Why is "Losing track of the imaginary unit $i$ during the binomial expansion" a common mistake in Applications of De Moivre’s Theorem in Trigonometry and Complex Numbers?

Why it happens: Treating i as a constant to factor out early, or forgetting i^2 = -1, i^3 = -i, i^4 = 1. How to avoid it: Keep i attached to every factor and simplify powers of i carefully; only separate real and imaginary parts at the very end. Source: 9231 Paper 2 Examiner Reports — recurring comment

Why is "Mishandling the $(2i)^n$ factor when converting $\sin^n\theta$" a common mistake in Applications of De Moivre’s Theorem in Trigonometry and Complex Numbers?

Why it happens: Forgetting that = 1/2i(z - \frac1z), so ^n carries a 1/(2i)^n that affects sign and (for odd n) leaves an i. How to avoid it: Compute (2i)^n explicitly: (2i)^2 = -4, (2i)^3 = -8i, (2i)^4 = 16. For odd n the residual i must cancel against the z^k - 1/z^k = 2i terms. Source: 9231 Paper 2 mark schemes

Why is "Failing to pair symmetric terms $z^k + \frac{1}{z^k}$ before converting" a common mistake in Applications of De Moivre’s Theorem in Trigonometry and Complex Numbers?

Why it happens: Trying to convert each z^k individually instead of recognising the 2 k pattern in the pairs. How to avoid it: After the binomial expansion, group terms equidistant from the centre into z^k + 1/z^k pairs; the unpaired middle term is the constant. Source: 9231 Paper 2 Examiner Reports

Why is "Guessing the closed form of a $C + iS$ series instead of factoring out half-angles" a common mistake in Applications of De Moivre’s Theorem in Trigonometry and Complex Numbers?

Why it happens: Knowing the target answer but not the technique, so writing it down without the geometric-series and half-angle steps. How to avoid it: Always sum z^k with the geometric formula, then factor out e^i/2 etc. The method marks are awarded for the working, not the final line. Source: 9231 Paper 2 Examiner Reports

Why is "Taking the wrong part (real instead of imaginary, or vice versa)" a common mistake in Applications of De Moivre’s Theorem in Trigonometry and Complex Numbers?

Why it happens: Not tracking which series is C (cosine, real) and which is S (sine, imaginary). How to avoid it: Remember: n and k are REAL parts; n and k are IMAGINARY parts. Label clearly before extracting. Source: 9231 Paper 2 mark schemes

Why is "Omitting the convergence condition $|r| < 1$ for an infinite series" a common mistake in Applications of De Moivre’s Theorem in Trigonometry and Complex Numbers?

Why it happens: Applying a/1-r automatically without checking the ratio's modulus. How to avoid it: For an infinite C + iS, state |z/c| < 1 (e.g. |z/2| = \frac12 < 1) before summing. It is a marked step. Source: 9231 Paper 2 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

Applications of De Moivre’s Theorem in Trigonometry and Complex Numbers

Q: Why is "Omitting the convergence condition $|r| < 1$ for an infinite series" a common mistake in Applications of De Moivre’s Theorem in Trigonometry and Complex Numbers?

Why it happens: Applying a/1-r automatically without checking the ratio's modulus. How to avoid it: For an infinite C + iS, state |z/c| < 1 (e.g. |z/2| = \frac12 < 1) before summing. It is a marked step. Source: 9231 Paper 2 Examiner Reports

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Detailed Study Notes

Detailed notes on Complex numbers for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Applications of De Moivre's Theorem in Trigonometry — Cambridge International AS & A Level Further Mathematics 9231 Paper 2 (Further Pure Mathematics 2, 2026-2027 syllabus)

De Moivre's theorem is a machine for trigonometric identities. Expand $(\cos\theta + i\sin\theta)^n$ to get $\cos n\theta$ and $\sin n\theta$ in powers of $\cos\theta, \sin\theta$ ; use $z + \frac1z = 2\cos\theta$ to do the reverse; and sum trigonometric series with the $C + iS$ method.

At a glance

Multiple-angle to powers: expand $(\cos\theta + i\sin\theta)^n$ with the binomial theorem and equate real/imaginary parts to get $\cos n\theta$ , $\sin n\theta$ .
Powers to multiple angles: with $z = e^{i\theta}$ , use $z + \frac1z = 2\cos\theta$ and $z - \frac1z = 2i\sin\theta$ , plus $z^n + \frac{1}{z^n} = 2\cos n\theta$ .
$C + iS$ method: a sum of $\cos$ (or $\sin$ ) terms is the real (or imaginary) part of a geometric series of complex exponentials — sum it, then take parts.
These methods are NOT in MF19 — the approach must be known and reproduced from scratch.
Always equate real with real, imaginary with imaginary — the heart of every derivation.

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

2.5d — Express $\cos n\theta$ and $\sin n\theta$ in terms of powers of $\cos\theta$ and $\sin\theta$ using de Moivre's theorem and the binomial expansion.
2.5e — Express powers such as $\cos^n\theta$ and $\sin^n\theta$ in terms of multiple angles using $z + \frac{1}{z} = 2\cos\theta$ and $z - \frac{1}{z} = 2i\sin\theta$ .
2.5f — Sum trigonometric series by recognising them as the real or imaginary part of a geometric series of complex exponentials (the $C + iS$ method).

cos nθ and sin nθ in powers of cos θ and sin θ

Expand (cos θ + i sin θ)ⁿ with the binomial theorem, then equate real and imaginary parts.

The idea. De Moivre says $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ . But the left side can also be expanded with the binomial theorem. Equating the two expressions, then matching real and imaginary parts, gives $\cos n\theta$ and $\sin n\theta$ as polynomials in $\cos\theta, \sin\theta$ .

