| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive tan/cot identities |
| Difficulty | Challenging +1.2 This is a standard Further Maths question requiring de Moivre's theorem to derive a multiple angle formula (routine technique) followed by solving a quartic by recognizing it matches the tan 4θ identity. While it requires multiple steps and the substitution insight in part (ii), this is a well-practiced exam technique at FP3 level, making it moderately above average difficulty but not exceptional. |
| Spec | 1.05l Double angle formulae: and compound angle formulae4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\cos 4\theta + i\sin 4\theta = (\cos\theta + i\sin\theta)^4 = c^4 + 4ic^3 s - 6c^2 s^2 - 4ics^3 + s^4\) | B1 | soi by at least |
| Taking re and im parts \(\cos 4\theta = c^4 - 6c^2 s^2 + s^4\), \(\sin 4\theta = 4c^3 s - 4cs^3\) | B1 | Can be broken down already but with \(i\)'s in place |
| \(\tan 4\theta = \frac{4\tan\theta - 4\tan^3\theta}{1 - 6\tan^2\theta + \tan^4\theta}\) | M1 | take real and imaginary parts AG. Must show division of numerator and denominator by \(c^4\) and must have been explicit about re and im |
| A1 | ||
| [4] | ||
| (ii) Rearranging polynomial gives \(1 - 6t^2 + t^4 = \sqrt{3}(4t - 4t^3)\) | M1 | |
| so \(\tan 4\theta = \frac{1}{\sqrt{3}}\) | A1 | |
| \(4\theta =\) their "\(\frac{1}{6}\pi" + n\pi\) | B1 | |
| \(t = \tan\theta = \tan\frac{\pi}{12}\), \(\tan\frac{7\pi}{12}\), \(\tan\frac{13\pi}{12}\), \(\tan\frac{19\pi}{12}\) | B1 | one correct |
| B1 | all correct or (4) equivalent | |
| [5] | condone all angles seen and no extras, but t not given as equal to \(\tan\theta\) |
**(i)** $\cos 4\theta + i\sin 4\theta = (\cos\theta + i\sin\theta)^4 = c^4 + 4ic^3 s - 6c^2 s^2 - 4ics^3 + s^4$ | B1 | soi by at least
Taking re and im parts $\cos 4\theta = c^4 - 6c^2 s^2 + s^4$, $\sin 4\theta = 4c^3 s - 4cs^3$ | B1 | Can be broken down already but with $i$'s in place
$\tan 4\theta = \frac{4\tan\theta - 4\tan^3\theta}{1 - 6\tan^2\theta + \tan^4\theta}$ | M1 | take real and imaginary parts AG. Must show division of numerator and denominator by $c^4$ and must have been explicit about re and im
| A1 |
| [4] |
**(ii)** Rearranging polynomial gives $1 - 6t^2 + t^4 = \sqrt{3}(4t - 4t^3)$ | M1 |
so $\tan 4\theta = \frac{1}{\sqrt{3}}$ | A1 |
$4\theta =$ their "$\frac{1}{6}\pi" + n\pi$ | B1 |
$t = \tan\theta = \tan\frac{\pi}{12}$, $\tan\frac{7\pi}{12}$, $\tan\frac{13\pi}{12}$, $\tan\frac{19\pi}{12}$ | B1 | one correct
| B1 | all correct or (4) equivalent
| [5] | condone all angles seen and no extras, but t not given as equal to $\tan\theta$7 (i) Use de Moivre's theorem to show that $\tan 4 \theta \equiv \frac { 4 \tan \theta - 4 \tan ^ { 3 } \theta } { 1 - 6 \tan ^ { 2 } \theta + \tan ^ { 4 } \theta }$.\\
(ii) Hence find the exact roots of $t ^ { 4 } + 4 \sqrt { 3 } t ^ { 3 } - 6 t ^ { 2 } - 4 \sqrt { 3 } t + 1 = 0$.
\hfill \mbox{\textit{OCR FP3 2015 Q7 [9]}}