| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Direct nth roots: roots with geometric or algebraic follow-up |
| Difficulty | Standard +0.3 This is a structured Further Maths question that guides students through finding exact trigonometric values using compound angle formulas, then applying De Moivre's theorem to find fourth roots. While it involves multiple steps, the path is clearly signposted and uses standard FP2 techniques without requiring novel insight or complex problem-solving. |
| Spec | 1.05l Double angle formulae: and compound angle formulae4.02q De Moivre's theorem: multiple angle formulae4.02r nth roots: of complex numbers |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin\left(\frac{\pi}{3}-\frac{\pi}{4}\right) = \sin\frac{\pi}{3}\cos\frac{\pi}{4} - \cos\frac{\pi}{3}\sin\frac{\pi}{4}\) | M1 | Correct expansion for sine, including surd values for all 4 trig functions. \(\frac{\sqrt{2}}{2}\) or \(\frac{1}{\sqrt{2}}\) accepted |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin\frac{\pi}{12} = \frac{\sqrt{6}-\sqrt{2}}{4} = \frac{1}{4}(\sqrt{6}-\sqrt{2})\)** | A1cso | Completion to given answer: No errors seen, cso |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos\left(\frac{\pi}{3}-\frac{\pi}{4}\right) = \cos\frac{\pi}{3}\cos\frac{\pi}{4} + \sin\frac{\pi}{3}\sin\frac{\pi}{4}\) | M1, NB A1 on e-PEN | Correct expansion for cosine, including surd values for all 4 trig functions. OR other complete method e.g. using \(\sin^2\theta+\cos^2\theta=1\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos\left(\frac{\pi}{12}\right) = \frac{\sqrt{2}+\sqrt{6}}{4} = \frac{1}{4}(\sqrt{6}+\sqrt{2})\)** | A1cso | Completion to given answer: No errors seen, cso |
| Answer | Marks | Guidance |
|---|---|---|
| \(z^4 = 4\left(\cos\left(2k\pi+\frac{\pi}{3}\right)+i\sin\left(2k\pi+\frac{\pi}{3}\right)\right)\) OR \(z^4 = 4e^{i\left(2k\pi+\frac{\pi}{3}\right)}\) | M1 | Use a valid method to generate at least 2 roots (eg use of \(2k\pi\) or rotate through \(\frac{\pi}{2}\), multiply by \(i\), symmetry) |
| \(z = 4^{\frac{1}{4}}\left(\cos\left(\frac{k\pi}{2}+\frac{\pi}{12}\right)+i\sin\left(\frac{k\pi}{2}+\frac{\pi}{12}\right)\right)\) OR \(z = 4^{\frac{1}{4}}e^{i\left(\frac{k\pi}{2}+\frac{\pi}{12}\right)}\) | M1 | Application of de Moivre's theorem resulting in at least 1 root being found. \((4 \to \sqrt{2}\) and arg divided by 4). \(4^{\frac{1}{4}}\) or \(\sqrt{2}\) accepted |
| Answer | Marks | Guidance |
|---|---|---|
| or \(\frac{1+\sqrt{3}}{2}+i\frac{-1+\sqrt{3}}{2}\) oe | B1 | Any correct root (this is the most likely one if only one found). Can be in any exact form (\(4^{\frac{1}{4}}\) or \(\sqrt{2}\) oe). Can be unsimplified using results from (a). Or simplified/calculator values ie \(\frac{1+\sqrt{3}}{2}+i\frac{-1+\sqrt{3}}{2}\) |