Worked derivation for $n = 3$ . Expand $(\cos\theta + i\sin\theta)^3$ using $(a+b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3$ with $a = \cos\theta$ , $b = i\sin\theta$ : $\cos^3\theta + 3\cos^2\theta(i\sin\theta) + 3\cos\theta(i\sin\theta)^2 + (i\sin\theta)^3.$

Using $i^2 = -1$ , $i^3 = -i$ : $= \bigl(\cos^3\theta - 3\cos\theta\sin^2\theta\bigr) + i\bigl(3\cos^2\theta\sin\theta - \sin^3\theta\bigr).$

Equate with $\cos 3\theta + i\sin 3\theta$ :

Real part: $\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta$ .
Imaginary part: $\sin 3\theta = 3\cos^2\theta\sin\theta - \sin^3\theta$ .

Tidy into a single function using $\sin^2\theta = 1 - \cos^2\theta$ (for $\cos$ ) or $\cos^2\theta = 1 - \sin^2\theta$ (for $\sin$ ): $\cos 3\theta = 4\cos^3\theta - 3\cos\theta, \qquad \sin 3\theta = 3\sin\theta - 4\sin^3\theta.$

The general recipe:

Expand $(\cos\theta + i\sin\theta)^n$ by the binomial theorem.
Use $i^2 = -1, i^3 = -i, i^4 = 1, \ldots$ to separate real and imaginary parts.
Equate real $\to \cos n\theta$ , imaginary $\to \sin n\theta$ .
If asked for a single-function form, substitute $\sin^2\theta = 1 - \cos^2\theta$ etc.

A neat by-product is $\tan n\theta$ : divide $\sin n\theta$ by $\cos n\theta$ and then divide top and bottom by $\cos^n\theta$ to get an expression purely in $\tan\theta$ .

Cambridge tip. Keep the imaginary unit $i$ attached to the $\sin$ terms throughout — only at the very end do you read off "imaginary part = coefficient of $i$ ". Dropping $i$ mid-expansion scrambles which terms are real.

$\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ , $\;\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ .
Real part of the binomial expansion $\to \cos n\theta$ ; imaginary part $\to \sin n\theta$ .
Use $\sin^2\theta = 1 - \cos^2\theta$ to reduce to a single function.

See the full worked example for applications of de moivre’s theorem in trigonometry and complex numbers →

Powers of cos θ and sin θ as multiple angles

Use z + 1/z = 2 cos θ and z − 1/z = 2i sin θ to turn cosⁿθ into a sum of cos kθ.

This is the reverse problem: express $\cos^n\theta$ or $\sin^n\theta$ as a sum of $\cos k\theta$ / $\sin k\theta$ — exactly the form you need to integrate such powers.

The key tools. Let $z = \cos\theta + i\sin\theta = e^{i\theta}$ . By de Moivre, $\frac{1}{z} = z^{-1} = \cos\theta - i\sin\theta$ . Adding and subtracting: $z + \frac{1}{z} = 2\cos\theta, \qquad z - \frac{1}{z} = 2i\sin\theta,$ and more generally $z^n + \frac{1}{z^n} = 2\cos n\theta, \qquad z^n - \frac{1}{z^n} = 2i\sin n\theta.$

The recipe (for $\cos^4\theta$ ):

Write $\cos^4\theta = \left(\dfrac{z + \frac1z}{2}\right)^4 = \dfrac{1}{16}\left(z + \dfrac1z\right)^4$ .
Expand with the binomial theorem: $\left(z + \tfrac1z\right)^4 = z^4 + 4z^2 + 6 + \frac{4}{z^2} + \frac{1}{z^4}.$
Pair symmetric terms $z^k + \frac{1}{z^k} = 2\cos k\theta$ : $= \left(z^4 + \tfrac{1}{z^4}\right) + 4\left(z^2 + \tfrac{1}{z^2}\right) + 6 = 2\cos 4\theta + 8\cos 2\theta + 6.$
Divide by $16$ : $\cos^4\theta = \frac{1}{8}\cos 4\theta + \frac{1}{2}\cos 2\theta + \frac{3}{8}.$

For odd powers of $\sin$ , remember the $(2i)$ factors: $\sin^n\theta = \left(\frac{z - 1/z}{2i}\right)^n$ , and the $(2i)^n$ in the denominator can introduce a $\pm$ sign and (for odd $n$ ) leave an $i$ that must cancel against the imaginary terms.

z and 1/z are mirror images in the real axis, so z + 1/z lies on the real axis with value 2 cos θ; z − 1/z is purely imaginary, 2i sin θ.

Cambridge tip. Always group the expansion into symmetric pairs $z^k + \frac{1}{z^k}$ before converting — this is where the $2\cos k\theta$ pattern appears cleanly. The constant middle term has no partner and becomes the $\cos 0\theta$ (constant) part.

$z + \frac1z = 2\cos\theta$ , $\;z - \frac1z = 2i\sin\theta$ , $\;z^n + \frac{1}{z^n} = 2\cos n\theta$ .
$\cos^4\theta = \frac18\cos 4\theta + \frac12\cos 2\theta + \frac38$ .
Pair symmetric powers before converting; the form is ideal for integration.

See the full worked example for applications of de moivre’s theorem in trigonometry and complex numbers →

Summing series — the C + iS method

A cosine/sine series is the real/imaginary part of a geometric series of complex exponentials.

Many Paper 2 questions ask you to sum a series such as $\cos\theta + \cos 2\theta + \cdots + \cos n\theta$ . The trick: bolt on the matching sine series with an $i$ , making a complex geometric series.

