| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2016 |
| Session | Specimen |
| Marks | 9 |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive tan/cot identities |
| Difficulty | Challenging +1.2 This is a standard Further Maths question using de Moivre's theorem to derive a multiple angle formula, followed by straightforward substitution and verification. While it requires knowledge of complex numbers and de Moivre's theorem (a Further Maths topic), the execution is methodical: expand (cos θ + i sin θ)^4, separate real/imaginary parts, divide to get tan 4θ, then substitute t = 1/5. The verification is arithmetic. This is harder than average A-level due to the Further Maths content and algebraic manipulation required, but it's a textbook application without novel insight. |
| Spec | 4.02q De Moivre's theorem: multiple angle formulae |
**(i)**
$\cos 4\theta + \mathrm{i}\sin 4\theta = (c + \mathrm{i}s)^4$ — Use of *de Moivre's Theorem* **M1**
$= c^4 + 4c^3.\mathrm{i}s + 6c^2.\mathrm{i}^2s^2 + 4c.\mathrm{i}^3s^3 + \mathrm{i}^4s^4$ — Binomial expansion attempted **M1**
$\cos 4\theta = c^4 - 6c^2s^2 + s^4$ and $\sin 4\theta = 4c^3s - 4cs^3$ — Equating Re & Im parts **M1**
$\tan 4\theta = \frac{\sin 4\theta}{\cos 4\theta} = \frac{4c^3s - 4cs^3}{c^4 - 6c^2s^2 + s^4}$ **M1**
Dividing throughout by $c^4$ to get $\frac{4t - 4t^3}{1 - 6t^2 + t^4}$ — legitimately **A1**
**(ii)**
$t = \frac{1}{5} \Rightarrow \tan 4\theta = \frac{120}{119}$ **B1**
$\tan\left(\frac{1}{4}\pi + \tan^{-1}\frac{1}{239}\right) = \frac{1 + \frac{1}{239}}{1 - \frac{1}{239}} = \frac{120}{119}$ **M1 A1**
Noting that this is $\tan(4\tan^{-1}\frac{1}{5})$ so that $4\tan^{-1}\frac{1}{5} = \frac{1}{4}\pi + \tan^{-1}\frac{1}{239}$ **A1**
**Total: 9 marks**7 (i) Use de Moivre's theorem to show that $\tan 4 \theta = \frac { 4 t \left( 1 - t ^ { 2 } \right) } { 1 - 6 t ^ { 2 } + t ^ { 4 } }$, where $\mathrm { t } = \tan \theta$.\\
(ii) Given that $\theta$ is the acute angle such that $\tan \theta = \frac { 1 } { 5 }$, express $\tan 4 \theta$ as a rational number in its simplest form, and verify that
$$\frac { 1 } { 4 } \pi + \tan ^ { - 1 } \left( \frac { 1 } { 239 } \right) = 4 \tan ^ { - 1 } \left( \frac { 1 } { 5 } \right) .$$
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2016 Q7 [9]}}