QLD 12 Specialist

SPECIALIST

Year 12 QLD Curriculum

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Title: Mathspace QLD Curriculum - 12 Specialist Mathematics: 2025 Edition ISBN: 978-1-963022-88-9 Editors and lead authors: Erin Gallagher, Jeremy Hunter, Elisha Templin, Alex Bunt, Kris Ellery, Anil Krishna, Neeti Meghani, Wassim Negroh, Isha Tiwari Writing and development team: Adam Humphreys, Lesley Ann Lopez, Regine Jara, Berna Barillo, Rona Josue, Maria Lou Ona, Adriane Abunda, Jady Lee Amarillas, Marielle Belen, Rosalita Campilla, Jelly Ann Candido, Macy Castillo, Sheramie Castillo, Christian Dinglasan, Adrian Doctolero, Dharent Fernandez, Dhave Fernandez, Jessa Lambinicio, Julie Ann Manzano, Kyra Joy Manzano, Mary Grace Manzano, Glady May Mejias, Mikaela Nicolas, and Paul Windell Platero.

Images and Design team: Scott Nolan, Chastine Marquez, Keith Gimeno, Jasper Jumawan, Mary Matheu, Jemark Orevillo, Jessa Ortega

Contents 1

2

3

Further complex numbers

2

1.01

Polar form identities

4

1.02

Square roots of complex numbers

14

1.03

De Moivre’s theorem

21

1.04

Complex roots of unity

27

1.05

Geometric complex roots of unity

37

1.06

Factor and remainder theorem

47

1.07

Conjugate root theorem

55

1.08

Solutions to polynomials

62

Chapter 1 review

68

Mathematical induction and trigonometric proofs

70

2.01

Introduction to proof by induction

72

2.02

Inductive proofs for sums of integers

78

2.03

Inductive proofs for divisibility results

83

2.04

Trigonometric identities and De Moivre’s theorem

88

2.05

Problems with trigonometric identities

94

Chapter 2 review

100

Vectors in two and three dimensions

102

3.01

Vectors in three dimensions

104

3.02

Magnitude and direction in three dimensions

112

3.03

Vectors and geometry in three dimensions

119

3.04

Vector addition and subtraction in three dimensions

127

3.05

Scalar multiplication and vector between two points

134

3.06

Dot product

142

3.07

Vector projections

148

3.08

Proofs of geometric results using vectors

154

Contents mathspace.co

iii

4

3.09

Equations of lines in 3D

162

3.10

Equations of planes in 3D

169

3.11

Vector and Cartesian equations of curves

175

3.12

Spheres and surfaces in 3D

181

3.13

Applications of vector equations

189

3.14

Cross product

195

3.15

Vector applications in mechanics and geometry

201

Chapter 3 review

208

Vector calculus 4.01

Position vectors and trajectories

212

4.02

Vector functions of curves

219

4.03

Particle motion and intersections

228

4.04

Differentiation of vector functions

234

4.05

Integration of vector functions

239

4.06

Equations of motion

246

4.07

Projectile and circular motion

254

4.08

5

iv

210

Applications of vector calculus

261

Chapter 4 review

269

Further matrices

272

5.01

Matrices and determinants

274

5.02

Matrix equations

278

5.03

Systems of linear equations with matrices

289

5.04

Solutions of systems and geometric interpretation

297

5.05

Dominance matrices

310

5.06

Leslie matrices

320

Investigation: Eigenvalues and eigenvectors

330

Investigation: Leontief, Markov, and Cryptology

333

Chapter 5 review

338

Mathspace QLD 12 Specialist Mathematics 2025 mathspace.co

6

7

Integration techniques 6.01

Basic trigonometric integrals

344

6.02

Trigonometric integrals

350

6.03

Substitution method

360

6.04

Integration using logarithms

368

6.05

Inverse trigonometric functions

377

6.06

Integration with inverse trigonometric results

385

6.07

Partial fractions

394

6.08

Partial fractions with quadratics

401

6.09

Integration by parts

409

Chapter 6 review

416

Applications of integral calculus

418

7.01

Area under curves and area between curves

7.02

Area bounded by the y-axis

431

7.03

Volumes of solids of revolution for one curve

439

7.04

Volumes of solids of revolution between two curves

446

7.05

Simpson’s rule

452

Investigation: Area estimation using numerical integration

458

Exponential random variables

461

Chapter 7 review

472

7.06

8

342

Rates of change and differential equations

420

474

8.01

