| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive tan/cot identities |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring de Moivre's theorem manipulation to derive a sec identity, then connecting it to a sextic equation. Part (i) demands algebraic dexterity with complex exponentials and trigonometric identities, while part (ii) requires recognizing the connection between the derived identity and the polynomial roots. The multi-step reasoning and non-standard application of de Moivre's theorem place it well above average difficulty, though it follows a recognizable FP1 pattern. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=14.02k Argand diagrams: geometric interpretation4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Write \(c = \cos\theta\), \(s = \sin\theta\). \(\cos 6\theta + i\sin 6\theta = (c+is)^6\) | M1 | Uses binomial theorem |
| \(\Rightarrow \cos 6\theta = c^6 - 15c^4s^2 + 15c^2s^4 - s^6\) | A1 | |
| \(c^6 - 15c^4s^2 + 15c^2s^4 - s^6 = c^6 - 15c^4(1-c^2) + 15c^2(1-c^2)^2 - (1-c^2)^3\) | M1 | Uses \(c^2 = 1-s^2\) |
| \(= c^6 - 15c^4(1-c^2) + 15c^2(1-2c^2+c^4) - (1-3c^2+3c^4-c^6)\) | A1 | |
| \(= 32c^6 - 48c^4 + 18c^2 - 1\) | M1 | Divides numerator and denominator by \(c^6\) |
| \(\Rightarrow \sec 6\theta = \frac{1}{32c^6-48c^4+18c^2-1} = \frac{\sec^6\theta}{32 - 48\sec^2\theta + 18\sec^4\theta - \sec^6\theta}\) | A1 | AG |
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x^6 = 2(32 - 48x^2 + 18x^4 - x^6) \Rightarrow \frac{x^6}{32-48x^2+18x^4-x^6} = 2\) | M1 A1 | Relates with equation in part (i) |
| \(\sec 6\theta = 2 \Rightarrow \cos 6\theta = \frac{1}{2}\) | M1 | Solves \(\cos 6\theta = \frac{1}{2}\) |
| \(x = \sec\frac{\pi}{18}\) | A1 | Gives one correct solution |
| \(x = \sec q\pi\), \(q = \frac{5}{18}, \frac{7}{18}, \frac{11}{18}, \frac{13}{18}, \frac{17}{18}\) | A1 | Gives five other solutions. Allow different values of \(q\) as long as all six solutions are found |
| 5 |
## Question 9(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Write $c = \cos\theta$, $s = \sin\theta$. $\cos 6\theta + i\sin 6\theta = (c+is)^6$ | M1 | Uses binomial theorem |
| $\Rightarrow \cos 6\theta = c^6 - 15c^4s^2 + 15c^2s^4 - s^6$ | A1 | |
| $c^6 - 15c^4s^2 + 15c^2s^4 - s^6 = c^6 - 15c^4(1-c^2) + 15c^2(1-c^2)^2 - (1-c^2)^3$ | M1 | Uses $c^2 = 1-s^2$ |
| $= c^6 - 15c^4(1-c^2) + 15c^2(1-2c^2+c^4) - (1-3c^2+3c^4-c^6)$ | A1 | |
| $= 32c^6 - 48c^4 + 18c^2 - 1$ | M1 | Divides numerator and denominator by $c^6$ |
| $\Rightarrow \sec 6\theta = \frac{1}{32c^6-48c^4+18c^2-1} = \frac{\sec^6\theta}{32 - 48\sec^2\theta + 18\sec^4\theta - \sec^6\theta}$ | A1 | AG |
| | **6** | |
## Question 9(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^6 = 2(32 - 48x^2 + 18x^4 - x^6) \Rightarrow \frac{x^6}{32-48x^2+18x^4-x^6} = 2$ | M1 A1 | Relates with equation in part (i) |
| $\sec 6\theta = 2 \Rightarrow \cos 6\theta = \frac{1}{2}$ | M1 | Solves $\cos 6\theta = \frac{1}{2}$ |
| $x = \sec\frac{\pi}{18}$ | A1 | Gives one correct solution |
| $x = \sec q\pi$, $q = \frac{5}{18}, \frac{7}{18}, \frac{11}{18}, \frac{13}{18}, \frac{17}{18}$ | A1 | Gives five other solutions. Allow different values of $q$ as long as all six solutions are found |
| | **5** | |9 (i) Use de Moivre's theorem to show that
$$\sec 6 \theta = \frac { \sec ^ { 6 } \theta } { 32 - 48 \sec ^ { 2 } \theta + 18 \sec ^ { 4 } \theta - \sec ^ { 6 } \theta }$$
(ii) Hence obtain the roots of the equation
$$3 x ^ { 6 } - 36 x ^ { 4 } + 96 x ^ { 2 } - 64 = 0$$
in the form sec $q \pi$, where $q$ is rational.\\
\hfill \mbox{\textit{CAIE FP1 2019 Q9 [11]}}