Question 3 - A-Level Maths

CAIE Further Paper 2 2023 June — Question 3 7 marks

Exam Board	CAIE
Module	Further Paper 2 (Further Paper 2)
Year	2023
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Complex numbers 2
Type	De Moivre to derive tan/cot identities
Difficulty	Challenging +1.2 This is a standard Further Maths question using a well-established technique (binomial expansion with z = e^(iθ) and de Moivre's theorem) to derive a trigonometric identity. While it requires multiple steps and careful algebraic manipulation, the method is routine for Further Maths students and follows a predictable pattern. The question explicitly guides the approach, making it moderately above average difficulty but not requiring novel insight.
Spec	1.04a Binomial expansion: (a+b)^n for positive integer n 4.02n Euler's formula: e^(itheta) = cos(theta) + isin(theta)4.02q De Moivre's theorem: multiple angle formulae

3 By considering the binomial expansions of $\left( z + \frac { 1 } { z } \right) ^ { 4 }$ and $\left( z - \frac { 1 } { z } \right) ^ { 4 }$, where $z = \cos \theta + i \sin \theta$, use de Moivre's theorem to show that $$\cot ^ { 4 } \theta = \frac { \cos 4 \theta + a \cos 2 \theta + b } { \cos 4 \theta - a \cos 2 \theta + b }$$ where $a$ and $b$ are integers to be determined.

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Question 3:

Answer	Marks	Guidance
$z - z^{-1} = 2i\sin\theta$ and $z + z^{-1} = 2\cos\theta$	B1	Use of $z - z^{-1} = 2i\sin\theta$ and $z + z^{-1} = 2\cos\theta$
$(z + z^{-1})^4 = (z^4 + z^{-4}) + 4(z^2 + z^{-2}) + 6$	M1 A1	Expands and groups
$(z - z^{-1})^4 = (z^4 + z^{-4}) - 4(z^2 + z^{-2}) + 6$	M1 A1	Expands and groups. Only withhold one A1 mark (this or the previous) for no clear grouping
$\frac{2^4\cos^4\theta}{2^4\sin^4\theta} = \frac{2\cos 4\theta + 4(2\cos 2\theta) + 6}{2\cos 4\theta - 4(2\cos 2\theta) + 6}$	M1	Substitutes $z^n + z^{-n} = 2\cos n\theta$ once in LHS
$\cot^4\theta = \frac{\cos 4\theta + 4\cos 2\theta + 3}{\cos 4\theta - 4\cos 2\theta + 3}$	A1

## Question 3:

| $z - z^{-1} = 2i\sin\theta$ and $z + z^{-1} = 2\cos\theta$ | B1 | Use of $z - z^{-1} = 2i\sin\theta$ and $z + z^{-1} = 2\cos\theta$ |
|---|---|---|
| $(z + z^{-1})^4 = (z^4 + z^{-4}) + 4(z^2 + z^{-2}) + 6$ | M1 A1 | Expands and groups |
| $(z - z^{-1})^4 = (z^4 + z^{-4}) - 4(z^2 + z^{-2}) + 6$ | M1 A1 | Expands and groups. Only withhold one A1 mark (this or the previous) for no clear grouping |
| $\frac{2^4\cos^4\theta}{2^4\sin^4\theta} = \frac{2\cos 4\theta + 4(2\cos 2\theta) + 6}{2\cos 4\theta - 4(2\cos 2\theta) + 6}$ | M1 | Substitutes $z^n + z^{-n} = 2\cos n\theta$ once in LHS |
| $\cot^4\theta = \frac{\cos 4\theta + 4\cos 2\theta + 3}{\cos 4\theta - 4\cos 2\theta + 3}$ | A1 | |

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3 By considering the binomial expansions of $\left( z + \frac { 1 } { z } \right) ^ { 4 }$ and $\left( z - \frac { 1 } { z } \right) ^ { 4 }$, where $z = \cos \theta + i \sin \theta$, use de Moivre's theorem to show that

$$\cot ^ { 4 } \theta = \frac { \cos 4 \theta + a \cos 2 \theta + b } { \cos 4 \theta - a \cos 2 \theta + b }$$

where $a$ and $b$ are integers to be determined.\\

\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q3 [7]}}

This paper (8 questions)

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Q1 6 Q2 7 Q3 7 Q4 8 Q5 11 Q6 11 Q7 11 Q8 14

Answer	Marks	Guidance
\(z - z^{-1} = 2i\sin\theta\) and \(z + z^{-1} = 2\cos\theta\)	B1	Use of \(z - z^{-1} = 2i\sin\theta\) and \(z + z^{-1} = 2\cos\theta\)
\((z + z^{-1})^4 = (z^4 + z^{-4}) + 4(z^2 + z^{-2}) + 6\)	M1 A1	Expands and groups
\((z - z^{-1})^4 = (z^4 + z^{-4}) - 4(z^2 + z^{-2}) + 6\)	M1 A1	Expands and groups. Only withhold one A1 mark (this or the previous) for no clear grouping
\(\frac{2^4\cos^4\theta}{2^4\sin^4\theta} = \frac{2\cos 4\theta + 4(2\cos 2\theta) + 6}{2\cos 4\theta - 4(2\cos 2\theta) + 6}\)	M1	Substitutes \(z^n + z^{-n} = 2\cos n\theta\) once in LHS
\(\cot^4\theta = \frac{\cos 4\theta + 4\cos 2\theta + 3}{\cos 4\theta - 4\cos 2\theta + 3}\)	A1