CAIE Further Paper 2 2024 November — Question 4 10 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2024
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeDe Moivre to derive tan/cot identities
DifficultyChallenging +1.8 This is a substantial Further Maths question requiring de Moivre's theorem manipulation to derive a multiple angle formula, then clever algebraic insight to connect the cotangent identity to polynomial roots. Part (a) involves extended algebraic manipulation with complex exponentials (6 marks suggests significant work), while part (b) requires recognizing that setting cot 6θ = 0 and rearranging yields the given polynomial. The techniques are standard for Further Maths but require careful execution and the connection between parts demands mathematical maturity.
Spec4.02q De Moivre's theorem: multiple angle formulae4.02r nth roots: of complex numbers
  1. Use de Moivre's theorem to show that $$\cot 6\theta = \frac{\cot^4 \theta - 15\cot^4 \theta + 15\cot^2 \theta - 1}{6\cot^5 \theta - 20\cot^3 \theta + 6\cot \theta}.$$ [6]
  2. Hence obtain the roots of the equation $$x^6 - 6x^5 - 15x^4 + 20x^3 + 15x^2 - 6x - 1 = 0$$ in the form \(\cot(q\pi)\), where \(q\) is a rational number. [4]
Question 4:
AnswerMarks Guidance
4(a)cos6=Re(cos+isin)6 =cos6−15cos4sin2+15cos2sin4−sin6 M1A1
sin6=Im(cos+isin)6 =6cos5sin−20cos3sin3+6cossin5M1A1 Takes imaginary part
cos6−15cos4sin2+15cos2sin4−sin6 sin−6
cot6= 
AnswerMarks Guidance
6cos5sin−20cos3sin3+6cossin5 sin−6M1 Divides numerator and denominator by
sin6.
cot6−15cot4+15cot2−1
cot6=
AnswerMarks Guidance
6cot5−20cot3+6cotA1 AG
6
AnswerMarks Guidance
4(b)x=cot, cot6=1 B1
6= 1+k
AnswerMarks Guidance
4M1 cot6=1
Solves
cot ( 1 π )
AnswerMarks Guidance
24A1 Gives one correct solution.
7.60
cot ( 5 π ) ,cot (3π ) ,cot (13π ) ,cot (17π ) ,cot (7π )
AnswerMarks Guidance
24 8 24 24 8A1 Gives other five solutions.
1.30, 0.41, –0.13, –0.77, –2.41
Withhold A1 mark for repeated roots.
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 4:
--- 4(a) ---
4(a) | cos6=Re(cos+isin)6 =cos6−15cos4sin2+15cos2sin4−sin6 | M1A1 | Expands and takes real part.
sin6=Im(cos+isin)6 =6cos5sin−20cos3sin3+6cossin5 | M1A1 | Takes imaginary part
cos6−15cos4sin2+15cos2sin4−sin6 sin−6
cot6= 
6cos5sin−20cos3sin3+6cossin5 sin−6 | M1 | Divides numerator and denominator by
sin6.
cot6−15cot4+15cot2−1
cot6=
6cot5−20cot3+6cot | A1 | AG
6
--- 4(b) ---
4(b) | x=cot, cot6=1 | B1 | Applies identity given in (b).
6= 1+k
4 | M1 | cot6=1
Solves
cot ( 1 π )
24 | A1 | Gives one correct solution.
7.60
cot ( 5 π ) ,cot (3π ) ,cot (13π ) ,cot (17π ) ,cot (7π )
24 8 24 24 8 | A1 | Gives other five solutions.
1.30, 0.41, –0.13, –0.77, –2.41
Withhold A1 mark for repeated roots.
4
Question | Answer | Marks | Guidance
\begin{enumerate}[label=(\alph*)]
\item Use de Moivre's theorem to show that
$$\cot 6\theta = \frac{\cot^4 \theta - 15\cot^4 \theta + 15\cot^2 \theta - 1}{6\cot^5 \theta - 20\cot^3 \theta + 6\cot \theta}.$$ [6]

\item Hence obtain the roots of the equation
$$x^6 - 6x^5 - 15x^4 + 20x^3 + 15x^2 - 6x - 1 = 0$$
in the form $\cot(q\pi)$, where $q$ is a rational number. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q4 [10]}}
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