| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Integration using De Moivre identities |
| Difficulty | Challenging +1.2 This is a standard FP2 De Moivre question following a well-established template: prove the identity, expand and simplify to express a power of cosine in terms of multiple angles, solve an equation, then integrate. While it requires multiple techniques (binomial expansion, trigonometric manipulation, integration), each step follows predictable patterns that students practice extensively. The multi-part structure guides students through the solution, making it more accessible than its length suggests. |
| Spec | 1.05o Trigonometric equations: solve in given intervals4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae |
8
\begin{enumerate}[label=(\alph*)]
\item Use De Moivre's Theorem to show that, if $z = \cos \theta + \mathrm { i } \sin \theta$, then
$$z ^ { n } + \frac { 1 } { z ^ { n } } = 2 \cos n \theta$$
\item \begin{enumerate}[label=(\roman*)]
\item Expand $\left( z ^ { 2 } + \frac { 1 } { z ^ { 2 } } \right) ^ { 4 }$.
\item Show that
$$\cos ^ { 4 } 2 \theta = A \cos 8 \theta + B \cos 4 \theta + C$$
where $A , B$ and $C$ are rational numbers.
\end{enumerate}\item Hence solve the equation
$$8 \cos ^ { 4 } 2 \theta = \cos 8 \theta + 5$$
for $0 \leqslant \theta \leqslant \pi$, giving each solution in the form $k \pi$.
\item Show that
$$\int _ { 0 } ^ { \frac { \pi } { 2 } } \cos ^ { 4 } 2 \theta d \theta = \frac { 3 \pi } { 16 }$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2012 Q8 [14]}}