Set up. Let $C = \sum_{k=1}^{n}\cos k\theta, \qquad S = \sum_{k=1}^{n}\sin k\theta.$ Then $C + iS = \sum_{k=1}^{n}(\cos k\theta + i\sin k\theta) = \sum_{k=1}^{n} z^k, \quad z = e^{i\theta}.$

This is a geometric series with first term $z$ , common ratio $z$ , $n$ terms: $C + iS = z\,\frac{z^n - 1}{z - 1} = \frac{z^{n+1} - z}{z - 1}.$

Extract real and imaginary parts. The standard manipulation factors $e^{i\theta/2}$ out of numerator and denominator to expose the $z^k - z^{-k} = 2i\sin$ pattern: $\frac{z^n - 1}{z - 1} = \frac{e^{in\theta/2}\left(e^{in\theta/2} - e^{-in\theta/2}\right)}{e^{i\theta/2}\left(e^{i\theta/2} - e^{-i\theta/2}\right)} = e^{i(n-1)\theta/2}\cdot\frac{2i\sin\frac{n\theta}{2}}{2i\sin\frac{\theta}{2}} = e^{i(n-1)\theta/2}\,\frac{\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}.$

Multiplying by the leading $z = e^{i\theta}$ gives a clean modulus-argument form, from which: $C = \sum_{k=1}^{n}\cos k\theta = \frac{\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}\cos\frac{(n+1)\theta}{2}, \qquad S = \sum_{k=1}^{n}\sin k\theta = \frac{\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}\sin\frac{(n+1)\theta}{2}.$

The recipe:

Form $C + iS$ and recognise $\sum z^k$ as geometric.
Sum it with the geometric formula $\frac{a(r^n - 1)}{r - 1}$ .
Factor out half-angle exponentials to expose $2i\sin$ in top and bottom.
Read off $C$ (real part) and $S$ (imaginary part).

Infinite version. If $|r| < 1$ (e.g. terms like $\frac{\cos k\theta}{2^k}$ ), the series converges and you sum $\frac{a}{1 - r}$ , then realise the denominator by multiplying by its conjugate.

Cambridge tip. The half-angle factor-out is the single technique examiners look for. Practise it until $\frac{z^n - 1}{z - 1} \to e^{i(n-1)\theta/2}\frac{\sin(n\theta/2)}{\sin(\theta/2)}$ is automatic — guessing the closed form loses the method marks.

$C + iS = \sum (\cos k\theta + i\sin k\theta) = \sum z^k$ is geometric.
Factor out half-angle exponentials to get $\frac{\sin(n\theta/2)}{\sin(\theta/2)}$ .
$C$ is the real part, $S$ the imaginary part of the summed series.

See the full worked example for applications of de moivre’s theorem in trigonometry and complex numbers →

Why these methods matter and when to reach for each

Choosing the right de Moivre tool for the question in front of you.

Three closely related techniques, each suited to a different request:

1. Want $\cos n\theta$ in powers of $\cos\theta$ ? Expand $(\cos\theta + i\sin\theta)^n$ binomially and equate real/imaginary parts. Direction: multiple angle $\to$ powers.

2. Want to integrate $\cos^n\theta$ or prove a power identity? Use $z + \frac1z = 2\cos\theta$ . Direction: powers $\to$ multiple angle (sums of $\cos k\theta$ are trivial to integrate).

3. Want to sum a $\cos$ or $\sin$ series? Use $C + iS$ and a geometric series.

A common combined question asks you to (a) show $\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta$ using de Moivre, then (b) solve an equation such as $16x^5 - 20x^3 + 5x = 0$ by setting $x = \cos\theta$ — the roots are then $\cos$ of specific angles. This links Topic 23.2 back to roots-and-coefficients ideas.

Match the tool to the direction the question asks: expand for powers, z + 1/z for multiple angles, C + iS for series.

Cambridge tip. Read the command word carefully: "express $\cos 4\theta$ in terms of $\cos\theta$ " $\to$ method 1; "express $\sin^5\theta$ in terms of sines of multiples of $\theta$ " $\to$ method 2; "find the sum of ... $\theta$ " $\to$ method 3.

Binomial expansion: multiple angle $\to$ powers.
$z + \frac1z$ : powers $\to$ multiple angle (for integration / power identities).
$C + iS$ : summing trigonometric series.

See the full worked example for applications of de moivre’s theorem in trigonometry and complex numbers →

Quick recap

Expand $(\cos\theta + i\sin\theta)^n$ and equate parts to get $\cos n\theta$ , $\sin n\theta$ in powers.
$\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ ; $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ .
$z + \frac1z = 2\cos\theta$ , $z - \frac1z = 2i\sin\theta$ , $z^n + \frac{1}{z^n} = 2\cos n\theta$ .
$\cos^4\theta = \frac18\cos 4\theta + \frac12\cos 2\theta + \frac38$ — pair symmetric powers, then convert.
$C + iS = \sum z^k$ is geometric; factor out half-angle exponentials, then read off real/imaginary parts.
None of these methods is in MF19 — the technique must be reproduced fully.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

$z + \frac{1}{z} = 2\cos\theta$ and $z - \frac{1}{z} = 2i\sin\theta$ (with $z = e^{i\theta}$ ).
$z^n + \frac{1}{z^n} = 2\cos n\theta$ , $z^n - \frac{1}{z^n} = 2i\sin n\theta$ .
$\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ , $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ (and how to derive them).
The $C + iS$ set-up: $\sum(\cos k\theta + i\sin k\theta) = \sum z^k$ , summed as a geometric series.
The half-angle factor-out $\frac{z^n - 1}{z - 1} = e^{i(n-1)\theta/2}\frac{\sin(n\theta/2)}{\sin(\theta/2)}$ .

How it’s examined

These applications are heavily examined in Paper 2 (FPM2), often worth 8-12 marks across a multi-part question. A typical structure: (a) use de Moivre to derive a multiple-angle identity (e.g. $\cos 5\theta$ in powers of $\cos\theta$ ), (b) use the identity to solve a polynomial equation or evaluate an integral, OR sum a trigonometric series via $C + iS$ . Marks are lost through (i) algebra slips in the binomial expansion or $i$ -powers, (ii) failing to factor out half-angle exponentials in the $C + iS$ method and instead guessing the closed form, and (iii) not stating which part is real and which imaginary. Because the methods are not in MF19, fluent reproduction of the technique is essential.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 2, Further Pure Mathematics 2); 9231/21, 9231/23 Further Pure Mathematics 2 — past papers and mark schemes. Last reviewed 2026-06-23.

