| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2007 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive trigonometric identities |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring proof by induction of De Moivre's theorem, then applying it through several connected steps involving complex number manipulation and algebraic identities. While each individual part uses standard techniques, the question requires sustained reasoning across four parts with the final part demanding insight to connect previous results. The induction proof and the 'hence' chain elevate this above routine exercises, but it remains a structured exam question with clear signposting. |
| Spec | 4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae |
5
\begin{enumerate}[label=(\alph*)]
\item Prove by induction that, if $n$ is a positive integer,
$$( \cos \theta + \mathrm { i } \sin \theta ) ^ { n } = \cos n \theta + \mathrm { i } \sin n \theta$$
\item Find the value of $\left( \cos \frac { \pi } { 6 } + \mathrm { i } \sin \frac { \pi } { 6 } \right) ^ { 6 }$.
\item Show that
$$( \cos \theta + \mathrm { i } \sin \theta ) ( 1 + \cos \theta - \mathrm { i } \sin \theta ) = 1 + \cos \theta + \mathrm { i } \sin \theta$$
\item Hence show that
$$\left( 1 + \cos \frac { \pi } { 6 } + i \sin \frac { \pi } { 6 } \right) ^ { 6 } + \left( 1 + \cos \frac { \pi } { 6 } - i \sin \frac { \pi } { 6 } \right) ^ { 6 } = 0$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2007 Q5 [14]}}