Implicit differentiation

476

8.02

Related rates

483

8.03

Differential equations

491

8.04

Separation of variables

501

8.05

Applications of separation of variables

507

8.06

Slope fields

515

8.07

Models with differential equations

524

Chapter 8 review

531

Contents mathspace.co

v

9

10

vi

Modelling motion

534

9.01

Force, momentum, and equilibrium

536

9.02

Motion on inclined planes

544

9.03

Friction

548

9.04

Motion with variable acceleration

553

9.05

Distance and displacement with variable acceleration

562

9.06

Forces that depend on velocity

568

9.07

Vertical motion

573

9.08

Forces that depend on displacement

579

9.09

Simple harmonic motion

584

9.10

Synthesis and advanced applications

590

Chapter 9 review

600

Statistical inference

604

10.01

Sample means and the central limit theorem

606

10.02

Distribution of sample means

612

10.03

Normal distribution of sample means

618

10.04

Repeated random sampling and distribution

624

10.05

Standard normal distribution and sample means

630

10.06

Applications of sample means

638

10.07

Confidence intervals

645

10.08

Construct confidence intervals

651

10.09

Margin of error

659

10.10

Error, confidence, and sample size

665

10.11

Variations in confidence intervals

672

10.12

Estimate population parameters

678

10.13

Applications of confidence intervals

683

Chapter 10 review

690

Answers

693

Mathspace QLD 12 Specialist Mathematics 2025 mathspace.co

2

Mathematical induction and trigonometric proofs Chapter outline 2.01 2.02 2.03 2.04 2.05

Introduction to proof by induction Inductive proofs for sums of integers Inductive proofs for divisibility results Trigonometric identities and De Moivre’s theorem Problems with trigonometric identities Chapter 2 review

72 78 83 88 94 100

Proofs don’t guess—they convince. It’s maths’ version of a courtroom drama.

2.04

Trigonometric identities and De Moivre’s theorem

After this lesson, you will be able to…

• prove multi-angle trigonometric identities up to angles of 4x • use the binomial expansion to prove trigonometric identities • use De Moivre’s theorem to prove trigonometric identities • generate examples of trigonometric identities for angles up to 4x

Proof of De Moivre’s theorem by induction Interactive exploration

Discover this concept in action online

mathspace.co

De Moivre’s theorem for complex numbers states that if z = r cis (θ ), then zn = rn cis(nθ ), n ∈ .

Proof by mathematical induction can be applied to prove De Moivre’s theorem for powers of positive integers: • Step 1: Determine the proposition.

Let z = r cis (θ ). The goal is to prove zn = rn cis(nθ ) for all positive integers n ≥ 1.

• Step 2: Initial statement.

Show the statement is true for n = 1: LHS = z1 Start with the left-hand side for n = 1 = r cis (θ )

By definition, z1 is z

RHS = r1 cis (1 × θ ) Start with the right-hand side for n = 1 = r cis (θ ) Simplify the expression Since LHS = RHS, the statement is true for n = 1.

• Step 3: Assumption statement.

For n = k ∈ +, assume that zk = rk cis (kθ ).

88

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• Step 4: Prove the inductive step.

Prove the statement is true for n = k + 1.

The goal is to prove: zk+1 = rk+1 cis ((k + 1)θ ) LHS = zk+1

Start with the left-hand side

k

1

=z ×z

Apply index laws

= (rk cis (kθ )) × (r cis (θ ))

Use the assumption statement

k+1

=r

(cos (kθ ) + i sin (kθ ))(cos (θ ) + i sin (θ ))

Expand the cis notation and combine r terms

= rk+1 [(cos (kθ ) cos (θ ) − sin (kθ ) sin (θ )) + i (sin (kθ ) cos (θ ) + cos (kθ ) sin (θ ))]

Expand brackets and group real and imaginary parts, using i2 = −1

= rk+1 (cos (kθ + θ ) + i sin (kθ + θ ))

Apply trigonometric sum identities

k+1

=r

(cos ((k + 1)θ ) + i sin ((k + 1)θ ))

Factorise out θ

= rk+1 cis ((k + 1)θ )

Write in cis notation

= RHS

This matches the right-hand side

Therefore, if the statement is true for n = k, then it is true for n = k + 1 ∈ +.

• Step 5: Determine the conclusion.

Therefore, using mathematical induction, De Moivre’s theorem is true for all positive integers n ≥ 1.

