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IB Mathematics AA Relationships in trigonometric functions Study Notes | New Syllabus

IB Mathematics AA Relationships in trigonometric functions

IB Mathematics AA Relationships in trigonometric functions Study Notes

IB Mathematics AA Relationships in trigonometric functions Study Notes Offer a clear explanation of Relationships in trigonometric functions, 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 Relationships in trigonometric functions.

Compound Angle Identities and Double-Angle Identities

Introduction

Compound angle identities are foundational tools in trigonometry, enabling the simplification of trigonometric functions involving sums or differences of angles. They also provide the basis for deriving double-angle identities, which are widely used in mathematics and physics.

Compound Angle Identities

Sine of Compound Angles

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

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

These formulas decompose the sine of a sum or difference of two angles into simpler trigonometric components.

Cosine of Compound Angles

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

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

Cosine formulas express the sum or difference of angles using products of cosine and sine.

Tangent of Compound Angles

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

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

The tangent formulas arise by dividing the sine compound formula by the cosine compound formula.

Double-Angle Identities

Sine Double Angle

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

Derived by applying the sine addition formula: \(\sin(2A) = \sin(A + A)\).

Cosine Double Angle

\(\cos(2A) = \cos^2 A – \sin^2 A\)

Alternate forms derived using the Pythagorean identity:

  • \(\cos(2A) = 2\cos^2 A – 1\)
  • \(\cos(2A) = 1 – 2\sin^2 A\)

Tangent Double Angle

\(\tan(2A) = \frac{2\tan A}{1 – \tan^2 A}\)

Applicable when \(|\tan A| \neq 1\).

Applications and Examples

Simplify \(\sin(75^\circ)\)

Rewrite \(75^\circ\) as \(45^\circ + 30^\circ\) and use the sine addition formula:

\(\sin(75^\circ) = \sin(45^\circ)\cos(30^\circ) + \cos(45^\circ)\sin(30^\circ)\)

Substitute values:

\(\sin(75^\circ) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}\)

Derive \(\cos(2A)\) from \(\cos(A + B)\)

Substitute \(B = A\) in \(\cos(A + B) = \cos A \cos B – \sin A \sin B\):

\(\cos(2A) = \cos^2 A – \sin^2 A\)

Using the Pythagorean identity, alternate forms are derived:

\(\cos(2A) = 2\cos^2 A – 1\) and \(\cos(2A) = 1 – 2\sin^2 A\).

Solve \(\tan(2x) = \sqrt{3}\)

Using the tangent double-angle formula:

\(\tan(2x) = \sqrt{3} \quad \implies \quad 2x = \frac{\pi}{3} + n\pi\)

\(x = \frac{\pi}{6} + \frac{n\pi}{2}, \, n \in \mathbb{Z}\)

Connections

  • Applications: Used in calculus, physics, and engineering, e.g., Fourier series, wave motion, and harmonic analysis.
  • De Moivre’s Theorem: Expanding \((\cos x + i\sin x)^n\) relies on compound angle identities.

IB Mathematics AA SL Relationships in trigonometric functions Exam Style Worked Out Questions

Question

In the diagram below, AD is perpendicular to BC.

CD  =  4, BD  =  2 and AD  =  3. \({\rm{C}}\hat {\rm{A}}{\rm{D}} = \alpha \) and \({\rm{B}}\hat {\rm{A}}{\rm{D}} = \beta \) .

 

Find the exact value of \(\cos (\alpha  – \beta )\) .

▶️Answer/Explanation

Markscheme

METHOD 1

\({\text{AC}} = 5\) and \(\text{AB} = \sqrt {13}\)   (may be seen on diagram)     (A1)

\(\cos \alpha = \frac{3}{5}\) and \(\sin \alpha = \frac{4}{5}\)     (A1)

\(\cos \beta = \frac{3}{{\sqrt {13} }}\) and \(\sin \beta = \frac{2}{{\sqrt {13} }}\)     (A1) 

Note: If only the two cosines are correctly given award (A1)(A1)(A0).