Take this whole topic with you

Step-by-step worked examples — Applications of De Moivre’s Theorem in Trigonometry and Complex Numbers

Step-by-step solutions to past-paper-style questions on applications of de moivre’s theorem in trigonometry and complex numbers, written exactly the way a tutor would explain them at the board.

Question type:

1cos 3θ in powers of cos θ
Getting startedShow that / prove• multiple-angle, binomial
Question
Use de Moivre's theorem to show that $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ .
Step-by-step solution
1. Step 1
  Start from de Moivre and expand the left side with the binomial theorem.
  $\cos 3\theta + i\sin 3\theta = (\cos\theta + i\sin\theta)^3$
2. Step 2
  Expand $(a + b)^3$ with $a = \cos\theta$ , $b = i\sin\theta$ , using $i^2 = -1$ , $i^3 = -i$ .
  $= \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta$
3. Step 3
  Take the real part (which equals $\cos 3\theta$ ).
  $\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta$
4. Step 4
  Replace $\sin^2\theta = 1 - \cos^2\theta$ and simplify.
  $= \cos^3\theta - 3\cos\theta(1 - \cos^2\theta) = 4\cos^3\theta - 3\cos\theta$
Answer
$\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ .
Examiner tip
Keep the $i$ attached to the sine terms during the expansion; only at the 'real part' step do you discard them. This is a 'show that', so the binomial expansion must be visible.
2sin 4θ in terms of sin θ and cos θ
Getting startedShow that / prove• multiple-angle, binomial
Question
Use de Moivre's theorem to express $\sin 4\theta$ in terms of $\sin\theta$ and $\cos\theta$ .
Step-by-step solution
1. Step 1
  De Moivre with $n = 4$ .
  $\cos 4\theta + i\sin 4\theta = (\cos\theta + i\sin\theta)^4$
2. Step 2
  Binomial expansion $(a+b)^4 = a^4 + 4a^3 b + 6a^2 b^2 + 4ab^3 + b^4$ , with $b = i\sin\theta$ .
  $= \cos^4\theta + 4i\cos^3\theta\sin\theta - 6\cos^2\theta\sin^2\theta - 4i\cos\theta\sin^3\theta + \sin^4\theta$
3. Step 3
  Take the imaginary part (the coefficient of $i$ ), which equals $\sin 4\theta$ .
  $\sin 4\theta = 4\cos^3\theta\sin\theta - 4\cos\theta\sin^3\theta$
Answer
$\sin 4\theta = 4\cos^3\theta\sin\theta - 4\cos\theta\sin^3\theta = 4\cos\theta\sin\theta(\cos^2\theta - \sin^2\theta)$ .
Examiner tip
The factorised form $4\cos\theta\sin\theta(\cos^2\theta - \sin^2\theta)$ is equivalent and often preferred. The $i^4 = 1$ and $i^3 = -i$ steps are where sign errors creep in.
3cos⁴θ as a sum of multiple angles
Building confidenceMulti-step problem• power-to-multiple, z-plus-1-over-z
Question
Express $\cos^4\theta$ in terms of cosines of multiples of $\theta$ .
Step-by-step solution
1. Step 1
  Use $z + \frac1z = 2\cos\theta$ with $z = e^{i\theta}$ , so $\cos\theta = \frac12\left(z + \frac1z\right)$ .
  $\cos^4\theta = \left[\tfrac12\left(z + \tfrac1z\right)\right]^4 = \frac{1}{16}\left(z + \tfrac1z\right)^4$
2. Step 2
  Expand $\left(z + \frac1z\right)^4$ by the binomial theorem.
  $\left(z + \tfrac1z\right)^4 = z^4 + 4z^2 + 6 + \frac{4}{z^2} + \frac{1}{z^4}$
3. Step 3
  Pair symmetric terms using $z^k + \frac{1}{z^k} = 2\cos k\theta$ .
  $= \left(z^4 + \tfrac{1}{z^4}\right) + 4\left(z^2 + \tfrac{1}{z^2}\right) + 6 = 2\cos 4\theta + 8\cos 2\theta + 6$
4. Step 4
  Divide by 16.
  $\cos^4\theta = \frac{1}{8}\cos 4\theta + \frac{1}{2}\cos 2\theta + \frac{3}{8}$
Answer
$\cos^4\theta = \frac{1}{8}\cos 4\theta + \frac{1}{2}\cos 2\theta + \frac{3}{8}$ .
Examiner tip
Group into $z^k + \frac{1}{z^k}$ pairs before converting; the central term $6$ has no partner and becomes the constant $\frac38$ . This form is exactly what you need to integrate $\cos^4\theta$ .
4Summing a cosine series with C + iS
Building confidenceMulti-step problem• series, c-plus-is, geometric
Question
Show that $\displaystyle\sum_{k=0}^{n-1}\cos k\theta = \frac{\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}\cos\frac{(n-1)\theta}{2}$ (for $\sin\frac{\theta}{2} \neq 0$ ).
Step-by-step solution
1. Step 1
  Form $C + iS$ with $z = e^{i\theta}$ ; the series is geometric with first term $1$ , ratio $z$ , $n$ terms.
  $C + iS = \sum_{k=0}^{n-1}(\cos k\theta + i\sin k\theta) = \sum_{k=0}^{n-1} z^k = \frac{z^n - 1}{z - 1}$
2. Step 2
  Factor out half-angle exponentials from numerator and denominator.
  $\frac{z^n - 1}{z - 1} = \frac{e^{in\theta/2}\left(e^{in\theta/2} - e^{-in\theta/2}\right)}{e^{i\theta/2}\left(e^{i\theta/2} - e^{-i\theta/2}\right)}$
3. Step 3
  Use $e^{ix} - e^{-ix} = 2i\sin x$ ; the $2i$ factors cancel.
  $= e^{i(n-1)\theta/2}\,\frac{\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}$
4. Step 4
  Take the real part (using $e^{i(n-1)\theta/2} = \cos\frac{(n-1)\theta}{2} + i\sin\frac{(n-1)\theta}{2}$ ) to read off $C$ .
  $C = \frac{\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}\cos\frac{(n-1)\theta}{2}$
Answer
$\displaystyle\sum_{k=0}^{n-1}\cos k\theta = \frac{\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}\cos\frac{(n-1)\theta}{2}$ .
Examiner tip
The half-angle factor-out is the examined skill. Note the imaginary part hands you the sine series for free: $S = \frac{\sin(n\theta/2)}{\sin(\theta/2)}\sin\frac{(n-1)\theta}{2}$ .
5Deriving cos 5θ and solving an equation
StretchMulti-step problem• multiple-angle, solve-equation
Question
(a) Show that $\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta$ . (b) Hence find, in exact form, the roots of $16x^5 - 20x^3 + 5x = 0$ .
Step-by-step solution
1. Step 1
  (a) Expand $(\cos\theta + i\sin\theta)^5$ binomially and take the real part. Using $\sin^2\theta = 1 - \cos^2\theta$ throughout.
  $\cos 5\theta = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta$
2. Step 2
  Substitute $\sin^2\theta = 1 - \cos^2\theta$ and simplify (let $c = \cos\theta$ ).
  $\cos 5\theta = 16c^5 - 20c^3 + 5c$
3. Step 3
  (b) The equation is $\cos 5\theta = 0$ with $x = \cos\theta$ . Solve $\cos 5\theta = 0$ : $5\theta = \frac{\pi}{2} + m\pi$ , so $\theta = \frac{(2m+1)\pi}{10}$ .
  $\theta = \frac{\pi}{10}, \frac{3\pi}{10}, \frac{5\pi}{10}, \frac{7\pi}{10}, \frac{9\pi}{10}$
4. Step 4
  The five roots are $x = \cos\theta$ for these angles (note $\cos\frac{5\pi}{10} = \cos\frac{\pi}{2} = 0$ ).
  $x = \cos\tfrac{\pi}{10},\; \cos\tfrac{3\pi}{10},\; 0,\; \cos\tfrac{7\pi}{10},\; \cos\tfrac{9\pi}{10}$
Answer
$x = \cos\frac{\pi}{10},\; \cos\frac{3\pi}{10},\; 0,\; \cos\frac{7\pi}{10},\; \cos\frac{9\pi}{10}$ .
Examiner tip
Recognising $16x^5 - 20x^3 + 5x = \cos 5\theta$ with $x = \cos\theta$ is the 'hence' link. The five distinct angles $\frac{(2m+1)\pi}{10}$ in $[0, \pi]$ give the five distinct roots.
6tan 3θ in terms of tan θ
StretchMulti-step problem• multiple-angle, tangent
Question
Use de Moivre's theorem to show that $\tan 3\theta = \dfrac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$ .
Step-by-step solution
1. Step 1
  From the expansion of $(\cos\theta + i\sin\theta)^3$ : real part gives $\cos 3\theta$ , imaginary part gives $\sin 3\theta$ .
  $\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta, \quad \sin 3\theta = 3\cos^2\theta\sin\theta - \sin^3\theta$
2. Step 2
  Form $\tan 3\theta = \frac{\sin 3\theta}{\cos 3\theta}$ .
  $\tan 3\theta = \frac{3\cos^2\theta\sin\theta - \sin^3\theta}{\cos^3\theta - 3\cos\theta\sin^2\theta}$
3. Step 3
  Divide every term by $\cos^3\theta$ to express in $\tan\theta$ .
  $\tan 3\theta = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$
Answer
$\tan 3\theta = \dfrac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$ .
Examiner tip
The trick is dividing numerator and denominator by $\cos^3\theta$ so that $\frac{\sin\theta}{\cos\theta} = \tan\theta$ appears throughout. A clean, frequently-examined derivation.

