| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Solve equations using trigonometric identities |
| Difficulty | Challenging +1.2 This is a standard Further Maths question combining de Moivre's theorem with solving equations. Part (a) is a routine derivation following a well-established method (expand (cos θ + i sin θ)^4, equate real/imaginary parts, divide to get tan). Part (b) requires recognizing that the polynomial matches the denominator/numerator pattern when x = tan θ, then solving tan 4θ = 1, which is straightforward. While it requires multiple techniques and is harder than typical A-level, it's a textbook-style Further Maths question without requiring novel insight. |
| Spec | 4.02j Cubic/quartic equations: conjugate pairs and factor theorem4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \((\cos\theta + i\sin\theta)^4 = \cos 4\theta + i\sin 4\theta\); attempt complete expansion | M1 | Correct use of de Moivre and attempt the complete expansion |
| \(\cos^4\theta + 4i\cos^3\theta\sin\theta + i^2 6\cos^2\theta\sin^2\theta + 4i^3\cos\theta\sin^3\theta + i^4\sin^4\theta\) | A1 | Correct expansion; coefficients to be single numbers but powers of \(i\) may still be present |
| \(\cos 4\theta = \cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta\) | M1 | Equate the real and imaginary parts |
| \(\sin 4\theta = 4\cos^3\theta\sin\theta - 4\cos\theta\sin^3\theta\) | A1 | Correct expressions for \(\cos 4\theta\) and \(\sin 4\theta\) |
| \(\tan 4\theta = \frac{\sin 4\theta}{\cos 4\theta}\); divide numerator and denominator by \(\cos^4\theta\) | M1 | Use \(\tan 4\theta = \frac{\sin 4\theta}{\cos 4\theta}\) and divide numerator and denominator by \(\cos^4\theta\). Only tangents now |
| \(\tan 4\theta = \frac{4\tan\theta - 4\tan^3\theta}{1 - 6\tan^2\theta + \tan^4\theta}\) | A1* | Correct given answer, no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\sin 4\theta = \frac{1}{2i}(z^4 - z^{-4}) = \frac{1}{2i}\left((\cos\theta - i\sin\theta)^4 - (\cos\theta + i\sin\theta)^{-4}\right)\) | M1 | For the expression derived from de Moivre for either \(\sin 4\theta\) or \(\cos 4\theta\) |
| Both expressions shown and correct | A1 | Both shown and correct |
| Attempt binomial expansion for either, reaching simplified expression | M1 | Attempt the binomial expansion for either |
| \(\sin 4\theta = 4\cos^3\theta\sin\theta - 4\cos\theta\sin^3\theta\); \(\cos 4\theta = \cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta\) | A1 | Both simplified expressions correct |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(x = \tan\theta\); \(\frac{2\tan\theta - 2\tan^3\theta}{1 - 6\tan^2\theta + \tan^4\theta} = \frac{1}{2}\tan 4\theta = 1\); \(\tan 4\theta = 2\) | M1 | Substitute \(x = \tan\theta\) and re-arrange to \(\tan 4\theta = \pm 2\) or \(\pm\frac{1}{2}\) |
| \(x = \tan\theta = 0.284,\ 1.79\) | A1A1 | A1 for either solution; A1 for both. Deduct one mark only for failing to round either or both to 3 sf |
# Question 7:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $(\cos\theta + i\sin\theta)^4 = \cos 4\theta + i\sin 4\theta$; attempt complete expansion | M1 | Correct use of de Moivre and attempt the complete expansion |
| $\cos^4\theta + 4i\cos^3\theta\sin\theta + i^2 6\cos^2\theta\sin^2\theta + 4i^3\cos\theta\sin^3\theta + i^4\sin^4\theta$ | A1 | Correct expansion; coefficients to be single numbers but powers of $i$ may still be present |
| $\cos 4\theta = \cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta$ | M1 | Equate the real and imaginary parts |
| $\sin 4\theta = 4\cos^3\theta\sin\theta - 4\cos\theta\sin^3\theta$ | A1 | Correct expressions for $\cos 4\theta$ and $\sin 4\theta$ |
| $\tan 4\theta = \frac{\sin 4\theta}{\cos 4\theta}$; divide numerator and denominator by $\cos^4\theta$ | M1 | Use $\tan 4\theta = \frac{\sin 4\theta}{\cos 4\theta}$ and divide numerator and denominator by $\cos^4\theta$. Only tangents now |
| $\tan 4\theta = \frac{4\tan\theta - 4\tan^3\theta}{1 - 6\tan^2\theta + \tan^4\theta}$ | A1* | Correct **given** answer, no errors seen |
**(6 marks)**
## Part (a) Alternative (first 4 marks):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\sin 4\theta = \frac{1}{2i}(z^4 - z^{-4}) = \frac{1}{2i}\left((\cos\theta - i\sin\theta)^4 - (\cos\theta + i\sin\theta)^{-4}\right)$ | M1 | For the expression derived from de Moivre for **either** $\sin 4\theta$ or $\cos 4\theta$ |
| Both expressions shown and correct | A1 | Both shown and correct |
| Attempt binomial expansion for either, reaching simplified expression | M1 | Attempt the binomial expansion for either |
| $\sin 4\theta = 4\cos^3\theta\sin\theta - 4\cos\theta\sin^3\theta$; $\cos 4\theta = \cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta$ | A1 | Both simplified expressions correct |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $x = \tan\theta$; $\frac{2\tan\theta - 2\tan^3\theta}{1 - 6\tan^2\theta + \tan^4\theta} = \frac{1}{2}\tan 4\theta = 1$; $\tan 4\theta = 2$ | M1 | Substitute $x = \tan\theta$ and re-arrange to $\tan 4\theta = \pm 2$ or $\pm\frac{1}{2}$ |
| $x = \tan\theta = 0.284,\ 1.79$ | A1A1 | A1 for either solution; A1 for both. Deduct one mark only for failing to round either or both to 3 sf |
**(3 marks) [9 total]**
---\begin{enumerate}
\item (a) Use de Moivre's theorem to show that
\end{enumerate}
$$\tan 4 \theta \equiv \frac { 4 \tan \theta - 4 \tan ^ { 3 } \theta } { 1 - 6 \tan ^ { 2 } \theta + \tan ^ { 4 } \theta }$$
(b) Use the identity given in part (a) to find the 2 positive roots of
$$x ^ { 4 } + 2 x ^ { 3 } - 6 x ^ { 2 } - 2 x + 1 = 0$$
giving your answers to 3 significant figures.\\
\includegraphics[max width=\textwidth, alt={}, center]{0d44aec7-a6e8-47fc-a215-7c8c4790e93f-29_2255_50_314_35}\\
\hfill \mbox{\textit{Edexcel F2 2021 Q7 [9]}}