CAIE FP1 2018 November — Question 7 10 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2018
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeExpress roots in trigonometric form
DifficultyChallenging +1.2 This is a standard Further Maths question applying de Moivre's theorem to derive a multiple angle formula, followed by a routine application to find roots. Part (i) requires systematic binomial expansion and algebraic manipulation but follows a well-practiced technique. Part (ii) involves recognizing that sin(8θ)/sin(2θ) = 0 leads to specific values of θ, then substituting x = sin(θ). While it requires multiple steps and careful algebra, this is a typical FP1 exam question with no novel insights needed—students who have practiced de Moivre's theorem applications will find this accessible.
Spec4.02q De Moivre's theorem: multiple angle formulae
  1. Use de Moivre's theorem to show that $$\sin 8\theta = 8\sin \theta \cos \theta(1 - 10\sin^2 \theta + 24\sin^4 \theta - 16\sin^6 \theta).$$ [6]
  2. Use the equation \(\frac{\sin 8\theta}{\sin 2\theta} = 0\) to find the roots of $$16x^6 - 24x^4 + 10x^2 - 1 = 0$$ in the form \(\sin k\pi\), where \(k\) is rational. [4]
Question 7:
AnswerMarks Guidance
7(i)Write c=cosθ, s=sinθ.
cos8θ+isin8θ=( c+is )8M1 Uses binomial theorem.
⇒sin8θ=8c7s−56c5s3+56c3s5 −8cs7A1
( )
AnswerMarks Guidance
=8cs c6 −7c4s2 +7c2s4 −s6M1 Factorises.
( )3 ( )2 ( )
AnswerMarks Guidance
c6 −7c4s2 +7c2s4 −s6 = 1−s2 −7 1−s2 s2 +7 1−s2 s4 −s6M1 Uses c2 =1−s2.
( ) ( ) ( )
= 1−3s2 +3s4 −s6 −7 s2 −2s4 +s6 +7 s4 −s6 −s6
AnswerMarks
=1−10s2 +24s4 −16s6A1
( )
AnswerMarks Guidance
⇒sin8θ=8cs 1−10s2 +24s4 −16s6A1 AG
6
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
7(ii)sin8θ ( )
sin2θ=2cs ⇒ =4 1−10s2 +24s4 −16s6
AnswerMarks Guidance
sin2θM1 Uses sin2θ=2cs to relate with equation in part (i)
nπ
sin8θ=0⇒θ=
AnswerMarks Guidance
8M1 Solves sin8θ=0
π
x=sin
AnswerMarks Guidance
8A1 Gives one correct solution
nπ
sin , n=±1,±2±3 or1 ,2,3,9,10,11
AnswerMarks Guidance
8A1 Gives five other solutions. Allow different values of k as
π
long as six distinct solutions are found. sin0 and sin
2
must be excluded.
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 7:
--- 7(i) ---
7(i) | Write c=cosθ, s=sinθ.
cos8θ+isin8θ=( c+is )8 | M1 | Uses binomial theorem.
⇒sin8θ=8c7s−56c5s3+56c3s5 −8cs7 | A1
( )
=8cs c6 −7c4s2 +7c2s4 −s6 | M1 | Factorises.
( )3 ( )2 ( )
c6 −7c4s2 +7c2s4 −s6 = 1−s2 −7 1−s2 s2 +7 1−s2 s4 −s6 | M1 | Uses c2 =1−s2.
( ) ( ) ( )
= 1−3s2 +3s4 −s6 −7 s2 −2s4 +s6 +7 s4 −s6 −s6
=1−10s2 +24s4 −16s6 | A1
( )
⇒sin8θ=8cs 1−10s2 +24s4 −16s6 | A1 | AG
6
Question | Answer | Marks | Guidance
--- 7(ii) ---
7(ii) | sin8θ ( )
sin2θ=2cs ⇒ =4 1−10s2 +24s4 −16s6
sin2θ | M1 | Uses sin2θ=2cs to relate with equation in part (i)
nπ
sin8θ=0⇒θ=
8 | M1 | Solves sin8θ=0
π
x=sin
8 | A1 | Gives one correct solution
nπ
sin , n=±1,±2±3 or1 ,2,3,9,10,11
8 | A1 | Gives five other solutions. Allow different values of k as
π
long as six distinct solutions are found. sin0 and sin
2
must be excluded.
4
Question | Answer | Marks | Guidance
\begin{enumerate}[label=(\roman*)]
\item Use de Moivre's theorem to show that
$$\sin 8\theta = 8\sin \theta \cos \theta(1 - 10\sin^2 \theta + 24\sin^4 \theta - 16\sin^6 \theta).$$ [6]

\item Use the equation $\frac{\sin 8\theta}{\sin 2\theta} = 0$ to find the roots of
$$16x^6 - 24x^4 + 10x^2 - 1 = 0$$
in the form $\sin k\pi$, where $k$ is rational. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP1 2018 Q7 [10]}}
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