Model Answers — Applications of De Moivre’s Theorem in Trigonometry and Complex Numbers

High-scoring sample answers for applications of de moivre’s theorem in trigonometry and complex numbers on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
3 marks
Use de Moivre's theorem to show that $\cos 2\theta = 2\cos^2\theta - 1$ .
Model answer
By de Moivre's theorem, $\cos 2\theta + i\sin 2\theta = (\cos\theta + i\sin\theta)^2.$

Expand the right-hand side: $(\cos\theta + i\sin\theta)^2 = \cos^2\theta + 2i\cos\theta\sin\theta + i^2\sin^2\theta = (\cos^2\theta - \sin^2\theta) + i(2\cos\theta\sin\theta).$

Equate real parts: $\cos 2\theta = \cos^2\theta - \sin^2\theta.$

Using $\sin^2\theta = 1 - \cos^2\theta$ : $\cos 2\theta = \cos^2\theta - (1 - \cos^2\theta) = 2\cos^2\theta - 1.$
Why this scores
M1 for the binomial expansion with $i^2 = -1$ , A1 for $\cos 2\theta = \cos^2\theta - \sin^2\theta$ , A1 for the final form using $\sin^2\theta = 1 - \cos^2\theta$ .
Question 2
4 marks
Show, using de Moivre's theorem, that $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ .
Model answer
By de Moivre's theorem, $\cos 3\theta + i\sin 3\theta = (\cos\theta + i\sin\theta)^3.$

