Sin Cos, Cos Sin” – Sine formula has Sine then Cosine plus Cosine then Sine.

For cosine: “Cos Cos, Sin Sin” – Cosine formula has Cosine then Cosine minus Sine then Sine.

Sign rule: Sine keeps the same sign (sin(A+B) has +, sin(A-B) has -). Cosine changes the sign (cos(A+B) has -, cos(A-B) has +).

Double angles: sin 2A = 2 sin A cos A (like “twice sin cos”)

Comparison Table: Key Formulae

Type	Formula	Special Notes	Common Uses
sin(A+B)	sin A cos B + cos A sin B	Sign between terms matches original	Finding sin of sums, expanding expressions
sin(A-B)	sin A cos B – cos A sin B	Sign between terms matches original	Finding sin of differences
cos(A+B)	cos A cos B – sin A sin B	Sign between terms opposite to original	Finding cos of sums, expanding expressions
cos(A-B)	cos A cos B + sin A sin B	Sign between terms opposite to original	Finding cos of differences
tan(A+B)	\(\frac{\tan A + \tan B}{1 – \tan A \tan B}\)	Valid when tan A tan B ≠ 1	Finding tan of sums, simplifying
tan(A-B)	\(\frac{\tan A – \tan B}{1 + \tan A \tan B}\)	Valid when tan A tan B ≠ -1	Finding tan of differences
sin 2A	2 sin A cos A	Simplest double angle formula	Solving equations, proving identities
cos 2A	cos² A – sin² A = 2cos² A – 1 = 1 – 2sin² A	Three equivalent forms	Choosing form based on given info
tan 2A	\(\frac{2\tan A}{1 – \tan^2 A}\)	Valid when tan A ≠ ±1	Solving equations, simplifying

Common Mistakes to Avoid: 1. Using wrong signs in compound angle formulae (remember: sine keeps, cosine changes)
2. Forgetting that double angle formulae are special cases of compound angles
3. Using the wrong form of cos 2A for the problem at hand
4. Not checking domains where formulae are valid (especially for tangent)
5. Forgetting to consider ± signs when using half-angle formulae
6. Confusing sin 2A with 2 sin A (they’re different: sin 2A = 2 sin A cos A, not 2 sin A)
7. Not simplifying expressions fully after applying formulae

Quiz: Test Your Understanding

Compound & Double Angle Formulae Quiz

Question 1: Find the exact value of sin 105° using compound angle formula.

Answer:
105° = 60° + 45°
sin(60° + 45°) = sin60° cos45° + cos60° sin45°
= \(\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2} + \frac{1}{2} \times \frac{\sqrt{2}}{2}\)
= \(\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}\)

Question 2: If sin A = 5/13 and A is acute, find the exact value of sin 2A.

Answer:
1. Find cos A: sin² A + cos² A = 1
(5/13)² + cos² A = 1 ⇒ 25/169 + cos² A = 1
cos² A = 144/169 ⇒ cos A = 12/13 (positive since A acute)
2. sin 2A = 2 sin A cos A = 2 × (5/13) × (12/13) = 120/169

Question 3: Prove the identity: cos 2A = 2cos² A – 1

Answer:
Starting from cos(A+B) = cos A cos B – sin A sin B
Set B = A: cos(A+A) = cos A cos A – sin A sin A
cos 2A = cos² A – sin² A
But sin² A = 1 – cos² A (from sin² A + cos² A = 1)
So cos 2A = cos² A – (1 – cos² A) = cos² A – 1 + cos² A = 2cos² A – 1

Question 4: Simplify the expression: sin 3A cos A – cos 3A sin A

Answer:
Recognize this as sin(A-B) form: sin P cos Q – cos P sin Q = sin(P-Q)
Here P = 3A, Q = A
So sin 3A cos A – cos 3A sin A = sin(3A – A) = sin 2A

Question 5: Solve for x in the interval 0 ≤ x ≤ 2π: cos 2x + 3 sin x – 2 = 0

Answer:
1. Use cos 2x = 1 – 2sin² x
2. Substitute: (1 – 2sin² x) + 3 sin x – 2 = 0
3. Simplify: -2sin² x + 3 sin x – 1 = 0
Multiply by -1: 2sin² x – 3 sin x + 1 = 0
4. Factor: (2 sin x – 1)(sin x – 1) = 0
5. Solve: sin x = 1/2 or sin x = 1
sin x = 1/2 ⇒ x = π/6, 5π/6
sin x = 1 ⇒ x = π/2
6. Solutions: x = π/6, π/2, 5π/6

🎯 Key Concepts Summary

Compound Angle Formulae:
- sin(A±B) = sin A cos B ± cos A sin B (sine keeps the sign)
- cos(A±B) = cos A cos B ∓ sin A sin B (cosine changes the sign)
- tan(A±B) = \(\frac{\tan A ± \tan B}{1 ∓ \tan A \tan B}\)
Double Angle Formulae:
- sin 2A = 2 sin A cos A
- cos 2A = cos² A – sin² A = 2cos² A – 1 = 1 – 2sin² A
- tan 2A = \(\frac{2\tan A}{1 – \tan^2 A}\)
Common Applications:
- Finding exact values for non-standard angles (15°, 75°, 105°, etc.)
- Simplifying trigonometric expressions
- Solving trigonometric equations
- Proving trigonometric identities
Memory Aids:
- “Sin Cos, Cos Sin” for sine formulae
- “Cos Cos, Sin Sin” for cosine formulae
- Sine keeps the sign, cosine changes it
Common CSEC Questions:
- Find exact values using compound angles
- Given sin A or cos A, find sin 2A, cos 2A, etc.
- Simplify expressions using formulae
- Solve equations involving double angles
- Prove identities using compound/double angles
Exam Strategy:
- Memorize the 6 compound angle formulae and 3 double angle formulae
- Practice recognizing which formula to use
- For cos 2A, choose the form that matches given information
- Check your work by testing with known angles (30°, 45°, 60°)
- Show all steps clearly in proofs and solutions

CSEC Exam Strategy: When working with compound/double angle formulae: (1) Write the appropriate formula first, (2) Substitute carefully, (3) Simplify step by step, (4) For proofs, work from one side to the other showing each step, (5) For equations, use identities to get a single trig function before solving. Remember: The double angle formulae are just special cases of compound angle formulae where B = A.