 

Use of \(\cos (\alpha – \beta ) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \)     (M1)

\( = \frac{3}{5} \times \frac{3}{{\sqrt {13} }} + \frac{4}{5} \times \frac{2}{{\sqrt {13} }}\)   (substituting)     M1

\( = \frac{{17}}{{5\sqrt {13} }}\)     \(\left( { = \frac{{17\sqrt {13} }}{{65}}} \right)\)     A1     N1

[6 marks]

 

METHOD 2

\({\text{AC}} = 5\) amd \({\text{AB}} = \sqrt {13} \)   (may be seen on diagram)     (A1)

Use of \(\cos (\alpha + \beta ) = \frac{{{\text{A}}{{\text{C}}^2} + {\text{A}}{{\text{B}}^2} – {\text{B}}{{\text{C}}^2}}}{{{\text{2(AC)(AB)}}}}\)     (M1)

\( = \frac{{25 + 13 – 36}}{{2 \times 5 \times \sqrt {13} }}\,\,\,\,\,\left( { = \frac{1}{{5\sqrt {13} }}} \right)\)     A1

Use of \(\cos (\alpha + \beta ) + \cos (\alpha – \beta ) = 2\cos \alpha \cos \beta \)     (M1)

\(\cos \alpha = \frac{3}{5}\) and \(\cos \beta = \frac{3}{{\sqrt {13} }}\)     (A1)

\(\cos (\alpha – \beta ) = \frac{{17}}{{5\sqrt {13} }}\,\,\,\,\,\left( { = 2 \times \frac{3}{5} \times \frac{3}{{\sqrt {13} }} – \frac{1}{{5\sqrt {13} }}} \right){\text{ }}\left( { = \frac{{17\sqrt {13} }}{{65}}} \right)\)     A1     N1

[6 marks]

 

Question

(a)     Show that \(\sin 2nx = \sin \left( {(2n + 1)x} \right)\cos x – \cos \left( {(2n + 1)x} \right)\sin x\).

(b)     Hence prove, by induction, that

\[\cos x + \cos 3x + \cos 5x +  \ldots  + \cos \left( {(2n – 1)x} \right) = \frac{{\sin 2nx}}{{2\sin x}},\]

for all \(n \in {\mathbb{Z}^ + }{\text{, }}\sin x \ne 0\).

(c)     Solve the equation \(\cos x + \cos 3x = \frac{1}{2},{\text{ }}0 < x < \pi \).

▶️Answer/Explanation

Markscheme

(a)     \(\sin (2n + 1)x\cos x – \cos (2n + 1)x\sin x = \sin (2n + 1)x – x\)     M1A1

\( = \sin 2nx\)     AG

[2 marks]

(b)     if n = 1     M1

\({\text{LHS}} = \cos x\)

\({\text{RHS}} = \frac{{\sin 2x}}{{2\sin x}} = \frac{{2\sin x\cos x}}{{2\sin x}} = \cos x\)     M1

so LHS = RHS and the statement is true for n = 1     R1

assume true for n = k     M1

Note: Only award M1 if the word true appears.

   Do not award M1 for ‘let n = k’ only.

   Subsequent marks are independent of this M1.

 

so \(\cos x + \cos 3x + \cos 5x +  \ldots  + \cos (2k – 1)x = \frac{{\sin 2kx}}{{2\sin x}}\)

if n = k + 1 then

\(\cos x + \cos 3x + \cos 5x +  \ldots  + \cos (2k – 1)x + \cos (2k + 1)x\)     M1

\( = \frac{{\sin 2kx}}{{2\sin x}} + \cos (2k + 1)x\)     A1

\( = \frac{{\sin 2kx + 2\cos (2k + 1)x\sin x}}{{2\sin x}}\)     M1

\( = \frac{{\sin (2k + 1)x\cos x – \cos (2k + 1)x\sin x + 2\cos (2k + 1)x\sin x}}{{2\sin x}}\)     M1

\( = \frac{{\sin (2k + 1)x\cos x + \cos (2k + 1)x\sin x}}{{2\sin x}}\)     A1

\( = \frac{{\sin (2k + 2)x}}{{2\sin x}}\)     M1

\( = \frac{{\sin 2(k + 1)x}}{{2\sin x}}\)     A1

so if true for n = k, then also true for n = k + 1

as true for n = 1 then true for all \(n \in {\mathbb{Z}^ + }\)     R1

Note: Final R1 is independent of previous work.

[12 marks]

(c)     \(\frac{{\sin 4x}}{{2\sin x}} = \frac{1}{2}\)     M1A1

\(\sin 4x = \sin x\)

\(4x = x \Rightarrow x = 0\) but this is impossible

\(4x = \pi – x \Rightarrow x = \frac{\pi }{5}\)     A1

\(4x = 2\pi + x \Rightarrow x = \frac{{2\pi }}{3}\)     A1

\(4x = 3\pi – x \Rightarrow x = \frac{{3\pi }}{5}\)     A1

for not including any answers outside the domain     R1

Note: Award the first M1A1 for correctly obtaining \(8{\cos ^3}x – 4\cos x – 1 = 0\) or equivalent and subsequent marks as appropriate including the answers \(\left( { – \frac{1}{2},\frac{{1 \pm \sqrt 5 }}{4}} \right)\).

[6 marks] Total [20 marks]

 

 

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