| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Solve equations using trigonometric identities |
| Difficulty | Standard +0.8 This is a standard Further Maths FP2 question requiring de Moivre's theorem, binomial expansion of (z + 1/z)^5, and equation solving. While it involves multiple steps and careful algebraic manipulation, it follows a well-established template that FP2 students practice extensively. The techniques are routine for this level, though the execution requires precision. |
| Spec | 1.05o Trigonometric equations: solve in given intervals4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(z^n + z^{-n} = e^{in\theta} + e^{-in\theta}\) | Using de Moivre's theorem | |
| \(= \cos n\theta + i\sin n\theta + \cos n\theta - i\sin n\theta\) | M1 | For showing \(z^n = \cos n\theta + i\sin n\theta\) or \(z^{-n} = \cos n\theta - i\sin n\theta\) |
| \(= 2\cos n\theta\) \(*\) | A1 | Completing to given result |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((z + z^{-1})^5 = 32\cos^5\theta\) | B1 | Using result in (a); need not be shown explicitly |
| \((z+z^{-1})^5 = z^5 + 5z^3 + 10z + 10z^{-1} + 5z^{-3} + z^{-5}\) | M1A1 | Attempting binomial expansion; \(^nC_r\) must be changed to numbers for marks |
| \(32\cos^5\theta = (z^5+z^{-5}) + 5(z^3+z^{-3}) + 10(z+z^{-1})\) | M1 | Replacing \((z^5+z^{-5})\), \((z^3+z^{-3})\), \((z+z^{-1})\) with \(2\cos 5\theta\), \(2\cos 3\theta\), \(2\cos\theta\) |
| \(= 2\cos 5\theta + 10\cos 3\theta + 20\cos\theta\) | ||
| \(\cos^5\theta = \frac{1}{16}(\cos 5\theta + 5\cos 3\theta + 10\cos\theta)\) \(*\) | A1cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\cos 5\theta + 5\cos 3\theta + 10\cos\theta = -2\cos\theta\) | M1 | Attempting re-arrangement with one side matching bracket in (b). "Hence" stated so no other method allowed |
| \(16\cos^5\theta = -2\cos\theta\) | A1 | |
| \(2\cos\theta(8\cos^4\theta + 1) = 0\) | ||
| \(8\cos^4\theta + 1 = 0\) no solution | B1 | For stating no solution; e.g. \(8\cos^4\theta+1\neq 0\), \(8\cos^4\theta+1>0\). "Ignore" without comment gets B0 |
| \(\cos\theta = 0\) | ||
| \(\theta = \frac{\pi}{2}, \frac{3\pi}{2}\) | A1 | Both values, no more in range. Must be in radians (decimals to 3 s.f. acceptable) |
## Question 6:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $z^n + z^{-n} = e^{in\theta} + e^{-in\theta}$ | | Using de Moivre's theorem |
| $= \cos n\theta + i\sin n\theta + \cos n\theta - i\sin n\theta$ | M1 | For showing $z^n = \cos n\theta + i\sin n\theta$ or $z^{-n} = \cos n\theta - i\sin n\theta$ |
| $= 2\cos n\theta$ $*$ | A1 | Completing to given result |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $(z + z^{-1})^5 = 32\cos^5\theta$ | B1 | Using result in (a); need not be shown explicitly |
| $(z+z^{-1})^5 = z^5 + 5z^3 + 10z + 10z^{-1} + 5z^{-3} + z^{-5}$ | M1A1 | Attempting binomial expansion; $^nC_r$ must be changed to numbers for marks |
| $32\cos^5\theta = (z^5+z^{-5}) + 5(z^3+z^{-3}) + 10(z+z^{-1})$ | M1 | Replacing $(z^5+z^{-5})$, $(z^3+z^{-3})$, $(z+z^{-1})$ with $2\cos 5\theta$, $2\cos 3\theta$, $2\cos\theta$ |
| $= 2\cos 5\theta + 10\cos 3\theta + 20\cos\theta$ | | |
| $\cos^5\theta = \frac{1}{16}(\cos 5\theta + 5\cos 3\theta + 10\cos\theta)$ $*$ | A1cso | |
### Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $\cos 5\theta + 5\cos 3\theta + 10\cos\theta = -2\cos\theta$ | M1 | Attempting re-arrangement with one side matching bracket in (b). "Hence" stated so no other method allowed |
| $16\cos^5\theta = -2\cos\theta$ | A1 | |
| $2\cos\theta(8\cos^4\theta + 1) = 0$ | | |
| $8\cos^4\theta + 1 = 0$ no solution | B1 | For stating no solution; e.g. $8\cos^4\theta+1\neq 0$, $8\cos^4\theta+1>0$. "Ignore" without comment gets B0 |
| $\cos\theta = 0$ | | |
| $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$ | A1 | Both values, no more in range. Must be in radians (decimals to 3 s.f. acceptable) |
---6. The complex number $z = \mathrm { e } ^ { \mathrm { i } \theta }$, where $\theta$ is real.
\begin{enumerate}[label=(\alph*)]
\item Use de Moivre's theorem to show that
$$z ^ { n } + \frac { 1 } { z ^ { n } } = 2 \cos n \theta$$
where $n$ is a positive integer.
\item Show that
$$\cos ^ { 5 } \theta = \frac { 1 } { 16 } ( \cos 5 \theta + 5 \cos 3 \theta + 10 \cos \theta )$$
\item Hence find all the solutions of
$$\cos 5 \theta + 5 \cos 3 \theta + 12 \cos \theta = 0$$
in the interval $0 \leqslant \theta < 2 \pi$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2013 Q6 [11]}}