| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive trigonometric identities |
| Difficulty | Challenging +1.2 This is a Pre-U Further Maths question combining de Moivre's theorem with recurrence relations and induction. Part (i) is a standard de Moivre application. Part (ii) requires pattern recognition to conjecture u_n = cos(nθ), then a straightforward induction proof using the recurrence relation and angle addition formulas. While it involves multiple techniques and is longer than typical A-level questions, the individual steps are well-signposted and follow standard procedures without requiring novel insight. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\cos 3\theta = \text{Re}(\cos 3\theta + i\sin 3\theta) = \text{Re}(c + is)^3\) | M1 | |
| \(= c^3 - 3cs^2 = c^3 - 3c(1-c^2) = 4c^3 - 3c\) ANSWER GIVEN | A1 | Use of de Moivre's Thm. |
| [2] | ||
| (ii) (a) \(u_2 = 2c^2 - 1\) and \(u_3 = 4c^3 - 3c\) | B1, B1 | |
| [2] | ||
| (b) Conjecture \(u_n = \cos(n\theta)\) | B1 | |
| Base case, "true for \(n = 0\) or 1 (or 2 or 3)", may be taken as read | M1 | |
| Assuming \(u_n = \cos(n\theta)\) for at least \(n = k\) and \(n = k-1\) | M1 | |
| Using given r.r. to generate \(u_{k+1} = 2\cos\theta\cos(k\theta) - \cos[(k-1)\theta]\) | M1 | |
| \(= 2\cos\theta\cos(k\theta) - \{\cos(k\theta)\cos\theta + \sin(k\theta)\sin\theta\}\) | ||
| \(= \cos\theta\cos(k\theta) - \sin\theta\sin(k\theta)\) | ||
| OR \(\{\cos(k+1)\theta - \cos(k-1)\theta\} - \cos(k-1)\theta\) | ||
| \(= \cos[(k+1)\theta]\) | A1 | |
| If statement is true for \(n = 0, 1, 2, \ldots, k\) (but allow assumption for one term only here) then it is also true for \(n = k+1\). Proof follows by (strong) induction | B1 | |
| [6] |
**(i)** $\cos 3\theta = \text{Re}(\cos 3\theta + i\sin 3\theta) = \text{Re}(c + is)^3$ | M1 |
$= c^3 - 3cs^2 = c^3 - 3c(1-c^2) = 4c^3 - 3c$ **ANSWER GIVEN** | A1 | Use of de Moivre's Thm.
| [2] |
**(ii)** **(a)** $u_2 = 2c^2 - 1$ and $u_3 = 4c^3 - 3c$ | B1, B1 |
| [2] |
**(b)** Conjecture $u_n = \cos(n\theta)$ | B1 |
Base case, "true for $n = 0$ or 1 (or 2 or 3)", may be taken as read | M1 |
Assuming $u_n = \cos(n\theta)$ for at least $n = k$ and $n = k-1$ | M1 |
Using given r.r. to generate $u_{k+1} = 2\cos\theta\cos(k\theta) - \cos[(k-1)\theta]$ | M1 |
$= 2\cos\theta\cos(k\theta) - \{\cos(k\theta)\cos\theta + \sin(k\theta)\sin\theta\}$ | |
$= \cos\theta\cos(k\theta) - \sin\theta\sin(k\theta)$ | |
OR $\{\cos(k+1)\theta - \cos(k-1)\theta\} - \cos(k-1)\theta$ | |
$= \cos[(k+1)\theta]$ | A1 |
If statement is true for $n = 0, 1, 2, \ldots, k$ (but allow assumption for one term only here) then it is also true for $n = k+1$. Proof follows by (strong) induction | B1 |
| [6] |
---\begin{enumerate}[label=(\roman*)]
\item Use de Moivre's theorem to show that $\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$. [2]
\item The sequence $\{u_n\}$ is such that $u_0 = 1$, $u_1 = \cos \theta$ and, for $n \geqslant 1$,
$$u_{n+1} = (2\cos \theta)u_n - u_{n-1}.$$
\begin{enumerate}[label=(\alph*)]
\item Determine $u_2$ and $u_3$ in terms of powers of $\cos \theta$ only. [2]
\item Suggest a simple expression for $u_n$, the $n$th term of the sequence, and prove it for all positive integers $n$ using induction. [6]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2011 Q10 [10]}}