Expand using $(a+b)^3$ with $a = \cos\theta$ , $b = i\sin\theta$ and $i^2 = -1$ , $i^3 = -i$ : $= \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta.$

Equate imaginary parts: $\sin 3\theta = 3\cos^2\theta\sin\theta - \sin^3\theta.$

Replace $\cos^2\theta = 1 - \sin^2\theta$ : $\sin 3\theta = 3(1 - \sin^2\theta)\sin\theta - \sin^3\theta = 3\sin\theta - 3\sin^3\theta - \sin^3\theta = 3\sin\theta - 4\sin^3\theta.$
Why this scores
M1 for the expansion, M1 for selecting the imaginary part, A1 for $3\cos^2\theta\sin\theta - \sin^3\theta$ , A1 for the final single-function form. 'Show that' means every step must appear.
Question 3
6 marks
Express $\sin^4\theta$ in terms of cosines of multiples of $\theta$ , and hence find $\displaystyle\int_0^{\pi/2}\sin^4\theta\,d\theta$ .
Model answer
Use $z - \frac1z = 2i\sin\theta$ , so $\sin\theta = \frac{1}{2i}\left(z - \frac1z\right)$ and $\sin^4\theta = \frac{1}{(2i)^4}\left(z - \tfrac1z\right)^4 = \frac{1}{16}\left(z - \tfrac1z\right)^4 \quad\text{(since } (2i)^4 = 16\text{)}.$

Expand: $\left(z - \tfrac1z\right)^4 = z^4 - 4z^2 + 6 - \frac{4}{z^2} + \frac{1}{z^4} = \left(z^4 + \tfrac{1}{z^4}\right) - 4\left(z^2 + \tfrac{1}{z^2}\right) + 6.$

Convert with $z^k + \frac{1}{z^k} = 2\cos k\theta$ : $= 2\cos 4\theta - 8\cos 2\theta + 6, \quad\text{so}\quad \sin^4\theta = \frac{1}{8}\cos 4\theta - \frac{1}{2}\cos 2\theta + \frac{3}{8}.$

Integrate from $0$ to $\frac{\pi}{2}$ : $\int_0^{\pi/2}\sin^4\theta\,d\theta = \left[\frac{\sin 4\theta}{32} - \frac{\sin 2\theta}{4} + \frac{3\theta}{8}\right]_0^{\pi/2} = (0 - 0 + \tfrac{3\pi}{16}) - (0) = \frac{3\pi}{16}.$
Why this scores
M1 for $\sin\theta = \frac{1}{2i}(z - \frac1z)$ and $(2i)^4 = 16$ , M1 expansion, A1 for $\sin^4\theta = \frac18\cos 4\theta - \frac12\cos 2\theta + \frac38$ ; M1 integrating, A1 limits, A1 for $\frac{3\pi}{16}$ . Watch the $(2i)^4 = 16$ (positive) factor.
Question 4
6 marks
(a) Show that $\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$ . (b) Hence solve $8\cos^4\theta - 8\cos^2\theta + 1 = 0$ for $0 \le \theta \le \frac{\pi}{2}$ .
Model answer
(a) By de Moivre, $\cos 4\theta$ is the real part of $(\cos\theta + i\sin\theta)^4$ . Expanding: $\cos 4\theta = \cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta.$ Replace $\sin^2\theta = 1 - \cos^2\theta$ (and $\sin^4\theta = (1-\cos^2\theta)^2$ ): $\cos 4\theta = \cos^4\theta - 6\cos^2\theta(1 - \cos^2\theta) + (1 - \cos^2\theta)^2 = 8\cos^4\theta - 8\cos^2\theta + 1.$

(b) The equation $8\cos^4\theta - 8\cos^2\theta + 1 = 0$ is exactly $\cos 4\theta = 0$ . Solving for $0 \le \theta \le \frac{\pi}{2}$ : $4\theta = \frac{\pi}{2}, \frac{3\pi}{2} \quad\Rightarrow\quad \theta = \frac{\pi}{8}, \frac{3\pi}{8}.$
Why this scores
(a) M1 expansion, M1 substitution of $\sin^2\theta$ , A1 for the stated identity. (b) M1 recognising the LHS as $\cos 4\theta$ , A1 for $\theta = \frac{\pi}{8}, \frac{3\pi}{8}$ . The 'hence' demands you use part (a), not solve the quartic in $\cos\theta$ directly.
Question 5
8 marks
By considering $\displaystyle\sum_{k=0}^{n-1} z^k$ where $z = \cos\theta + i\sin\theta$ , show that $\displaystyle\sum_{k=0}^{n-1}\sin k\theta = \frac{\sin\frac{n\theta}{2}\sin\frac{(n-1)\theta}{2}}{\sin\frac{\theta}{2}}$ .
Model answer
Set up $C + iS$ . With $z = \cos\theta + i\sin\theta = e^{i\theta}$ , $C + iS = \sum_{k=0}^{n-1}(\cos k\theta + i\sin k\theta) = \sum_{k=0}^{n-1} z^k.$

This is a geometric series with first term $1$ , common ratio $z$ and $n$ terms (valid since $z \neq 1$ , i.e. $\theta \neq 0$ ): $C + iS = \frac{z^n - 1}{z - 1}.$

Factor out half-angle exponentials: $\frac{z^n - 1}{z - 1} = \frac{e^{in\theta/2}\left(e^{in\theta/2} - e^{-in\theta/2}\right)}{e^{i\theta/2}\left(e^{i\theta/2} - e^{-i\theta/2}\right)} = e^{i(n-1)\theta/2}\cdot\frac{2i\sin\frac{n\theta}{2}}{2i\sin\frac{\theta}{2}} = e^{i(n-1)\theta/2}\,\frac{\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}.$

Write the exponential in trig form: $C + iS = \left[\cos\frac{(n-1)\theta}{2} + i\sin\frac{(n-1)\theta}{2}\right]\frac{\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}.$