Note that De Moivre’s theorem holds for powers of all positive and negative integers, and zero.

Idea summary De Moivre’s theorem states that for a complex number z = r (cos (θ ) + i sin (θ )), its nth power is given by zn = rn (cos(nθ ) + i sin (nθ )), where n is an integer.

For positive integers, the theorem can be formally proven using the method of mathematical induction.

Derive multi-angle trigonometric identities De Moivre’s theorem provides a powerful link between complex numbers and trigonometry. It can be combined with the binomial expansion to derive various multi-angle trigonometric identities, such as expressions for cos (nθ ) and sin (nθ ) in terms of powers of cos (θ ) and sin (θ ).

The method involves expanding the expression (cos (θ ) + i sin (θ ))n in 2 different ways: • Method 1: De Moivre’s theorem — applying De Moivre’s theorem directly gives cos (nθ ) + i sin (nθ ).

• Method 2: Binomial theorem — applying the binomial theorem, (cos (θ ) + i sin(θ ))n =

(cos(θ ))n−k (i sin(θ ))k.

By equating the real parts of the 2 resulting expressions, an identity for cos (nθ ) can be determined. Similarly, equating the imaginary parts yields an identity for sin (nθ ). This process often requires the use of the Pythagorean identity, cos2 (θ ) + sin2 (θ ) =1, to simplify the derived expressions into desired forms (e.g., entirely in terms of sin (θ ) or cos (θ )).

2.04 Trigonometric identities and De Moivre’s theorem mathspace.co

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Example 1 Use De Moivre’s theorem and the binomial expansion to prove these trigonometric identities: • cos (2θ ) = cos2 (θ ) − sin2 (θ ) • sin (2θ ) = 2 sin (θ ) cos (θ )

Create a strategy Consider the expression (cos (θ ) + i sin (θ ))2. First, apply De Moivre’s theorem. Next, expand using the binomial theorem for n = 2, which is (a + b)2 = a2 + 2ab + b2. Equate the real and imaginary parts of both expansions to derive the identities for cos (2θ ) and sin (2θ ).

Apply the idea

By De Moivre’s theorem: (cos (θ ) + i sin (θ ))2 = cos (2θ ) + i sin (2θ ) By binomial expansion: (cos (θ ) + i sin (θ ))2 = cos2 (θ ) + 2 cos (θ )(i sin (θ )) + (i sin (θ ))2 2

2

2

= cos (θ ) + 2i cos (θ ) sin (θ ) + i sin (θ ) 2

2

Apply the binomial theorem Simplify powers of brackets

= cos (θ ) + 2i cos (θ ) sin (θ ) − sin (θ )

Evaluate i2 = −1

= (cos2 (θ ) − sin2 (θ )) + i (2 cos (θ ) sin (θ ))

Group real and imaginary parts

Equate the results from De Moivre’s theorem and the binomial expansion: cos (2θ ) + i sin (2θ ) = (cos2 (θ ) − sin2 (θ )) + i (2 cos (θ ) sin (θ )) • Proof for the cosine identity Equate the real parts: cos (2θ ) = cos2 (θ ) − sin2 (θ )

Equate real components

• Proof for the sine identity Equate the imaginary parts: sin (2θ ) = 2 sin (θ ) cos (θ )

Equate imaginary components

Reflect and check The identities derived are the standard double-angle formulas for cosine and sine, confirming the accuracy of the method. For example, for θ = : •

. Also,

. (Matches)

•

. Also,

. (Matches)

90

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Example 2 Use De Moivre’s theorem and the binomial expansion to prove these trigonometric identities: • cos (3θ ) = 4 cos3 (θ ) − 3 cos (θ ) • sin (3θ ) = 3 sin (θ ) − 4 sin3 (θ )

Create a strategy Consider the expression (cos (θ ) + i sin (θ ))3. Firstly, apply De Moivre’s theorem. Secondly, expand the expression using the binomial theorem for n = 3, which is (a + b)3 = a3 + 3a2b + 3ab2 + b3.

Then, equate the two resulting expressions. Group the real and imaginary parts of the expanded form. Equate the real parts to determine the identity for cos (3θ ) and equate the imaginary parts to determine the identity for sin (3θ ). Use the Pythagorean identity to express the results in terms of only cos (θ ) or sin (θ ).

