IB Mathematics AA Compound angle identities Study Notes | New Syllabus

IB Mathematics AA Compound angle identities Study Notes | New Syllabus

IB Mathematics AA Compound angle identities Study Notes

IB Mathematics AA Compound angle identities Study Notes Offer a clear explanation of Compound angle identities , including various formula, rules, exam style questions as example to explain the topics. Worked Out  examples and common problem types provided here will be sufficient to cover for topic Compound angle identities.

Compound Angle Identities

Compound Angle Identities

These identities express the trigonometric functions of sums or differences of two angles in terms of the functions of the individual angles.

Sine of sum:

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

Sine of difference:

\( \sin (A – B) = \sin A \cos B – \cos A \sin B \)

Cosine of sum:

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

Cosine of difference:

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

Tangent of sum:

\( \tan (A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \quad (\text{when } 1 – \tan A \tan B \ne 0) \)

Tangent of difference:

\( \tan (A – B) = \frac{\tan A – \tan B}{1 + \tan A \tan B} \quad (\text{when } 1 + \tan A \tan B \ne 0) \)

 Note: These identities are fundamental in simplifying expressions, solving equations, and proving other identities.

Example

Find exact values using compound angle identities for:

  • \( \sin(75^\circ) \)
  • \( \cos(75^\circ) \)
  • \( \tan(75^\circ) \)
▶️Answer/Explanation: Sine
  1. Express \( 75^\circ = 45^\circ + 30^\circ \)
  2. Apply identity: \( \sin(75^\circ) = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ \)
  3. Substitute: \( = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4} \)
  1. Express \( 75^\circ = 45^\circ + 30^\circ \)
  2. Apply identity: \( \cos(75^\circ) = \cos 45^\circ \cos 30^\circ – \sin 45^\circ \sin 30^\circ \)
  3. Substitute: \( = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} – \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} – \frac{\sqrt{2}}{4} = \frac{\sqrt{6} – \sqrt{2}}{4} \)
  1. Express \( 75^\circ = 45^\circ + 30^\circ \)
  2. Apply identity: \( \tan(75^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 – \tan 45^\circ \tan 30^\circ} \)
  3. Substitute: \( = \frac{1 + \frac{1}{\sqrt{3}}}{1 – 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} – 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} – 1} \)
  4. Rationalize: \( = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 – (1)^2} = \frac{3 + 2\sqrt{3} + 1}{3 – 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \)

Derivation of Double Angle Identities from Compound Angle Identities

Derivation of Double Angle Identities from Compound Angle Identities

We can derive the double angle identities for sine, cosine, and tangent by applying the compound angle formulas where both angles are equal.

Sine Double Angle:

Start with: \( \sin(2\theta) = \sin(\theta + \theta) \)

Apply the sine compound angle formula: \( = \sin \theta \cos \theta + \cos \theta \sin \theta \)

Simplify: \( = 2 \sin \theta \cos \theta \)

Cosine Double Angle:

Start with: \( \cos(2\theta) = \cos(\theta + \theta) \)

Apply the cosine compound angle formula: \( = \cos^2 \theta – \sin^2 \theta \)

Alternative forms using Pythagoras: \( = 2\cos^2 \theta – 1 \) \( = 1 – 2\sin^2 \theta \)

Tangent Double Angle:

Start with: \( \tan(2\theta) = \tan(\theta + \theta) \)

Apply the tangent compound angle formula: \( = \frac{2 \tan \theta}{1 – \tan^2 \theta} \) where \( 1 – \tan^2 \theta \ne 0 \)

Example

Given that \( \sin \theta = \frac{3}{5} \) and \( \theta \) is in the first quadrant, find the exact value of \( \cos(2\theta) \).

▶️Answer/Explanation

Since \( \sin \theta = \frac{3}{5} \), we use \( \sin^2 \theta + \cos^2 \theta = 1 \).

\( \cos^2 \theta = 1 – \sin^2 \theta = 1 – \left( \frac{3}{5} \right)^2 = 1 – \frac{9}{25} = \frac{16}{25} \)

So \( \cos \theta = \frac{4}{5} \) (positive in first quadrant).

\( \cos(2\theta) = \cos^2 \theta – \sin^2 \theta = \frac{16}{25} – \frac{9}{25} = \frac{7}{25} \)

Example

Given that \( \tan \theta = \frac{2}{3} \) and \( \theta \) is in the first quadrant, find the exact value of \( \tan(2\theta) \).

▶️Answer/Explanation

double angle identity for tangent: \( \tan(2\theta) = \frac{2 \tan \theta}{1 – \tan^2 \theta} \)

Substitute \( \tan \theta = \frac{2}{3} \): \( \tan(2\theta) = \frac{2 \times \frac{2}{3}}{1 – \left( \frac{2}{3} \right)^2} \)

denominator: \( = \frac{ \frac{4}{3} }{ 1 – \frac{4}{9} } = \frac{ \frac{4}{3} }{ \frac{5}{9} } \)

\( = \frac{4}{3} \times \frac{9}{5} = \frac{12}{5} \)

Connection between Double Angle Identities and De Moivre’s Theorem

De Moivre’s Theorem states:

\( (\cos \theta + i \sin \theta)^n = \cos (n \theta) + i \sin (n \theta) \)

If we set \( n = 2 \), we have:

\( (\cos \theta + i \sin \theta)^2 = \cos (2\theta) + i \sin (2\theta) \)

Expanding the left-hand side:

\( (\cos \theta + i \sin \theta)^2 = \cos^2 \theta + 2 i \sin \theta \cos \theta – \sin^2 \theta \)

Match real and imaginary parts:

  • Real part: \( \cos (2\theta) = \cos^2 \theta – \sin^2 \theta \)
  • Imaginary part: \( \sin (2\theta) = 2 \sin \theta \cos \theta \)

 This shows that the double angle identities for sine and cosine are direct consequences of De Moivre’s Theorem for \( n = 2 \).

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