| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2012 |
| Session | June |
| 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 derivation follows a well-established method, and part (ii) involves only arithmetic substitution. It's moderately above average due to the Further Maths content and multi-step nature, but not particularly challenging for students at this level. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs4.02q 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**
$= c^4 + 4c^3\cdot\mathrm{i}s + 6c^2\cdot\mathrm{i}^2s^2 + 4c\cdot\mathrm{i}^3s^3 + \mathrm{i}^4s^4$ 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**
**[5]**
**(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**
**[4]**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 $t = \tan \theta$.\\
(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 2012 Q7 [9]}}