Compound Angle Formulae (AQA A-Level Mathematics): Revision Notes

5.6.1 Compound Angle Formulae

Compound angle formulae are trigonometric identities that express the sine, cosine, and tangent of the sum or difference of two angles in terms of the sine, cosine, and tangent of the individual angles. These formulae are extremely useful in solving trigonometric equations, simplifying expressions, and proving other trigonometric identities.

1. Sine of a Sum/Difference:

• Sine of the Sum of Two Angles:

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

• This formula expresses the sine of the sum of two angles $A$ and $B$as a combination of the sines and cosines of the individual angles.
• Sine of the Difference of Two Angles:

$\sin(A - B) = \sin A \cos B - \cos A \sin B$

• This formula expresses the sine of the difference of two angles A and B similarly.

2. Cosine of a Sum/Difference:

• Cosine of the Sum of Two Angles:

$\cos(A + B) = \cos A \cos B - \sin A \sin B$

• This formula expresses the cosine of the sum of two angles $A$ and $B$ in terms of the cosines and sines of the individual angles.
• Cosine of the Difference of Two Angles:

$\cos(A - B) = \cos A \cos B + \sin A \sin B$

• This formula expresses the cosine of the difference of two angles $A$ and $B.$

3. Tangent of a Sum/Difference:

• Tangent of the Sum of Two Angles:

$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

• This formula expresses the tangent of the sum of two angles $A$ and $B$ in terms of the tangents of the individual angles.
• Tangent of the Difference of Two Angles:

$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$

• This formula expresses the tangent of the difference of two angles $A$ and $B.$

4. Derivation of Compound Angle Formulae:

The compound angle formulae can be derived using the unit circle, trigonometric identities, and geometric methods. Here's a brief outline of how you might derive them:

• Sine and Cosine Formulae:
• Consider the angles A and B placed on the unit circle. Use the coordinates of the points where the terminal sides of these angles intersect the circle to express $sin(A + B)$ and $\cos(A + B).$
• Use the fact that the length of the hypotenuse in the unit circle is $1$, and apply the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ to derive the formulae.
• Tangent Formula:
• The tangent formulae can be derived by dividing the sine formula by the cosine formula:

$\tan(A \pm B) = \frac{\sin(A \pm B)}{\cos(A \pm B)}$

5. Example Problems Using Compound Angle Formulae:

6. Applications of Compound Angle Formulae:

• Solving Trigonometric Equations: Compound angle formulae are essential when solving trigonometric equations involving sums or differences of angles.
• Proving Identities: These formulae are frequently used in proving more complex trigonometric identities.
• Simplifying Expressions: They help in simplifying trigonometric expressions to make them easier to evaluate or integrate.
• Geometry and Physics: These formulae are useful in applications involving rotations, wave interference, and oscillations.

Summary:

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Expanding Trig Brackets: Compound Angle Formulae

When expanding trigonometric functions, e.g., $\sin(A \pm B)$, the ordinary rules of algebra do not apply.

Proposition:

$$\sin(A \pm B) = \sin A \cos B \pm \sin B \cos A$$

Proof:

Finding the lengths of the four coloured sides:

1. For the side corresponding to $\cos A$:

$$\cos A = \frac{\text{Red segment}}{1} \Rightarrow \text{Red segment} = \cos A$$

2. For the side corresponding to $\sin A$:

$$\sin A = \frac{\text{Blue segment}}{1} \Rightarrow \text{Blue segment} = \sin A$$

3. For the side corresponding to $\sin B \cos A$:

$$\cos B = \frac{\text{Green segment}}{\sin A} \Rightarrow \text{Green segment} = \sin A \cos B$$

4. For the side corresponding to $\sin B$:

$$\sin B = \frac{\text{Orange segment}}{\sin A} \Rightarrow \text{Orange segment} = \sin B \cos A$$

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For the equation $\sin(A + B)$:

The calculation in the diagram confirms that:

$$\sin(A + B) = \frac{\text{purple segment}}{1} \Rightarrow \text{purple segment} = \sin(A + B)$$

From the diagram, we can also see that:

$$\therefore \quad \sin(A + B) \equiv \sin(A)\cos(B) + \sin(B)\cos(A) \quad \text{(:success[proven by the diagram])}$$

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Proposition:

$$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$$

Proof:

Diagram Explanation:

Compound Angle Formulae

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Solutions:

$a) \tan A = \frac{\text{OPP}}{\text{ADJ}} = \frac{4}{3}$

b) $\sin B = \frac{\text{OPP}}{\text{HYP}} = \frac{\sqrt{5}}{3}$

c) $\cos(A + B) = \cos A \cos B - \sin A \sin B$

• We can now find each of these individually from the triangles to evaluate the expression. From above triangle, $\cos A = \frac{\text{ADJ}}{\text{HYP}} = \frac{3}{5}$

$$\therefore \cos(A + B) = \left(\frac{\cancel3}{5}\right)\left(\frac{2}{\cancel3}\right) - \left(\frac{4}{5}\right)\left(\frac{\sqrt{5}}{3}\right) = \frac{2}{5} - \frac{4\sqrt{5}}{15} = \frac{6 - 4\sqrt{5}}{15}$$

d) $\sin(A + B) = \sin A \cos B + \cos A \sin B$

$$= \left(\frac{4}{5}\right)\left(\frac{2}{3}\right) + \left(\frac{\cancel3}{5}\right)\left(\frac{\sqrt{5}}{\cancel3}\right) = \frac{8}{15} + \frac{\sqrt{5}}{5} = \frac{8 + 3\sqrt{5}}{15}$$

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Solving Equations With Compound Angle Formulae

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