CAIE FP1 2019 November — Question 9 11 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2019
SessionNovember
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeExpress roots in trigonometric form
DifficultyChallenging +1.8 This is a challenging Further Maths question requiring de Moivre's theorem manipulation to derive a non-trivial identity, then reverse-engineering to find roots in a specific form. Part (i) demands algebraic stamina converting between cos/sec and expanding (cos θ + i sin θ)^6, while part (ii) requires recognizing the connection between the derived identity and the given polynomial. The multi-step reasoning, algebraic complexity, and need for insight into the relationship between parts elevates this significantly above standard A-level, though it follows a recognizable FM1 pattern.
Spec4.02q De Moivre's theorem: multiple angle formulae4.02r nth roots: of complex numbers
  1. Use de Moivre's theorem to show that $$\sec 6\theta = \frac{\sec^6 \theta}{32 - 48\sec^2 \theta + 18\sec^4 \theta - \sec^6 \theta}.$$ [6]
  2. Hence obtain the roots of the equation $$3t^6 - 36t^4 + 96t^2 - 64 = 0$$ in the form \(\sec q\pi\), where \(q\) is rational. [5]
Question 9:
AnswerMarks
9(i)c=cosθ, s=sinθ.
Write
AnswerMarks Guidance
cos6θ+isin6θ=( c+is )6M1 Uses binomial theorem.
⇒cos6θ=c6 −15c4s2 +15c2s4 −s6A1
( ) ( ) ( )
AnswerMarks Guidance
c6 −15c4s2 +15c2s4 −s6 =c6 −15c4 1−c2 +15c2 1−c2 2 − 1−c2 3M1 Uses c2 =1−s2.
( ) ( ) ( )
AnswerMarks Guidance
=c6 −15c4 1−c2 +15c2 1−2c2 +c4 − 1−3c2 +3c4 −c6A1
=32c6 −48c4 +18c2 −1M1 Divides numerator and denominator by c6.
1 sec6θ
⇒sec6θ= =
AnswerMarks Guidance
32c6 −48c4 +18c2 −1 32−48sec2θ+18sec4θ−sec6θA1 AG
6
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
9(ii)( ) x6
x6 =2 32−48x2 +18x4 −x6 ⇒ =2
AnswerMarks Guidance
32−48x2 +18x4 −x6M1 A1 Relates with equation in part (i).
1
sec6θ=2⇒cos6θ=
AnswerMarks Guidance
2M1 1
cos6θ=
Solves .
2
π
x=sec
AnswerMarks Guidance
18A1 Gives one correct solution.
5 7 11 13 17
x=secqπ, q= , , , ,
AnswerMarks Guidance
18 18 18 18 18A1 Gives five other solutions. Allow different values of q as long
as all six solutions are found.
5
AnswerMarks Guidance
QuestionAnswer Marks 
Question 9:
--- 9(i) ---
9(i) | c=cosθ, s=sinθ.
Write
cos6θ+isin6θ=( c+is )6 | M1 | Uses binomial theorem.
⇒cos6θ=c6 −15c4s2 +15c2s4 −s6 | A1
( ) ( ) ( )
c6 −15c4s2 +15c2s4 −s6 =c6 −15c4 1−c2 +15c2 1−c2 2 − 1−c2 3 | M1 | Uses c2 =1−s2.
( ) ( ) ( )
=c6 −15c4 1−c2 +15c2 1−2c2 +c4 − 1−3c2 +3c4 −c6 | A1
=32c6 −48c4 +18c2 −1 | M1 | Divides numerator and denominator by c6.
1 sec6θ
⇒sec6θ= =
32c6 −48c4 +18c2 −1 32−48sec2θ+18sec4θ−sec6θ | A1 | AG
6
Question | Answer | Marks | Guidance
--- 9(ii) ---
9(ii) | ( ) x6
x6 =2 32−48x2 +18x4 −x6 ⇒ =2
32−48x2 +18x4 −x6 | M1 A1 | Relates with equation in part (i).
1
sec6θ=2⇒cos6θ=
2 | M1 | 1
cos6θ=
Solves .
2
π
x=sec
18 | A1 | Gives one correct solution.
5 7 11 13 17
x=secqπ, q= , , , ,
18 18 18 18 18 | A1 | Gives five other solutions. Allow different values of q as long
as all six solutions are found.
5
Question | Answer | Marks | Guidance
\begin{enumerate}[label=(\roman*)]
\item Use de Moivre's theorem to show that
$$\sec 6\theta = \frac{\sec^6 \theta}{32 - 48\sec^2 \theta + 18\sec^4 \theta - \sec^6 \theta}.$$ [6]

\item Hence obtain the roots of the equation
$$3t^6 - 36t^4 + 96t^2 - 64 = 0$$
in the form $\sec q\pi$, where $q$ is rational. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP1 2019 Q9 [11]}}
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