Apply the idea

By De Moivre’s theorem: (cos (θ ) + i sin (θ ))3 = cos (3θ ) + i sin (3θ ) By binomial expansion: (cos (θ ) + i sin (θ ))3 = cos3 (θ ) + 3 cos2 (θ )(i sin (θ )) + 3 cos (θ ) (i sin (θ ))2 + (i sin (θ ))3 = c os3 (θ ) + 3i cos2 (θ ) sin (θ ) + 3 cos (θ ) i2 sin2 (θ ) + i3 sin3 (θ ) = c os3 (θ ) + 3i cos2 (θ ) sin (θ ) − 3 cos (θ ) sin2 (θ ) − i sin3 (θ ) = ( cos3 (θ ) − 3 cos (θ ) sin2 (θ )) + i (3 cos2 (θ ) sin (θ ) − sin3 (θ ))

Apply the binomial theorem Simplify powers of brackets Evaluate powers of i, where i2 = −1 and i3 = −i Group real and imaginary parts

Now, equate the results from De Moivre’s theorem and the binomial expansion: cos (3θ ) + i sin (3θ ) = (cos3 (θ ) − 3 cos (θ ) sin2 (θ )) + i (3 cos2 (θ ) sin (θ ) − sin3 (θ )) • Proof for the cosine identity Equate the real parts: cos (3θ ) = cos3 (θ ) − 3 cos (θ ) sin2 (θ )

Equate real components

= cos (θ ) − 3 cos (θ ) (1 − cos (θ ))

Substitute sin2 (θ ) = 1 − cos2 (θ )

= cos3 (θ ) − 3 cos (θ ) + 3 cos3 (θ )

Expand the brackets

3

2

3

= 4 cos (θ ) − 3 cos (θ )

Collect like terms

• Proof for the sine identity Equate the imaginary parts: sin (3θ ) = 3 cos2 (θ ) sin (θ )− sin3 (θ )

Equate imaginary components

= 3 (1 − sin (θ )) sin (θ ) − sin (θ ) 2

3

3

3

Substitute cos2 (θ ) = 1 − sin2 (θ )

= 3 sin (θ ) − 3 sin (θ ) − sin (θ )

Expand the brackets

= 3 sin (θ ) − 4 sin3 (θ )

Collect like terms

2.04 Trigonometric identities and De Moivre’s theorem mathspace.co

91

Reflect and check The derived identities are the standard triple-angle formulas for cosine and sine. To check, substitute a value for θ, e.g., θ = (or 30°): 6 • For cos (3θ ):

. Also, . (Matches)

• For sin (3θ ):

. Also, . (Matches)

Idea summary Trigonometric identities for multi-angles like cos (nθ ) and sin (nθ ) can be derived by combining De Moivre’s theorem with the binomial theorem.

The process involves expanding (cos (θ ) + i sin(θ ))n in 2 ways (using De Moivre’s theorem and the binomial theorem), then equating the real and imaginary parts of the resulting expressions.

Pythagorean identities (cos2 (θ ) + sin2 (θ ) = 1) are often used to simplify the derived identities into their most common forms.

2.04 Practice questions Simple familiar 1

Determine whether each statement is true or false: a

De Moivre’s theorem can be directly applied to a complex number in Cartesian form (e.g., (a + bi)n) without first converting it to polar form.

b

When using De Moivre’s theorem for zn, the angle of the resulting power is determined by multiplying the original angle θ by n.

2

If z = r (cos(θ ) + i sin (θ )), determine the general form of zn according to De Moivre’s theorem.

Identify the modulus and argument of zn.

3

Expand the expression (cos (θ ) + i sin (θ ))2 using the binomial theorem.

4

Given a complex number z = 3 cis

92

:

a

Use De Moivre’s theorem to determine z2 in polar form.

b

Use De Moivre’s theorem to determine z4 in polar form.

Mathspace QLD 12 Specialist Mathematics 2025 mathspace.co

Complex familiar 5

Use De Moivre’s theorem and the binomial expansion to prove these trigonometric identities: • cos (2θ ) = 2 cos2 (θ ) − 1 • sin (2θ ) = 2 sin (θ ) cos (θ )

Ex 1

6

Use the principle of mathematical induction to prove De Moivre’s theorem, (cos (θ ) + i sin (θ ))n = cos (nθ ) + i sin (nθ ) for all positive integers n.

7

8

Given the complex number z = −1 + i: a

Express z in polar form.

b

Use De Moivre’s theorem to evaluate z8 in polar form.

c

Convert your answer from part (b) to Cartesian form (a + bi).

