Edexcel FP2 — Question 7 11 marks

Exam BoardEdexcel
ModuleFP2 (Further Pure Mathematics 2)
Marks11
PaperDownload PDF ↗
TopicComplex numbers 2
TypeSolve equations using trigonometric identities
DifficultyStandard +0.8 Part (a) requires applying de Moivre's theorem to derive a multiple angle formula, involving binomial expansion of (cos θ + i sin θ)^5 and separating real/imaginary parts—a standard FP2 technique but algebraically intensive. Part (b) requires substituting both formulas to create a quintic equation in sin θ, factoring it, and solving multiple trigonometric equations in a given interval—this combines algebraic manipulation with systematic case analysis, going beyond routine exercises.
Spec1.05o Trigonometric equations: solve in given intervals4.02q De Moivre's theorem: multiple angle formulae

  1. Use de Moivre's theorem to show that $$\sin 5\theta = 16 \sin^5 \theta - 20 \sin^3 \theta + 5 \sin \theta$$ [5]
Hence, given also that \(\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta\),
  1. find all the solutions of $$\sin 5\theta = 5 \sin 3\theta$$ in the interval \(0 \leq \theta < 2\pi\). Give your answers to 3 decimal places. [6]

\begin{enumerate}[label=(\alph*)]
\item Use de Moivre's theorem to show that
$$\sin 5\theta = 16 \sin^5 \theta - 20 \sin^3 \theta + 5 \sin \theta$$ [5]
\end{enumerate}

Hence, given also that $\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta$,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find all the solutions of
$$\sin 5\theta = 5 \sin 3\theta$$
in the interval $0 \leq \theta < 2\pi$. Give your answers to 3 decimal places. [6]
\end{enumerate}

\hfill \mbox{\textit{Edexcel FP2  Q7 [11]}}