| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2020 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive trigonometric identities |
| Difficulty | Challenging +1.8 Part (a) is a standard Further Maths application of de Moivre's theorem requiring binomial expansion of (cos θ + i sin θ)^4 and algebraic manipulation to isolate sin^4 θ. Part (b) is a first-order linear ODE requiring integrating factor method combined with the identity from (a), plus applying initial conditions. While both parts are multi-step and require synthesis of techniques, they follow well-established procedures taught in Further Maths courses without requiring novel insight. |
| Spec | 4.02q De Moivre's theorem: multiple angle formulae4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z - z^{-1} = 2i\sin\theta\) | B1 | Use of \(z - z^{-1} = 2i\sin\theta\) |
| \((z - z^{-1})^4 = (z^4 + z^{-4}) - 4(z^2 + z^{-2}) + 6\) | M1 A1 | Expands and groups |
| \((2i\sin\theta)^4 = (2\cos 4\theta) - 4(2\cos 2\theta) + 6\) | M1 | Substitutes \(z^n + z^{-n} = 2\cos n\theta\) |
| \(\sin^4\theta = \frac{1}{8}(\cos 4\theta - 4\cos 2\theta + 3)\) | A1 | AG |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(e^{\int \cot\theta\,d\theta} = e^{\ln\sin\theta} = \sin\theta\) | M1 A1 | Finds integrating factor |
| \(\frac{d}{d\theta}(y\sin\theta) = \sin^4\theta = \frac{1}{8}(\cos 4\theta - 4\cos 2\theta + 3)\) | M1 | Correct form on LHS and uses identity given in (a) |
| \(y\sin\theta = \frac{1}{8}\left(\frac{1}{4}\sin 4\theta - 2\sin 2\theta + 3\theta\right) + C\) | A1 | |
| \(0 = \frac{1}{8}\left(\frac{3}{2}\pi\right) + C\) | M1 | Substitutes initial conditions |
| \(y\sin\theta = \frac{1}{8}\left(\frac{1}{4}\sin 4\theta - 2\sin 2\theta + 3\theta - \frac{3}{2}\pi\right)\) | A1 | OE |
| Total: 6 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z - z^{-1} = 2i\sin\theta$ | B1 | Use of $z - z^{-1} = 2i\sin\theta$ |
| $(z - z^{-1})^4 = (z^4 + z^{-4}) - 4(z^2 + z^{-2}) + 6$ | M1 A1 | Expands and groups |
| $(2i\sin\theta)^4 = (2\cos 4\theta) - 4(2\cos 2\theta) + 6$ | M1 | Substitutes $z^n + z^{-n} = 2\cos n\theta$ |
| $\sin^4\theta = \frac{1}{8}(\cos 4\theta - 4\cos 2\theta + 3)$ | A1 | AG |
| **Total: 5** | | |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $e^{\int \cot\theta\,d\theta} = e^{\ln\sin\theta} = \sin\theta$ | M1 A1 | Finds integrating factor |
| $\frac{d}{d\theta}(y\sin\theta) = \sin^4\theta = \frac{1}{8}(\cos 4\theta - 4\cos 2\theta + 3)$ | M1 | Correct form on LHS and uses identity given in (a) |
| $y\sin\theta = \frac{1}{8}\left(\frac{1}{4}\sin 4\theta - 2\sin 2\theta + 3\theta\right) + C$ | A1 | |
| $0 = \frac{1}{8}\left(\frac{3}{2}\pi\right) + C$ | M1 | Substitutes initial conditions |
| $y\sin\theta = \frac{1}{8}\left(\frac{1}{4}\sin 4\theta - 2\sin 2\theta + 3\theta - \frac{3}{2}\pi\right)$ | A1 | OE |
| **Total: 6** | | |6
\begin{enumerate}[label=(\alph*)]
\item Use de Moivre's theorem to show that $\sin ^ { 4 } \theta = \frac { 1 } { 8 } ( \cos 4 \theta - 4 \cos 2 \theta + 3 )$.
\item Find the solution of the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} \theta } + y \cot \theta = \sin ^ { 3 } \theta$$
for which $y = 0$ when $\theta = \frac { 1 } { 2 } \pi$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q6 [11]}}