## Question 3:
### Part (a)(i)
$\sin\left(\frac{\pi}{3}-\frac{\pi}{4}\right) = \sin\frac{\pi}{3}\cos\frac{\pi}{4} - \cos\frac{\pi}{3}\sin\frac{\pi}{4}$ | M1 | Correct expansion for sine, including surd values for all 4 trig functions. $\frac{\sqrt{2}}{2}$ or $\frac{1}{\sqrt{2}}$ accepted
$\sin\frac{\pi}{12} = \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} - \frac{1}{2}\cdot\frac{\sqrt{2}}{2}$
$\sin\frac{\pi}{12} = \frac{\sqrt{6}-\sqrt{2}}{4} = \frac{1}{4}(\sqrt{6}-\sqrt{2})$** | A1cso | Completion to given answer: No errors seen, cso
### Part (a)(ii)
$\cos\left(\frac{\pi}{3}-\frac{\pi}{4}\right) = \cos\frac{\pi}{3}\cos\frac{\pi}{4} + \sin\frac{\pi}{3}\sin\frac{\pi}{4}$ | M1, NB A1 on e-PEN | Correct expansion for cosine, including surd values for all 4 trig functions. OR other complete method e.g. using $\sin^2\theta+\cos^2\theta=1$
$\cos\frac{\pi}{12} = \frac{1}{2}\cdot\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}$
$\cos\left(\frac{\pi}{12}\right) = \frac{\sqrt{2}+\sqrt{6}}{4} = \frac{1}{4}(\sqrt{6}+\sqrt{2})$** | A1cso | Completion to given answer: No errors seen, cso
### Part (b)
**Allow all marks using EXACT calculator values for the trig functions. Decimal answers qualify for M marks only.**
$z^4 = 4\left(\cos\left(2k\pi+\frac{\pi}{3}\right)+i\sin\left(2k\pi+\frac{\pi}{3}\right)\right)$ OR $z^4 = 4e^{i\left(2k\pi+\frac{\pi}{3}\right)}$ | M1 | Use a valid method to generate at least 2 roots (eg use of $2k\pi$ or rotate through $\frac{\pi}{2}$, multiply by $i$, symmetry)
$z = 4^{\frac{1}{4}}\left(\cos\left(\frac{k\pi}{2}+\frac{\pi}{12}\right)+i\sin\left(\frac{k\pi}{2}+\frac{\pi}{12}\right)\right)$ OR $z = 4^{\frac{1}{4}}e^{i\left(\frac{k\pi}{2}+\frac{\pi}{12}\right)}$ | M1 | Application of de Moivre's theorem resulting in at least 1 root being found. $(4 \to \sqrt{2}$ and arg divided by 4). $4^{\frac{1}{4}}$ or $\sqrt{2}$ accepted
$(k=0 \to)$
$z = \sqrt{2}\left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)$ or $\sqrt{2}e^{i\left(\frac{\pi}{12}\right)}$
or $\frac{\sqrt{2}}{4}(\sqrt{6}+\sqrt{2})+\frac{i\sqrt{2}}{4}(\sqrt{6}-\sqrt{2})$
or $\frac{1+\sqrt{3}}{2}+i\frac{-1+\sqrt{3}}{2}$ oe | B1 | Any correct root (this is the most likely one if only one found). Can be in any exact form ($4^{\frac{1}{4}}$ or $\sqrt{2}$ oe). Can be unsimplified using results from (a). Or simplified/calculator values ie $\frac{1+\sqrt{3}}{2}+i\frac{-1+\sqrt{3}}{2}$3. (a) By writing $\frac { \pi } { 12 } = \frac { \pi } { 3 } - \frac { \pi } { 4 }$, show that
\begin{enumerate}[label=(\roman*)]
\item $\sin \left( \frac { \pi } { 12 } \right) = \frac { 1 } { 4 } ( \sqrt { 6 } - \sqrt { 2 } )$
\item $\cos \left( \frac { \pi } { 12 } \right) = \frac { 1 } { 4 } ( \sqrt { 6 } + \sqrt { 2 } )$\\
(b) Hence find the exact values of $z$ for which
$$z ^ { 4 } = 4 \left( \cos \frac { \pi } { 3 } + i \sin \frac { \pi } { 3 } \right)$$
Give your answers in the form $z = a + i b$ where $a , b \in \mathbb { R }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2018 Q3 [9]}}