| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Session | Specimen |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive trigonometric identities |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring a formal induction proof of de Moivre's theorem followed by binomial expansion and manipulation to express sin(5θ) in terms of powers of sin(θ). While technically demanding with multiple steps, it follows a standard template that Further Maths students practice extensively. The limit evaluation at the end is straightforward once the expansion is obtained. The question is harder than typical A-level due to the proof requirement and algebraic manipulation, but represents a well-rehearsed Further Maths technique rather than requiring novel insight. |
| Spec | 4.01a Mathematical induction: construct proofs4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Putting \(n=1\), the proposition gives \(\cos\theta + i\sin\theta = \cos\theta + i\sin\theta\) which is true | B1 | AO2 |
| Let the proposition be true for \(n=k\), ie \([\cos\theta + i\sin\theta]^k = \cos k\theta + i\sin k\theta\) | M1 | AO2 |
| Answer | Marks | Guidance |
|---|---|---|
| \((\cos\theta + i\sin\theta)^{k+1} = (\cos\theta + i\sin\theta)^k(\cos\theta + i\sin\theta)\) | M1 | AO2 |
| \(= (\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta)\) | A1 | AO2 |
| \(= \cos k\theta \cos\theta - \sin k\theta \sin\theta + i(\sin k\theta \cos\theta + \sin\theta \cos k\theta)\) | A1 | AO2 |
| \(= \cos(k+1)\theta + i\sin(k+1)\theta\) | A1 | AO2 |
| Therefore true for \(n=k \Rightarrow\) true for \(n=k+1\) and since true for \(n=1\) the proposition is proved by induction. | A1 | AO2 |
| Answer | Marks | Guidance |
|---|---|---|
| Consider \(\cos 5\theta + i\sin 5\theta = (\cos\theta + i\sin\theta)^5\) | M1 | AO2 |
| \(= i(5\cos^4\theta \sin\theta - 10\cos^2\theta \sin^3\theta + \sin^5\theta)\) | A1 | AO2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin 5\theta = 5\cos^4\theta \sin\theta - 10\cos^2\theta \sin^3\theta + \sin^5\theta\) | A1 | AO2 |
| \(= 5(1-\sin^2\theta)^2\sin\theta - 10(1-\sin^2\theta)\sin^3\theta + \sin^5\theta\) | A1 | AO2 |
| \(= 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta\) | A1 | AO2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\sin 5\theta}{\sin\theta} = 16\sin^4\theta - 20\sin^2\theta + 5\) | M1 | AO1 |
| \(\to 5\) as \(\theta \to 0\) | A1 | AO1 |
## Part (a)
Putting $n=1$, the proposition gives $\cos\theta + i\sin\theta = \cos\theta + i\sin\theta$ which is true | B1 | AO2
Let the proposition be true for $n=k$, ie $[\cos\theta + i\sin\theta]^k = \cos k\theta + i\sin k\theta$ | M1 | AO2
Consider (for $n = k+1$):
$(\cos\theta + i\sin\theta)^{k+1} = (\cos\theta + i\sin\theta)^k(\cos\theta + i\sin\theta)$ | M1 | AO2
$= (\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta)$ | A1 | AO2
$= \cos k\theta \cos\theta - \sin k\theta \sin\theta + i(\sin k\theta \cos\theta + \sin\theta \cos k\theta)$ | A1 | AO2
$= \cos(k+1)\theta + i\sin(k+1)\theta$ | A1 | AO2
Therefore true for $n=k \Rightarrow$ true for $n=k+1$ and since true for $n=1$ the proposition is proved by induction. | A1 | AO2
## Part (b)(i)
Consider $\cos 5\theta + i\sin 5\theta = (\cos\theta + i\sin\theta)^5$ | M1 | AO2
$= i(5\cos^4\theta \sin\theta - 10\cos^2\theta \sin^3\theta + \sin^5\theta)$ | A1 | AO2
$\quad + \text{real terms}$
It follows equating imaginary terms that:
$\sin 5\theta = 5\cos^4\theta \sin\theta - 10\cos^2\theta \sin^3\theta + \sin^5\theta$ | A1 | AO2
$= 5(1-\sin^2\theta)^2\sin\theta - 10(1-\sin^2\theta)\sin^3\theta + \sin^5\theta$ | A1 | AO2
$= 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta$ | A1 | AO2
## Part (b)(ii)
$\frac{\sin 5\theta}{\sin\theta} = 16\sin^4\theta - 20\sin^2\theta + 5$ | M1 | AO1
$\to 5$ as $\theta \to 0$ | A1 | AO1
**Total: [14]**
---\begin{enumerate}[label=(\alph*)]
\item Use mathematical induction to prove de Moivre's Theorem, namely that
$$(\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta,$$
where $n$ is a positive integer. [7]
\item \begin{enumerate}[label=(\roman*)]
\item Use this result to show that
$$\sin 5\theta = a \sin^5 \theta - b \sin^3 \theta + c \sin \theta,$$
where $a$, $b$ and $c$ are positive integers to be found.
\item Hence determine the value of $\lim_{\theta \to 0} \frac{\sin 5\theta}{\sin \theta}$. [7]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 Q9 [14]}}