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Vector calculus
Further matrices
Rates
inference
2 Mathematical induction and trigonometric proofs
Chapter outline
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
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De Moivre’s theorem for complex numbers states that if z = r cis (θ), then z n = r n 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 z n = r n cis(nθ) for all positive integers n ≥ 1.
• Step 2: Initial statement.
Show the statement is true for n = 1:
LHS = z 1
= r cis (θ)
RHS = r 1 cis (1 × θ)
= r cis (θ)
Since LHS = RHS, the statement is true for n = 1.
• Step 3: Assumption statement.
For n = k ∈ +, assume that z k = r k cis (kθ).
Start with the left-hand side for n = 1
By definition, z 1 is z
Start with the right-hand side for n = 1
Simplify the expression
• Step 4: Prove the inductive step.
Prove the statement is true for n = k + 1.
The goal is to prove: z k+1 = r k+1 cis ((k + 1)θ)
LHS = z k+1
= z k × z 1
= (r k cis(kθ)) × (r cis(θ))
= r k+1 (cos(kθ) + i sin(kθ))(cos(θ) + i sin(θ))
= r k+1 [(cos(kθ) cos(θ) sin(kθ) sin(θ)) + i (sin(kθ) cos(θ) + cos(kθ) sin(θ))]
= r k+1 (cos(kθ + θ) + i sin(kθ + θ))
= r k+1 (cos((k + 1)θ) + i sin((k + 1)θ))
= r k+1 cis((k + 1)θ)
= RHS
Start with the left-hand side
Apply index laws
Use the assumption statement
Expand the cis notation and combine r terms
Expand brackets and group real and imaginary parts, using i2 =−1
Apply trigonometric sum identities
Factorise out θ
Write in cis notation
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 z n = r n (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
Example 1
Use De Moivre’s theorem and the binomial expansion to prove these trigonometric identities:
• cos (2θ) = cos 2 (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 = a 2 + 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 (
By binomial expansion:
Apply the binomial theorem
Simplify powers of brackets
Evaluate i2 = 1
Group real and imaginary parts
Equate the results from De Moivre’s theorem and the binomial expansion:
• Proof for the cosine identity
Equate the real parts:
• Proof for the sine identity
Equate the imaginary parts:
Equate real components
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)
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 = a 3 + 3a 2b + 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 cos 2 (θ)(i sin (θ)) + 3 cos (θ) (i sin (θ))2 + (i sin (θ))3
Apply the binomial theorem = cos3 (θ) + 3i cos 2 (θ) sin (θ) + 3 cos (θ) i2 sin2 (θ) + i3 sin3 (θ)
Simplify powers of brackets = cos3 (θ) + 3i cos 2 (θ) sin (θ) 3 cos (θ) sin2 (θ) i sin3 (θ)
= (cos3 (θ) 3 cos (θ) sin2 (θ)) + i (3 cos 2 (θ) sin (θ) sin3 (θ))
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 cos 2 (θ) sin (θ) sin3 (θ))
• Proof for the cosine identity
Equate the real parts:
cos (3θ) = cos3 (θ) 3 cos (θ) sin2 (θ)
= cos3 (θ) 3 cos (θ) (1 cos 2 (θ))
= cos3 (θ) 3 cos (θ) + 3 cos3 (θ)
= 4 cos3 (θ) 3 cos (θ)
• Proof for the sine identity
Equate the imaginary parts:
sin (3θ) = 3 cos 2 (θ) sin (θ) sin3 (θ)
= 3 (1 sin2 (θ)) sin (θ) sin3 (θ)
= 3 sin (θ) 3 sin3 (θ) sin3 (θ)
= 3 sin (θ) 4 sin3 (θ)
Equate real components
Substitute sin2 (θ) = 1 cos 2 (θ)
Expand the brackets
Collect like terms
Equate imaginary components
Substitute cos2 (θ) = 1 sin2 (θ)
Expand the brackets
Collect like terms
2.04 Trigonometric identities and De Moivre’s theorem
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 = 6 (or 30°):
• 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 (cos 2 (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 z n, 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 z n according to De Moivre’s theorem. Identify the modulus and argument of z n
3 Expand the expression (cos(θ) + i sin(θ))2 using the binomial theorem.
4 Given a complex number z = 3cis :
a Use De Moivre’s theorem to determine z 2 in polar form.
b Use De Moivre’s theorem to determine z 4 in polar form.
Complex familiar
5 Use De Moivre’s theorem and the binomial expansion to prove these trigonometric identities:
• cos (2θ) = 2 cos 2 (θ) 1
• sin (2θ) = 2 sin (θ) cos (θ)
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 Given the complex number z = 1 + i:
a Express z in polar form.
b Use De Moivre’s theorem to evaluate z 8 in polar form.
c Convert your answer from part (b) to Cartesian form (a + bi).