Given the complex number z = 1 + i

:

a

Express z in polar form.

b

Use De Moivre’s theorem to evaluate z5 in polar form.

c

Convert your answer from part (b) to Cartesian form (a + bi).

Complex unfamiliar 9

Use De Moivre’s theorem and the binomial expansion to prove the identity for cos (3θ ): cos (θ ) − 4 cos (θ ) sin2 (θ )

10

Consider the implications of De Moivre’s theorem when extending its use beyond positive integers: a

Prove by mathematical induction that for n ∈ +, if z = cos θ + i sin θ, then z−n = cos (−nθ ) + i sin (−nθ ).

b

Briefly explain how this result, along with the proof for positive integers, supports the statement that De Moivre’s theorem holds for all integers n.

2.04 Trigonometric identities and De Moivre’s theorem mathspace.co

93

2.05

Problems with trigonometric identities

After this lesson, you will be able to…

• prove multi-angle trigonometric identities up to angles of 4x using the binomial expansion • prove multi-angle trigonometric identities up to angles of 4x using De Moivre’s theorem • solve problems involving trigonometric identities derived from De Moivre’s theorem • use examples such as cos (3x) = 4 cos3 (x) - 3 cos (x) and sin (3x) = 3 sin (x) - 4 sin3 (x) to demonstrate proofs

Derive trigonometric identities using De Moivre’s theorem De Moivre’s theorem provides a powerful method to derive multi-angle trigonometric identities for cos (nθ ) and sin (nθ ) by combining it with binomial expansion.

For a complex number z = cos (θ ) + i sin (θ ), De Moivre’s theorem states: (cos (θ ) + i sin (θ ))n = cos (nθ ) + i sin (nθ ) The binomial expansion of (cos (θ ) + i sin (θ ))n is: (cos (θ ) + i sin (θ ))n = The binomial coefficients

cosn−r θ (i sin (θ ))r

are determined in the nth row of Pascal’s triangle, where each number

is the sum of the 2 numbers directly above it (starting with row 0: 1; row 1: 1, 1; row 2: 1, 2, 1; etc.).

By equating the real and imaginary parts of the binomial expansion to cos (nθ ) and sin (nθ ), respectively, the trigonometric identities can be derived. During simplification, use: • Powers of i: i2 = −1, i3 = −i, i4 = 1, and so on.

• The Pythagorean identity: cos2 (θ ) + sin2 (θ ) = 1 to express the result in terms of a single trigonometric function.

Common trigonometric identities that may also be used includes: Pythagorean identity Double-angle identities

94

sin2 (θ ) + cos2 (θ ) = 1 sin (2θ ) = 2 sin (θ ) cos (θ ) cos (2θ ) = cos2 (θ ) − sin2 (θ ) = 2 cos2 (θ ) − 1 = 1 − 2 sin2 (θ )

Mathspace QLD 12 Specialist Mathematics 2025 mathspace.co

Example 1 Use De Moivre’s theorem and the binomial expansion to prove these trigonometric identities: • cos (3θ ) = 4 cos3 (θ ) − 3 cos (θ ) • sin (3θ ) = 3 sin (θ ) − 4 sin3 (θ )

Create a strategy Consider the expression (cos (θ ) + i sin (θ ))3.

Firstly, apply De Moivre’s theorem. Secondly, expand the expression using the binomial theorem for n = 3, which is (a + b)3 = a3 + 3a2b + 3ab2 + b3 . Then, equate the 2 resulting expressions. Group the real and imaginary parts of the expanded form. Equate the real parts to determine the identity for cos (3θ ) and the imaginary parts to determine the identity for sin (3θ ). Use the Pythagorean identity to express the results in terms of only cos (θ ) or sin (θ ).