Take the imaginary part to obtain $S$ : $\sum_{k=0}^{n-1}\sin k\theta = S = \frac{\sin\frac{n\theta}{2}\sin\frac{(n-1)\theta}{2}}{\sin\frac{\theta}{2}}.$
Why this scores
M1 forming $C + iS$ as a geometric series, M1 summing to $\frac{z^n - 1}{z - 1}$ , M1 factoring out half-angle exponentials, A1 reaching $e^{i(n-1)\theta/2}\frac{\sin(n\theta/2)}{\sin(\theta/2)}$ , M1 converting the exponential to trig form, A1 taking the imaginary part, A1 the stated result, B1 for noting the restriction $\theta \neq 0$ . The half-angle factor-out is the decisive technique.
Question 6
7 marks
Find the sum to infinity of the series $\displaystyle\sum_{k=1}^{\infty}\frac{\cos k\theta}{2^k}$ , where $\theta$ is a fixed angle.
Model answer
Form $C + iS$ with $z = e^{i\theta}$ : $C + iS = \sum_{k=1}^{\infty}\frac{\cos k\theta + i\sin k\theta}{2^k} = \sum_{k=1}^{\infty}\left(\frac{z}{2}\right)^k.$

This is an infinite geometric series with first term $\frac{z}{2}$ and ratio $\frac{z}{2}$ , where $\left|\frac{z}{2}\right| = \frac12 < 1$ , so it converges: $C + iS = \frac{z/2}{1 - z/2} = \frac{z}{2 - z}.$

Substitute $z = \cos\theta + i\sin\theta$ and realise the denominator by multiplying by its conjugate $\,(2 - \cos\theta) + i\sin\theta$ : $C + iS = \frac{\cos\theta + i\sin\theta}{(2 - \cos\theta) - i\sin\theta}\cdot\frac{(2 - \cos\theta) + i\sin\theta}{(2 - \cos\theta) + i\sin\theta}.$

The denominator becomes $(2 - \cos\theta)^2 + \sin^2\theta = 4 - 4\cos\theta + \cos^2\theta + \sin^2\theta = 5 - 4\cos\theta$ .

The real part of the numerator is $\cos\theta(2 - \cos\theta) - \sin\theta(-\sin\theta)$ ... taking the real part of the whole quotient: $C = \frac{2\cos\theta - \cos^2\theta + \sin^2\theta}{5 - 4\cos\theta} = \frac{2\cos\theta - \cos 2\theta}{5 - 4\cos\theta}.$

Hence $\sum_{k=1}^{\infty}\frac{\cos k\theta}{2^k} = \frac{2\cos\theta - \cos 2\theta}{5 - 4\cos\theta}.$
Why this scores
M1 forming $\sum(z/2)^k$ , M1 confirming $|z/2| = \frac12 < 1$ and summing $\frac{z/2}{1 - z/2}$ , A1 for $\frac{z}{2-z}$ ; M1 multiplying by the conjugate, M1 simplifying the denominator to $5 - 4\cos\theta$ , A1 the real part, A1 the final simplified form (using $\cos^2\theta - \sin^2\theta = \cos 2\theta$ ). Stating convergence is required.

Key Formulae — Applications of De Moivre’s Theorem in Trigonometry and Complex Numbers

The formulae you need to memorise for applications of de moivre’s theorem in trigonometry and complex numbers on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

z + 1/z and z − 1/z
$z + \frac{1}{z} = 2\cos\theta, \qquad z - \frac{1}{z} = 2i\sin\theta \quad (z = e^{i\theta})$
$z$
$e^{i\theta} = \cos\theta + i\sin\theta$ , a unit complex number
$\theta$
the angle (radians)
When to use
Converting powers of $\cos\theta$ or $\sin\theta$ into sums of multiple angles.
Powers: zⁿ ± 1/zⁿ
$z^n + \frac{1}{z^n} = 2\cos n\theta, \qquad z^n - \frac{1}{z^n} = 2i\sin n\theta$
$n$
the power / multiple of the angle
When to use
Pairing symmetric terms after a binomial expansion of $\left(z \pm \frac1z\right)^m$ .
Multiple-angle expansion
$\cos n\theta + i\sin n\theta = (\cos\theta + i\sin\theta)^n = \sum_{r=0}^{n}\binom{n}{r}\cos^{n-r}\theta\,(i\sin\theta)^r$
$\binom{n}{r}$
binomial coefficient
$n$
the multiple of the angle
When to use
Expressing $\cos n\theta$ (real part) or $\sin n\theta$ (imaginary part) in powers of $\cos\theta, \sin\theta$ .
Standard triple-angle identities
$\cos 3\theta = 4\cos^3\theta - 3\cos\theta, \qquad \sin 3\theta = 3\sin\theta - 4\sin^3\theta$
$\theta$
the base angle
When to use
Quotable results that the de Moivre expansion produces; also derive on demand for 'show that'.
C + iS series method
$C + iS = \sum_{k}(\cos k\theta + i\sin k\theta) = \sum_k z^k, \quad z = e^{i\theta}$
$C$
the cosine series $\sum\cos k\theta$
$S$
the sine series $\sum\sin k\theta$
When to use
Summing a finite or infinite trigonometric series via a geometric series of complex exponentials.
Half-angle factor-out
$\frac{z^n - 1}{z - 1} = e^{i(n-1)\theta/2}\,\frac{\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}$
$z$
$e^{i\theta}$
$n$
number of terms
When to use
The key step in $C + iS$ : exposes the $\frac{\sin(n\theta/2)}{\sin(\theta/2)}$ factor for reading off real/imaginary parts.

