| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Topic | Complex numbers 2 |
| Type | Express roots in trigonometric form |
| Difficulty | Standard +0.3 This is a standard Further Maths question combining de Moivre's theorem with solving cubic equations. Part (i) is a routine proof using binomial expansion, part (ii) is straightforward trigonometric equation solving, and part (iii) requires recognizing the connection between the cubic and the triple angle formula—a common exam technique. While it requires multiple steps and some insight to connect parts, the individual components are well-practiced Further Maths material with clear signposting between parts. |
| Spec | 1.05o Trigonometric equations: solve in given intervals4.02q De Moivre's theorem: multiple angle formulae |
\begin{enumerate}[label=(\roman*)]
\item Use de Moivre's theorem to prove that $\cos 3\theta = 4c^3 - 3c$, where $c = \cos\theta$. [3]
\item Solve the equation $2\cos 3\theta - \sqrt{3} = 0$ for $0 < \theta < \pi$, giving each answer in an exact form. [2]
\item Deduce, in trigonometric form, the three roots of the equation $x^3 - 3x - \sqrt{3} = 0$. [3]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2018 Q9 [8]}}