| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive tan/cot identities |
| Difficulty | Challenging +1.3 This is a standard FP2 de Moivre's theorem question with typical progression: expand using binomial theorem, derive tan formula, find roots, and use factor theorem properties. While it requires multiple techniques and careful algebra across 16 marks, each step follows a well-established pattern taught in FP2 courses. The final part requires recognizing that roots come in pairs and using Vieta's formulas, which is moderately challenging but within expected FP2 problem-solving scope. |
| Spec | 1.05l Double angle formulae: and compound angle formulae4.02q De Moivre's theorem: multiple angle formulae |
\begin{enumerate}[label=(\alph*)]
\item
\begin{enumerate}[label=(\roman*)]
\item Use de Moivre's Theorem to show that
$$\cos 5\theta = \cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta$$
and find a similar expression for $\sin 5\theta$. [5 marks]
\item Deduce that
$$\tan 5\theta = \frac{\tan \theta(5 - 10 \tan^2 \theta + \tan^4 \theta)}{1 - 10 \tan^2 \theta + 5 \tan^4 \theta}$$ [3 marks]
\end{enumerate}
\item Explain why $t = \tan \frac{\pi}{5}$ is a root of the equation
$$t^4 - 10t^2 + 5 = 0$$
and write down the three other roots of this equation in trigonometrical form. [3 marks]
\item Deduce that
$$\tan \frac{\pi}{5} \tan \frac{2\pi}{5} = \sqrt{5}$$ [5 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2011 Q7 [16]}}