| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Integration using De Moivre identities |
| Difficulty | Challenging +1.2 This is a standard Further Maths question testing De Moivre's theorem and binomial expansion technique. Part (i) requires routine manipulation (showing z - 1/z = 2i sin θ is straightforward, then expanding (2i sin θ)^5 using binomial theorem and De Moivre). Part (ii) is direct integration of the resulting expression. While it requires multiple steps and is Further Maths content, it follows a well-established algorithmic procedure with no novel insight needed. |
| Spec | 4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02o Loci in Argand diagram: circles, half-lines4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z - z^{-1} = \cos\theta + i\sin\theta - \cos(-\theta) - i\sin(-\theta) = 2i\sin\theta\) | B1 | |
| \(\left(z - \frac{1}{z}\right)^5 = \left(z^5 - \frac{1}{z^5}\right) - 5\left(z^3 - \frac{1}{z^3}\right) + 10\left(z - \frac{1}{z}\right)\) | M1A1 | |
| \(\Rightarrow 32\sin^5\theta \, i = 2i\sin5\theta - 10i\sin3\theta + 20i\sin\theta\) | M1A1 | Grouping not required at this stage |
| \(\Rightarrow 16\sin^5\theta = \sin5\theta - 5\sin3\theta + 10\sin\theta\) | A1 | M1 for grouping and applying initial result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^{\frac{1}{3}\pi} 16\sin^5\theta \, d\theta = \left[-\frac{\cos5\theta}{5} + \frac{5\cos3\theta}{3} - 10\cos\theta\right]_0^{\frac{1}{3}\pi}\) | M1A1 FT | |
| \(= \left[-\frac{1}{10} - \frac{5}{3} - 5\right] - \left[-\frac{1}{5} + \frac{5}{3} - 10\right] = \frac{53}{30}\) | A1 |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z - z^{-1} = \cos\theta + i\sin\theta - \cos(-\theta) - i\sin(-\theta) = 2i\sin\theta$ | B1 | |
| $\left(z - \frac{1}{z}\right)^5 = \left(z^5 - \frac{1}{z^5}\right) - 5\left(z^3 - \frac{1}{z^3}\right) + 10\left(z - \frac{1}{z}\right)$ | M1A1 | |
| $\Rightarrow 32\sin^5\theta \, i = 2i\sin5\theta - 10i\sin3\theta + 20i\sin\theta$ | M1A1 | Grouping not required at this stage |
| $\Rightarrow 16\sin^5\theta = \sin5\theta - 5\sin3\theta + 10\sin\theta$ | A1 | **M1** for grouping and applying initial result |
## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^{\frac{1}{3}\pi} 16\sin^5\theta \, d\theta = \left[-\frac{\cos5\theta}{5} + \frac{5\cos3\theta}{3} - 10\cos\theta\right]_0^{\frac{1}{3}\pi}$ | M1A1 FT | |
| $= \left[-\frac{1}{10} - \frac{5}{3} - 5\right] - \left[-\frac{1}{5} + \frac{5}{3} - 10\right] = \frac{53}{30}$ | A1 | |8 (i) Let $z = \cos \theta + \mathrm { i } \sin \theta$. Show that $z - \frac { 1 } { z } = 2 \mathrm { i } \sin \theta$ and hence express $16 \sin ^ { 5 } \theta$ in the form $\sin 5 \theta + p \sin 3 \theta + q \sin \theta$, where $p$ and $q$ are integers to be determined.\\
(ii) Hence find the exact value of $\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } 16 \sin ^ { 5 } \theta \mathrm {~d} \theta$.\\
\hfill \mbox{\textit{CAIE FP1 2017 Q8 [9]}}