Key Definitions and Keywords — Applications of De Moivre’s Theorem in Trigonometry and Complex Numbers

Definitions to memorise and the exact keywords mark schemes credit for applications of de moivre’s theorem in trigonometry and complex numbers answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Multiple-angle identity
Examiner keyword
An identity expressing $\cos n\theta$ or $\sin n\theta$ as a polynomial in $\cos\theta$ and $\sin\theta$ , derived by expanding $(\cos\theta + i\sin\theta)^n$ .
Equating real and imaginary parts
Examiner keyword
The principle that if $a + ib = c + id$ (with $a,b,c,d$ real) then $a = c$ and $b = d$ . The engine of every de Moivre identity derivation.
C + iS method
Examiner keyword
Summing $\sum\cos k\theta$ and $\sum\sin k\theta$ together as the real and imaginary parts of $\sum e^{ik\theta}$ , a geometric series.
Geometric series
Examiner keyword
A series $a + ar + ar^2 + \cdots$ with constant ratio $r$ ; finite sum $\frac{a(r^n - 1)}{r - 1}$ , infinite sum $\frac{a}{1 - r}$ when $|r| < 1$ .
Binomial theorem
$(a + b)^n = \sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$ ; used to expand $(\cos\theta + i\sin\theta)^n$ and $\left(z \pm \frac1z\right)^n$ .
Euler's relation
$e^{i\theta} = \cos\theta + i\sin\theta$ ; lets de Moivre be treated as an index law and underpins $z + \frac1z = 2\cos\theta$ .

Common Mistakes and Misconceptions — Applications of De Moivre’s Theorem in Trigonometry and Complex Numbers

The traps other students keep falling into on applications of de moivre’s theorem in trigonometry and complex numbers questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Losing track of the imaginary unit $i$ during the binomial expansion
9231 Paper 2 Examiner Reports — recurring comment
Why it happens
Treating $i$ as a constant to factor out early, or forgetting $i^2 = -1$ , $i^3 = -i$ , $i^4 = 1$ .
How to avoid it
Keep $i$ attached to every $\sin$ factor and simplify powers of $i$ carefully; only separate real and imaginary parts at the very end.
✕Mishandling the $(2i)^n$ factor when converting $\sin^n\theta$
9231 Paper 2 mark schemes
Why it happens
Forgetting that $\sin\theta = \frac{1}{2i}(z - \frac1z)$ , so $\sin^n\theta$ carries a $\frac{1}{(2i)^n}$ that affects sign and (for odd $n$ ) leaves an $i$ .
How to avoid it
Compute $(2i)^n$ explicitly: $(2i)^2 = -4$ , $(2i)^3 = -8i$ , $(2i)^4 = 16$ . For odd $n$ the residual $i$ must cancel against the $z^k - \frac{1}{z^k} = 2i\sin$ terms.
✕Failing to pair symmetric terms $z^k + \frac{1}{z^k}$ before converting
9231 Paper 2 Examiner Reports
Why it happens
Trying to convert each $z^k$ individually instead of recognising the $2\cos k\theta$ pattern in the pairs.
How to avoid it
After the binomial expansion, group terms equidistant from the centre into $z^k + \frac{1}{z^k}$ pairs; the unpaired middle term is the constant.
✕Guessing the closed form of a $C + iS$ series instead of factoring out half-angles
9231 Paper 2 Examiner Reports
Why it happens
Knowing the target answer but not the technique, so writing it down without the geometric-series and half-angle steps.
How to avoid it
Always sum $\sum z^k$ with the geometric formula, then factor out $e^{i\theta/2}$ etc. The method marks are awarded for the working, not the final line.
✕Taking the wrong part (real instead of imaginary, or vice versa)
9231 Paper 2 mark schemes
Why it happens
Not tracking which series is $C$ (cosine, real) and which is $S$ (sine, imaginary).
How to avoid it
Remember: $\cos n\theta$ and $\sum\cos k\theta$ are REAL parts; $\sin n\theta$ and $\sum\sin k\theta$ are IMAGINARY parts. Label clearly before extracting.
✕Omitting the convergence condition $|r| < 1$ for an infinite series
9231 Paper 2 Examiner Reports
Why it happens
Applying $\frac{a}{1-r}$ automatically without checking the ratio's modulus.
How to avoid it
For an infinite $C + iS$ , state $\left|\frac{z}{c}\right| < 1$ (e.g. $\left|\frac{z}{2}\right| = \frac12 < 1$ ) before summing. It is a marked step.

Applications of De Moivre’s Theorem in Trigonometry and Complex Numbers

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1cos 3θ in powers of cos θ

2sin 4θ in terms of sin θ and cos θ

3cos⁴θ as a sum of multiple angles

4Summing a cosine series with C + iS

5Deriving cos 5θ and solving an equation

6tan 3θ in terms of tan θ

z + 1/z and z − 1/z

Powers: zⁿ ± 1/zⁿ

Multiple-angle expansion

Standard triple-angle identities

C + iS series method

Half-angle factor-out

Multiple-angle identity

Equating real and imaginary parts

C + iS method

Geometric series

Binomial theorem

Euler's relation

✕Losing track of the imaginary unit iii during the binomial expansion

✕Mishandling the (2i)n(2i)^n(2i)n factor when converting sin⁡nθ\sin^n\thetasinnθ

✕Failing to pair symmetric terms zk+1zkz^k + \frac{1}{z^k}zk+zk1​ before converting

✕Guessing the closed form of a C+iSC + iSC+iS series instead of factoring out half-angles

✕Taking the wrong part (real instead of imaginary, or vice versa)

✕Omitting the convergence condition ∣r∣<1|r| < 1∣r∣<1 for an infinite series

✕Losing track of the imaginary unit $i$ during the binomial expansion

✕Mishandling the $(2i)^n$ factor when converting $\sin^n\theta$

✕Failing to pair symmetric terms $z^k + \frac{1}{z^k}$ before converting

✕Guessing the closed form of a $C + iS$ series instead of factoring out half-angles

✕Omitting the convergence condition $|r| < 1$ for an infinite series