| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive tan/cot identities |
| Difficulty | Challenging +1.3 This is a standard Further Maths FP3 question on using de Moivre's theorem to derive multiple angle formulas. Part (i) is a routine proof following a well-established method (expand (cos θ + i sin θ)^5, equate real/imaginary parts, divide to get tan). Parts (ii) and (iii) require solving for specific angles and connecting roots to tan values, which involves some algebraic manipulation but follows predictable patterns. While this requires more sophistication than typical A-level questions, it's a textbook Further Maths exercise without novel insight required. |
| Spec | 1.05o Trigonometric equations: solve in given intervals4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos 5\theta + i\sin 5\theta = c^5 + 5ic^4x - 10c^3x^2 - 10ic^2x^3 + 5cx^4 + ix^5\) | B1 | For explicit use of de Moivre with \(n = 5\) |
| \(\Rightarrow \tan 5\theta = \frac{\sin 5\theta}{\cos 5\theta} = \frac{5c^4x - 10c^2x^3 + x^5}{c^5 - 10c^3x^2 + 5cx^4}\) | M1 | For correct expressions for \(\sin 5\theta\) and \(\cos 5\theta\) |
| Division of numerator & denominator by \(c^5\). \(\Rightarrow \tan 5\theta = \frac{5\tan\theta - 10\tan^3\theta + \tan^5\theta}{1 - 10\tan^2\theta + 5\tan^4\theta}\) | M1 | For \(\frac{\sin 5\theta}{\cos 5\theta}\) in terms of \(c\) and \(x\) |
| A1 | For simplifying to AG, www with explicit mention of division by \(c^5\) | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(5\theta = \{1, 5, 9, 13, 17\}\frac{1}{4}\pi\) | M1 | For at least 2 of given values and no extras. |
| \(\theta = \{1, 5, 9, 13, 17\}\frac{1}{20}\pi\) | A1 | For at least 3 values of \(\theta\) and no extras in range |
| A1 | For all 5 values and no extras outside range | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan 5\theta = 1 \Rightarrow t^5 - 5t^4 - 10t^3 + 10t^2 + 5t - 1 = 0\) | M1* | For \(\tan 5\theta = 1\) and equation in \(t\) |
| \(\Rightarrow (t - 1)(t^4 - 4t^3 - 14t^2 - 4t + 1) = 0\) | A1 | For correct factors |
| \(\tan\alpha = 1\) OR \(\alpha = \frac{1}{4}\pi\) is not included in roots of the quartic | B1 | For solution rejected (may be implied by \(\frac{3\pi}{20}\) not appearing in set of solutions) |
| \(\Rightarrow t = \tan\alpha\) for \(\alpha = \{1, 9, 13, 17\}\frac{3\pi}{20}\pi\) | M1*dep | For 2 correct values of \(t\) |
| A1 | For all 4 values and no more in range | |
| [5] |
**(i)**
| $\cos 5\theta + i\sin 5\theta = c^5 + 5ic^4x - 10c^3x^2 - 10ic^2x^3 + 5cx^4 + ix^5$ | B1 | For explicit use of de Moivre with $n = 5$ |
| $\Rightarrow \tan 5\theta = \frac{\sin 5\theta}{\cos 5\theta} = \frac{5c^4x - 10c^2x^3 + x^5}{c^5 - 10c^3x^2 + 5cx^4}$ | M1 | For correct expressions for $\sin 5\theta$ and $\cos 5\theta$ |
| Division of numerator & denominator by $c^5$. $\Rightarrow \tan 5\theta = \frac{5\tan\theta - 10\tan^3\theta + \tan^5\theta}{1 - 10\tan^2\theta + 5\tan^4\theta}$ | M1 | For $\frac{\sin 5\theta}{\cos 5\theta}$ in terms of $c$ and $x$ |
| | A1 | For simplifying to AG, www with explicit mention of division by $c^5$ |
| | [4] | |
**(ii)**
| $5\theta = \{1, 5, 9, 13, 17\}\frac{1}{4}\pi$ | M1 | For at least 2 of given values and no extras. |
| $\theta = \{1, 5, 9, 13, 17\}\frac{1}{20}\pi$ | A1 | For at least 3 values of $\theta$ and no extras in range |
| | A1 | For all 5 values and no extras outside range |
| | [3] | |
**(iii)**
| $\tan 5\theta = 1 \Rightarrow t^5 - 5t^4 - 10t^3 + 10t^2 + 5t - 1 = 0$ | M1* | For $\tan 5\theta = 1$ and equation in $t$ |
| $\Rightarrow (t - 1)(t^4 - 4t^3 - 14t^2 - 4t + 1) = 0$ | A1 | For correct factors |
| $\tan\alpha = 1$ OR $\alpha = \frac{1}{4}\pi$ is not included in roots of the quartic | B1 | For solution rejected (may be implied by $\frac{3\pi}{20}$ not appearing in set of solutions) |
| $\Rightarrow t = \tan\alpha$ for $\alpha = \{1, 9, 13, 17\}\frac{3\pi}{20}\pi$ | M1*dep | For 2 correct values of $t$ |
| | A1 | For all 4 values and no more in range |
| | [5] | |8 (i) Use de Moivre's theorem to prove that
$$\tan 5 \theta \equiv \frac { 5 \tan \theta - 10 \tan ^ { 3 } \theta + \tan ^ { 5 } \theta } { 1 - 10 \tan ^ { 2 } \theta + 5 \tan ^ { 4 } \theta } .$$
(ii) Solve the equation $\tan 5 \theta = 1$, for $0 \leqslant \theta < \pi$.\\
(iii) Show that the roots of the equation
$$t ^ { 4 } - 4 t ^ { 3 } - 14 t ^ { 2 } - 4 t + 1 = 0$$
may be expressed in the form $\tan \alpha$, stating the exact values of $\alpha$, where $0 \leqslant \alpha < \pi$.
\section*{THERE ARE NO QUESTIONS WRITTEN ON THIS PAGE}
\hfill \mbox{\textit{OCR FP3 2012 Q8 [12]}}