| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | November |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Express roots in trigonometric form |
| Difficulty | Challenging +1.2 This is a standard Further Maths question applying de Moivre's theorem to derive a multiple angle formula, then using it to solve a polynomial equation. Part (i) follows a well-established procedure (expand (cos θ + i sin θ)^8, equate imaginary parts, factor). Part (ii) requires recognizing the connection between the derived formula and the polynomial, then solving sin 8θ/sin 2θ = 0 systematically. While it requires multiple steps and careful algebra, the techniques are routine for FP1 students with no novel insight needed. |
| Spec | 4.02q De Moivre's theorem: multiple angle formulae4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Write \(c = \cos\theta\), \(s = \sin\theta\). \(\cos 8\theta + i\sin 8\theta = (c+is)^8\) | M1 | Uses binomial theorem |
| \(\Rightarrow \sin 8\theta = 8c^7s - 56c^5s^3 + 56c^3s^5 - 8cs^7\) | A1 | |
| \(= 8cs(c^6 - 7c^4s^2 + 7c^2s^4 - s^6)\) | M1 | Factorises |
| \(c^6 - 7c^4s^2 + 7c^2s^4 - s^6 = (1-s^2)^3 - 7(1-s^2)^2s^2 + 7(1-s^2)s^4 - s^6\) | M1 | Uses \(c^2 = 1 - s^2\) |
| \(= (1 - 3s^2 + 3s^4 - s^6) - 7(s^2 - 2s^4 + s^6) + 7(s^4 - s^6) - s^6\) \(= 1 - 10s^2 + 24s^4 - 16s^6\) | A1 | |
| \(\Rightarrow \sin 8\theta = 8cs(1 - 10s^2 + 24s^4 - 16s^6)\) | A1 | AG |
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sin 2\theta = 2cs \Rightarrow \frac{\sin 8\theta}{\sin 2\theta} = 4(1 - 10s^2 + 24s^4 - 16s^6)\) | M1 | Uses \(\sin 2\theta = 2cs\) to relate with equation in part (i) |
| \(\sin 8\theta = 0 \Rightarrow \theta = \frac{n\pi}{8}\) | M1 | Solves \(\sin 8\theta = 0\) |
| \(x = \sin\frac{\pi}{8}\) | A1 | Gives one correct solution |
| \(\sin\frac{n\pi}{8}\), \(n = \pm1, \pm2, \pm3\) or \(1, 2, 3, 9, 10, 11\) | A1 | Gives five other solutions. Allow different values of \(k\) as long as six distinct solutions are found. \(\sin 0\) and \(\sin\frac{\pi}{2}\) must be excluded |
| 4 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Write $c = \cos\theta$, $s = \sin\theta$. $\cos 8\theta + i\sin 8\theta = (c+is)^8$ | M1 | Uses binomial theorem |
| $\Rightarrow \sin 8\theta = 8c^7s - 56c^5s^3 + 56c^3s^5 - 8cs^7$ | A1 | |
| $= 8cs(c^6 - 7c^4s^2 + 7c^2s^4 - s^6)$ | M1 | Factorises |
| $c^6 - 7c^4s^2 + 7c^2s^4 - s^6 = (1-s^2)^3 - 7(1-s^2)^2s^2 + 7(1-s^2)s^4 - s^6$ | M1 | Uses $c^2 = 1 - s^2$ |
| $= (1 - 3s^2 + 3s^4 - s^6) - 7(s^2 - 2s^4 + s^6) + 7(s^4 - s^6) - s^6$ $= 1 - 10s^2 + 24s^4 - 16s^6$ | A1 | |
| $\Rightarrow \sin 8\theta = 8cs(1 - 10s^2 + 24s^4 - 16s^6)$ | A1 | AG |
| | **6** | |
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin 2\theta = 2cs \Rightarrow \frac{\sin 8\theta}{\sin 2\theta} = 4(1 - 10s^2 + 24s^4 - 16s^6)$ | M1 | Uses $\sin 2\theta = 2cs$ to relate with equation in part (i) |
| $\sin 8\theta = 0 \Rightarrow \theta = \frac{n\pi}{8}$ | M1 | Solves $\sin 8\theta = 0$ |
| $x = \sin\frac{\pi}{8}$ | A1 | Gives one correct solution |
| $\sin\frac{n\pi}{8}$, $n = \pm1, \pm2, \pm3$ or $1, 2, 3, 9, 10, 11$ | A1 | Gives five other solutions. Allow different values of $k$ as long as six distinct solutions are found. $\sin 0$ and $\sin\frac{\pi}{2}$ must be excluded |
| | **4** | |7 (i) Use de Moivre's theorem to show that
$$\sin 8 \theta = 8 \sin \theta \cos \theta \left( 1 - 10 \sin ^ { 2 } \theta + 24 \sin ^ { 4 } \theta - 16 \sin ^ { 6 } \theta \right) .$$
(ii) Use the equation $\frac { \sin 8 \theta } { \sin 2 \theta } = 0$ to find the roots of
$$16 x ^ { 6 } - 24 x ^ { 4 } + 10 x ^ { 2 } - 1 = 0$$
in the form $\sin k \pi$, where $k$ is rational.\\
\hfill \mbox{\textit{CAIE FP1 2018 Q7}}