8 Given the complex number z = 1 + i :
a Express z in polar form.
b Use De Moivre’s theorem to evaluate z 5 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
2.05 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 = cosn−r θ(i sin (θ))r
The binomial coefficients 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 sin2 (θ) + cos 2 (θ) = 1
Double-angle identities sin (2θ) = 2 sin (θ) cos (θ) cos (2θ) = cos 2 (θ) sin2 (θ) = 2 cos 2 (θ) 1 = 1 2 sin2 (θ)
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 = a 3 + 3a 2b + 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: (cos (θ) + i sin (θ))3 = cos3 (θ) + 3 cos 2 (θ) (i sin (θ)) + 3 cos (θ) (i sin (θ))2 + (i sin (θ))3
Apply the binomial theorem = cos3 (θ) + 3i cos 2 (θ) sin (θ) + 3 cos (θ) (i2 sin2 (θ)) + i3 sin3 (θ)
Simplify powers of brackets = cos3 (θ) + 3i cos 2 (θ) sin (θ) 3 cos (θ) sin2 (θ) i sin3 (θ)
= (cos3 (θ) 3 cos (θ) sin2 (θ)) + i (3 cos 2 (θ) sin (θ) sin3 (θ))
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 cos 2 (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 = cos3 (θ) 3 cos (θ) (1 cos 2 (θ))
= cos3 (θ) 3 cos (θ) + 3 cos3 (θ)
= 4 cos3(θ) 3 cos (θ)
• Proof for the sine identity
Equate the imaginary parts:
sin (3θ) = 3 cos2(θ) sin (θ) sin3 (θ)
= 3 (1 - sin2(θ)) sin (θ) sin3 (θ)
= 3 sin (θ) 3 sin3 (θ) sin3 (θ)
= 3 sin (θ) 4 sin3 (θ)
Substitute sin2 (θ) = 1 cos 2 (θ)
Expand the brackets
Collect like terms
Equate imaginary components
Substitute cos2 (θ) = 1 sin2 (θ)
Expand the brackets
Collect like terms
2.05 Problems with trigonometric identities
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 cos 2 (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 z 4 = (1 cis (θ))4 = (cos (θ) + i sin (θ))4.
Expanding using the fourth row of Pascal’s triangle (1 4 6 4 1):
z 4 = cos 4 (θ) + 4 cos3 (θ) (i sin (θ)) + 6 cos 2 (θ) (i sin (θ))2 + 4 cos (θ) (i sin (θ))3 + (i sin (θ))4
= cos 4 (θ) + 4i cos3 (θ) sin (θ) + 6 cos 2 (θ) i2 sin2 (θ) + 4 cos (θ) i3 sin3 (θ) + i4 sin4 (θ)
= cos 4 (θ) + 4i cos3 (θ) sin (θ) 6 cos 2 (θ) sin2 (θ) 4i cos (θ) sin3 (θ) + sin4 (θ)
= (cos 4 (θ) 6 cos 2 (θ) sin2 (θ) + sin4 (θ)) + i (4 cos3 (θ) sin (θ) 4 cos (θ) sin3 (θ))
Using De Moivre’s theorem: z 4 = cos (4θ) + i sin (4θ)
Apply the binomial theorem
Simplify powers of brackets
Evaluate powers of i
Group real and imaginary parts
Therefore, equating the two expressions for z 4 :
1 cos (4θ) + i sin (4θ) = (cos 4 (θ) 6 cos 2 (θ) sin2 (θ) + sin4 (θ)) + i (4 cos3 (θ) sin (θ) 4 cos (θ) sin3 (θ))
Equating the real parts of 1 gives:
Equate the results from De Moivre’s theorem and the binomial expansion
Applying the Pythagorean trigonometric identity: sin2 (θ) = 1 cos 2 (θ) cos (4θ) = cos 4 (θ) 6 cos 2 (θ) (1 cos (θ))2 (1
Substitute sin2 (θ) = 1 cos 2 (θ) = cos 4 (θ) 6 cos 2 (θ) + 6 cos 4 (θ) + (1 2 cos 2 (θ) + cos 4 (θ))
Expand the brackets = cos 4 (θ
Group like terms
Simplify the expression
Therefore, cos (4θ) = 8 cos 4 (θ) 8 cos 2 (θ) + 1.
Similarly, equating the imaginary parts of 1 gives: sin (4θ) = 4 cos3 (θ) sin (θ) 4 cos (θ) sin3 (θ)
Equate imaginary components = 4 cos (θ) sin (θ) (cos 2 (θ) − sin2 (θ))
Factor out the common term 4 cos (θ) sin (θ) = 4 sin (θ) cos (θ) (2 cos 2 (θ) − 1)
Substitute sin2 (θ) = 1 cos 2 (θ) and simplify
Therefore, sin (4θ) = 4 sin (θ) cos (θ) (2 cos 2 (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.
2.05 Problems with trigonometric identities
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 z 5 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) = cos 2 (θ) sin2 (θ) b sin(2θ) = 2sin(θ) cos(θ)
Complex
familiar
6 Use the binomial expansion and De Moivre’s theorem to prove: cos(3θ) = 4cos3 (θ) 3cos(θ)
7 Use the binomial expansion and De Moivre’s theorem to prove sin(3θ) = 3sin(θ) 4sin3 (θ).
8 Use the binomial expansion and De Moivre’s theorem to prove: sin(4θ) = 4cos3 (θ) sin(θ) 4cos(θ) sin3 (θ)
9 Use the binomial expansion and De Moivre’s theorem to prove: cos(4θ) = 8cos4 (θ) 8cos2 (θ) + 1
10 By considering the expansion of (cos (θ) + i sin (θ))3, prove that tan (3θ)
11
Consider the equation 8x 3 6x 1 = 0:
a By substituting x = cos (θ), show that the equation can be written in the form cos (3θ)
b Determine the 3 real roots of the equation 8x 3 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.
