| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Express roots in trigonometric form |
| Difficulty | Challenging +1.2 Part (i) is a standard de Moivre's theorem application requiring binomial expansion and equating real parts—routine for Further Maths students. Part (ii) requires recognizing the substitution x = cos θ and solving cos 4θ = 0, then converting to tan form, which involves several steps but follows a well-established pattern for this topic. |
| Spec | 1.05l Double angle formulae: and compound angle formulae4.02q De Moivre's theorem: multiple angle formulae4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((c+is)^4 = c^4 + 4c^3(is) + 6c^2(-s^2) + 4c(is)^3 + (is)^4\) | M1 | Uses binomial theorem to expand \((c+is)^4\) |
| \(\Rightarrow \cos 4\theta = c^4 - 6c^2s^2 + s^4\) | M1 A1 | Takes real part, AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\cos 4\theta}{\cos^4\theta} = \tan^4\theta - 6\tan^2\theta + 1\) | M1 A1 | Divides through by \(\cos^4\theta\) |
| Let \(x = \tan\theta\), then \(x^4 - 6x^2 + 1 = 0 \Rightarrow \cos 4\theta = 0\) | dM1 | Following on from finding correct quartic; Solves \(\cos 4\theta = 0\) |
| \(\Rightarrow 4\theta = \pm\frac{\pi}{2} + 2m\pi,\ m \in \mathbb{Z}\) | ||
| Roots are \(\tan q\pi\) where \(q = \frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}\) \((q > 0)\) | A1 A1 | Alt methods: Solves \(\tan(4\theta) = (4t-t^3)/(t^4-6t^2+1)\) and \(\tan(4\theta) = \infty\); Solves equation in \(\cot(\theta)\) after dividing by \(\sin^4\theta\) |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(c+is)^4 = c^4 + 4c^3(is) + 6c^2(-s^2) + 4c(is)^3 + (is)^4$ | M1 | Uses binomial theorem to expand $(c+is)^4$ |
| $\Rightarrow \cos 4\theta = c^4 - 6c^2s^2 + s^4$ | M1 A1 | Takes real part, AG |
## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\cos 4\theta}{\cos^4\theta} = \tan^4\theta - 6\tan^2\theta + 1$ | M1 A1 | Divides through by $\cos^4\theta$ |
| Let $x = \tan\theta$, then $x^4 - 6x^2 + 1 = 0 \Rightarrow \cos 4\theta = 0$ | dM1 | Following on from finding correct quartic; Solves $\cos 4\theta = 0$ |
| $\Rightarrow 4\theta = \pm\frac{\pi}{2} + 2m\pi,\ m \in \mathbb{Z}$ | | |
| Roots are $\tan q\pi$ where $q = \frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}$ $(q > 0)$ | A1 A1 | Alt methods: Solves $\tan(4\theta) = (4t-t^3)/(t^4-6t^2+1)$ and $\tan(4\theta) = \infty$; Solves equation in $\cot(\theta)$ after dividing by $\sin^4\theta$ |
---3 (i) Use de Moivre's theorem to show that
$$\cos 4 \theta = \cos ^ { 4 } \theta - 6 \cos ^ { 2 } \theta \sin ^ { 2 } \theta + \sin ^ { 4 } \theta$$
(ii) Hence find all the roots of the equation
$$x ^ { 4 } - 6 x ^ { 2 } + 1 = 0$$
in the form $\tan q \pi$, where $q$ is a positive rational number.\\
\hfill \mbox{\textit{CAIE FP1 2018 Q3 [8]}}