| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive tan/cot identities |
| Difficulty | Challenging +1.2 This is a standard Further Maths FP2 question following a well-established template: apply de Moivre's theorem, derive a tan identity, then use roots of a polynomial. Part (a) is routine bookwork, part (b) requires recognizing the connection between tan(π/16) and tan(4θ)=1, and part (c) uses Vieta's formulas—all standard techniques for this module. While multi-step, it requires no novel insight beyond following the established method. |
| Spec | 1.05l Double angle formulae: and compound angle formulae4.02q De Moivre's theorem: multiple angle formulae4.05a Roots and coefficients: symmetric functions |
8
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Use de Moivre's theorem to show that
$$\cos 4 \theta = \cos ^ { 4 } \theta - 6 \cos ^ { 2 } \theta \sin ^ { 2 } \theta + \sin ^ { 4 } \theta$$
and find a similar expression for $\sin 4 \theta$.
\item Deduce that
$$\tan 4 \theta = \frac { 4 \tan \theta - 4 \tan ^ { 3 } \theta } { 1 - 6 \tan ^ { 2 } \theta + \tan ^ { 4 } \theta }$$
\end{enumerate}\item Explain why $t = \tan \frac { \pi } { 16 }$ is a root of the equation
$$t ^ { 4 } + 4 t ^ { 3 } - 6 t ^ { 2 } - 4 t + 1 = 0$$
and write down the three other roots in trigonometric form.
\item Hence show that
$$\tan ^ { 2 } \frac { \pi } { 16 } + \tan ^ { 2 } \frac { 3 \pi } { 16 } + \tan ^ { 2 } \frac { 5 \pi } { 16 } + \tan ^ { 2 } \frac { 7 \pi } { 16 } = 28$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2013 Q8 [17]}}