Apply the idea

By De Moivre’s theorem: (cos (θ ) + i sin (θ ))3 = cos (3θ ) + i sin (3θ ) By binomial expansion: 3 (θ ) + 3 cos2 (θ ) (i sin (θ )) (cos (θ ) + i sin (θ ))3 = cos

+ 3 cos (θ ) (i sin (θ ))2 + (i sin (θ ))3

Apply the binomial theorem

cos3 (θ ) + 3i cos2 (θ ) sin (θ ) + 3 cos (θ ) (i2 sin2 (θ )) + i3 sin3 (θ )

Simplify powers of brackets

=

(θ ) + 3i cos (θ ) sin (θ )

= cos - 3 cos (θ ) sin2 (θ ) - i sin3 (θ )

3

2

valuate powers of i, where i2 = -1 E and i3 = -i

3 cos (θ ) sin (θ )) = (cos (θ ) + i (3 cos2 (θ ) sin (θ ) - sin3 (θ ))

Group real and imaginary parts

3

2

Now, equate the results from De Moivre’s theorem and the binomial expansion: cos (3θ ) + i sin (3θ ) = (cos3 (θ ) − 3 cos (θ ) sin2 (θ )) + i (3 cos2 (θ ) sin (θ ) − sin3 (θ )) • Proof for the cosine identity Equate the real parts: cos (3θ ) = cos3 (θ ) − 3 cos (θ ) sin2 (θ )

Equate real components

= cos (θ ) − 3 cos (θ ) (1 - cos (θ ))

Substitute sin2 (θ ) = 1 − cos2 (θ )

= cos3 (θ ) − 3 cos (θ ) + 3 cos3 (θ )

Expand the brackets

3

2

3

= 4 cos (θ ) − 3 cos (θ )

Collect like terms

• Proof for the sine identity Equate the imaginary parts: sin (3θ ) = 3 cos2(θ ) sin (θ ) − sin3 (θ )

Equate imaginary components

= 3 (1 - sin (θ )) sin (θ ) − sin (θ )

Substitute cos2 (θ ) = 1 − sin2 (θ )

= 3 sin (θ ) − 3 sin3 (θ ) - sin3 (θ )

Expand the brackets

2

3

3

= 3 sin (θ ) − 4 sin (θ )

Collect like terms

2.05 Problems with trigonometric identities mathspace.co

95

Reflect and check To build confidence in the result, test the identity for cos (3θ ) with a specific value, such as θ = • LHS:

:

=0 • RHS

=0 The identity holds true, supporting the proof.

Example 2 Use binomial expansion and De Moivre’s theorem to prove cos (4θ ) = 8 cos4(θ ) − 8 cos2(θ ) + 1 and sin (4θ ) = 4 sin (θ ) cos (θ ) (2 cos2 (θ ) − 1).

Create a strategy The trigonometric identities to be proved involve the angle 4θ. Thus, z4, where z = 1 cis (θ ), and the 4th row of Pascal’s triangle are required in the proof.

Apply the idea Let z = 1 cis (θ ).

Consider z4 = (1 cis (θ ))4 = (cos (θ ) + i sin (θ ))4.

Expanding using the fourth row of Pascal’s triangle (1 4 6 4 1): z4 = cos4 (θ ) + 4 cos3 (θ ) (i sin (θ )) + 6 cos2 (θ ) (i sin (θ ))2 + 4 cos (θ ) (i sin (θ ))3 + (i sin (θ ))4 4

3

2

2

2

= c os (θ ) + 4i3 cos3 (θ ) sin4(θ ) 4+ 6 cos (θ ) i sin (θ ) + 4 cos (θ ) i sin (θ ) + i sin (θ ) 4

3

4

2

2

2

(θ ) - 6 cos (θ ) sin (θ ) = c os (θ ) + 4i3cos (θ ) sin 4i cos (θ ) sin (θ ) + sin4 (θ ) 2

Simplify powers of brackets Evaluate powers of i

4

(θ )) + = ( cos (θ3) - 6 cos (θ ) sin (θ ) + sin 3 i (4 cos (θ ) sin (θ ) - 4 cos (θ ) sin (θ )) Using De Moivre’s theorem: z4 = cos (4θ ) + i sin (4θ )

96

Apply the binomial theorem

Mathspace QLD 12 Specialist Mathematics 2025 mathspace.co

Group real and imaginary parts

Therefore, equating the two expressions for z4: 1

cos (4θ ) + i sin (4θ ) = ( cos4 (θ ) - 6 cos2 (θ ) sin2 (θ ) + sin4 (θ )) + i (4 cos3 (θ ) sin (θ ) 4 cos (θ ) sin3 (θ ))

Equate the results from De Moivre’s theorem and the binomial expansion

Equating the real parts of 1 gives: cos (4θ ) = cos4 (θ ) − 6 cos2 (θ ) sin2 (θ ) + sin4 (θ ) Applying the Pythagorean trigonometric identity: sin2 (θ ) = 1 − cos2 (θ ) cos (4θ ) = c os4 (θ ) - 6 cos2 (θ ) (1 - cos (θ ))2 (1 - cos2 (θ ))2 Substitute sin2 (θ ) = 1 − cos2 (θ ) 4

