OCR FP3 — Question 4 9 marks

Exam BoardOCR
ModuleFP3 (Further Pure Mathematics 3)
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeSolve equations using trigonometric identities
DifficultyStandard +0.8 This FP3 question requires using De Moivre's theorem and binomial expansion to derive a multiple angle identity, then solving a trigonometric equation using that result. Part (i) is a guided derivation requiring careful algebraic manipulation of complex exponentials (5 marks suggests multiple steps), while part (ii) requires recognizing that the equation simplifies to cos θ = -1/2 and finding solutions in the given range. The techniques are standard for FP3 but require precision and multiple steps, placing it moderately above average difficulty.
Spec1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02o Loci in Argand diagram: circles, half-lines

  1. By expressing \(\cos \theta\) in terms of \(e^{i\theta}\) and \(e^{-i\theta}\), show that $$\cos^5 \theta \equiv \frac{1}{16}(\cos 5\theta + 5\cos 3\theta + 10\cos \theta).$$ [5]
  2. Hence solve the equation \(\cos 5\theta + 5\cos 3\theta + 9\cos \theta = 0\) for \(0 \leqslant \theta \leqslant \pi\). [4]

\begin{enumerate}[label=(\roman*)]
\item By expressing $\cos \theta$ in terms of $e^{i\theta}$ and $e^{-i\theta}$, show that
$$\cos^5 \theta \equiv \frac{1}{16}(\cos 5\theta + 5\cos 3\theta + 10\cos \theta).$$ [5]
\item Hence solve the equation $\cos 5\theta + 5\cos 3\theta + 9\cos \theta = 0$ for $0 \leqslant \theta \leqslant \pi$. [4]
\end{enumerate}

\hfill \mbox{\textit{OCR FP3  Q4 [9]}}