2

4

= cos (θ ) -26 cos (θ )4+ 6 cos (θ ) + (1 - 2 cos (θ ) + cos (θ )) 4

4

4

Expand the brackets 2

= cos (2θ ) + 6 cos (θ ) + cos (θ ) - 6 cos (θ ) 2 cos (θ ) + 1

Group like terms

= 8 cos4 (θ ) - 8 cos2 (θ ) + 1

Simplify the expression

Therefore, cos (4θ ) = 8 cos4 (θ ) − 8 cos2 (θ ) + 1.

Similarly, equating the imaginary parts of 1 gives: cos3 (θ ) sin (θ ) - 4 cos (θ ) sin3 (θ ) sin (4θ ) = 4

Equate imaginary components

= 4 cos (θ ) sin (θ ) (cos2 (θ ) - sin2 (θ ))

Factor out the common term 4 cos (θ ) sin (θ )

= 4 sin (θ ) cos (θ ) (2 cos2 (θ ) - 1)

Substitute sin2 (θ ) = 1 − cos2 (θ ) and simplify

Therefore, sin (4θ ) = 4 sin (θ ) cos (θ ) (2 cos2 (θ ) − 1).

Reflect and check To verify the result, test the identity with a specific value, such as θ • LHS:

:

= -1 • RHS

=2-4+1 = -1 The identity holds true, supporting the proof.

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Idea summary Multi-angle trigonometric identities can be derived by applying De Moivre’s theorem and binomial expansion to (cos (θ ) + i sin (θ ))n. Equate the real parts to determine cos (nθ ) and the imaginary parts for sin (nθ ). Use binomial coefficients from Pascal’s triangle, evaluate powers of i, and apply the Pythagorean identity to simplify.

2.05 Practice questions Simple familiar 1

Use the binomial theorem to expand (a + b)4.

2

Let z = cos (θ ) + i sin (θ ). Use De Moivre’s theorem to write z5 in the form cos ( A) + i sin ( A).

3

Consider the expansion of (cos (θ ) + isin (θ ))3: a

Determine the term(s) that are purely real.

b

Determine the term(s) that are purely imaginary.

4

Using the identity cos2 (θ ) + sin2 (θ ) = 1, express cos2 (θ ) in terms of sin (θ ).

5

By considering the expansion of (cos (θ ) + i sin (θ ))2, prove the double-angle identities: a

cos (2θ ) = cos2 (θ ) − sin2 (θ )

b

sin (2θ ) = 2 sin (θ ) cos (θ )

Complex familiar 6

Use the binomial expansion and De Moivre’s theorem to prove:

Ex 1

cos (3θ ) = 4 cos3 (θ ) − 3 cos (θ )

7

Use the binomial expansion and De Moivre’s theorem to prove sin (3θ ) = 3 sin (θ ) − 4 sin3 (θ ).

8

Use the binomial expansion and De Moivre’s theorem to prove:

Ex 2

9

sin (4θ ) = 4 cos3 (θ ) sin (θ ) − 4 cos (θ ) sin3 (θ ) Use the binomial expansion and De Moivre’s theorem to prove: cos (4θ ) = 8 cos4 (θ ) − 8 cos2 (θ ) + 1

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Mathspace QLD 12 Specialist Mathematics 2025 mathspace.co

Complex unfamiliar 10

By considering the expansion of (cos (θ ) + i sin (θ ))3, prove that tan (3θ )

11

Consider the equation 8x3 − 6x − 1 = 0: a b

.

By substituting x = cos (θ ), show that the equation can be written in the form cos (3θ )

.

3

Determine the 3 real roots of the equation 8x − 6x − 1 = 0 in the form cos (θ ).

12

By using the identities for sin (4θ ) and cos (4θ ), prove that tan (4θ )

13

a

Prove that sin (5θ ) = 16 sin5 (θ ) − 20 sin3 (θ ) + 5 sin (θ ).

b

By letting θ = 36°, determine the exact value of sin (36°).

.

Did you know?

Infinitely complex fractals, like the Koch snowflake, are created using a process similar to proof by induction! You start with a simple base shape, such as a single triangle, then apply a rule to each line segment over and over again to create a shape with a finite area but an endlessly long perimeter.

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