| Exam Board | AQA |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive trigonometric identities |
| Difficulty | Challenging +1.8 This is a sophisticated Further Maths question requiring multiple techniques: solving trigonometric equations, applying de Moivre's theorem to expand cos 6θ, using roots to find factors, and manipulating nested radicals. While each individual step follows standard FM procedures, the multi-part structure requiring connections between parts (especially using solutions to find factors, then evaluating a specific value) demands strong problem-solving and algebraic manipulation skills beyond typical A-level questions. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.05g Exact trigonometric values: for standard angles1.05o Trigonometric equations: solve in given intervals4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(6\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2}, \frac{11\pi}{2}\) | M1 (AO1.1a) | Obtains at least two correct solutions |
| \(\theta = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{11\pi}{12}\) (in addition to given solutions) | A1 (AO1.1b) | Obtains all correct solutions |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Expands \((\cos\theta + i\sin\theta)^6\) | M1 (AO1.1a) | |
| By de Moivre's theorem: \(\cos 6\theta + i\sin 6\theta = (\cos\theta + i\sin\theta)^6\) | M1 (AO3.1a) | Equates real parts |
| \(\cos 6\theta = \cos^6\theta - 15\cos^4\theta\sin^2\theta + 15\cos^2\theta\sin^4\theta - \sin^6\theta\) | A1 (AO1.1b) | Correct expression for real part in terms of powers of \(\cos\theta\) and \(\sin\theta\) |
| \(= \cos^6\theta - 15\cos^4\theta(1-\cos^2\theta) + 15\cos^2\theta(1-\cos^2\theta)^2 - (1-\cos^2\theta)^3\) | M1 (AO3.1a) | Uses trig identity to express real part in terms of \(\cos\theta\) |
| \(\cos 6\theta = 32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1\) | R1 (AO2.1) | Completes rigorous argument |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\theta = \frac{\pi}{4} \Rightarrow \cos 6\theta = 0 \Rightarrow 32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1 = 0\) | M1 (AO2.2a) | Uses fact that \(\theta = \frac{\pi}{4}\) or \(\theta = \frac{3\pi}{4}\) is a solution to deduce it is also solution to second equation |
| \(\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}\), so \(\left(\cos\theta - \frac{1}{\sqrt{2}}\right)\) is a factor; similarly \(\left(\cos\theta + \frac{1}{\sqrt{2}}\right)\) is also a factor | M1 (AO3.1a) | Uses factor theorem |
| \(\left(\cos\theta - \frac{1}{\sqrt{2}}\right)\left(\cos\theta + \frac{1}{\sqrt{2}}\right) = \left(\cos^2\theta - \frac{1}{2}\right)\), so \((2\cos^2\theta - 1)\) is a factor | M1 (AO1.1a) | Multiplies linear factors together |
| Correct result obtained | R1 (AO2.1) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Let \(c = \cos\theta\); \(32c^6 - 48c^4 + 18c^2 - 1 = (2c^2-1)(16c^4 - 16c^2 + 1)\) | M1 (AO3.1a) | Divides polynomial by quadratic factor |
| \(16c^4 - 16c^2 + 1 = 0\) solved as quadratic in \(c^2\): \(c^2 = \frac{2\pm\sqrt{3}}{4}\) | M1 (AO1.1a) | Solves quartic as quadratic in \(c^2\) |
| The roots of the quartic correspond to cosines of the angles found in part (a) | E1 (AO2.4) | Explains that quartic roots correspond to cosines of angles from part (a) |
| \(c = \pm\sqrt{\frac{2\pm\sqrt{3}}{4}}\) | A1 (AO1.1b) | Obtains all correct roots of quartic |
| Of angles \(\frac{\pi}{12}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{11\pi}{12}\): \(\frac{11\pi}{12}\) has negative cosine of greatest magnitude, therefore \(\cos\left(\frac{11\pi}{12}\right) = -\sqrt{\frac{2+\sqrt{3}}{4}}\) | R1 (AO2.1) | Rigorous argument including reason why that root corresponds to \(\cos\left(\frac{11\pi}{12}\right)\) |
## Question 13(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $6\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2}, \frac{11\pi}{2}$ | M1 (AO1.1a) | Obtains at least two correct solutions |
| $\theta = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{11\pi}{12}$ (in addition to given solutions) | A1 (AO1.1b) | Obtains all correct solutions |
---
## Question 13(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Expands $(\cos\theta + i\sin\theta)^6$ | M1 (AO1.1a) | |
| By de Moivre's theorem: $\cos 6\theta + i\sin 6\theta = (\cos\theta + i\sin\theta)^6$ | M1 (AO3.1a) | Equates real parts |
| $\cos 6\theta = \cos^6\theta - 15\cos^4\theta\sin^2\theta + 15\cos^2\theta\sin^4\theta - \sin^6\theta$ | A1 (AO1.1b) | Correct expression for real part in terms of powers of $\cos\theta$ and $\sin\theta$ |
| $= \cos^6\theta - 15\cos^4\theta(1-\cos^2\theta) + 15\cos^2\theta(1-\cos^2\theta)^2 - (1-\cos^2\theta)^3$ | M1 (AO3.1a) | Uses trig identity to express real part in terms of $\cos\theta$ |
| $\cos 6\theta = 32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1$ | R1 (AO2.1) | Completes rigorous argument |
---
## Question 13(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\theta = \frac{\pi}{4} \Rightarrow \cos 6\theta = 0 \Rightarrow 32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1 = 0$ | M1 (AO2.2a) | Uses fact that $\theta = \frac{\pi}{4}$ or $\theta = \frac{3\pi}{4}$ is a solution to deduce it is also solution to second equation |
| $\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$, so $\left(\cos\theta - \frac{1}{\sqrt{2}}\right)$ is a factor; similarly $\left(\cos\theta + \frac{1}{\sqrt{2}}\right)$ is also a factor | M1 (AO3.1a) | Uses factor theorem |
| $\left(\cos\theta - \frac{1}{\sqrt{2}}\right)\left(\cos\theta + \frac{1}{\sqrt{2}}\right) = \left(\cos^2\theta - \frac{1}{2}\right)$, so $(2\cos^2\theta - 1)$ is a factor | M1 (AO1.1a) | Multiplies linear factors together |
| Correct result obtained | R1 (AO2.1) | |
---
## Question 13(d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Let $c = \cos\theta$; $32c^6 - 48c^4 + 18c^2 - 1 = (2c^2-1)(16c^4 - 16c^2 + 1)$ | M1 (AO3.1a) | Divides polynomial by quadratic factor |
| $16c^4 - 16c^2 + 1 = 0$ solved as quadratic in $c^2$: $c^2 = \frac{2\pm\sqrt{3}}{4}$ | M1 (AO1.1a) | Solves quartic as quadratic in $c^2$ |
| The roots of the quartic correspond to cosines of the angles found in part (a) | E1 (AO2.4) | Explains that quartic roots correspond to cosines of angles from part (a) |
| $c = \pm\sqrt{\frac{2\pm\sqrt{3}}{4}}$ | A1 (AO1.1b) | Obtains all correct roots of quartic |
| Of angles $\frac{\pi}{12}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{11\pi}{12}$: $\frac{11\pi}{12}$ has negative cosine of greatest magnitude, therefore $\cos\left(\frac{11\pi}{12}\right) = -\sqrt{\frac{2+\sqrt{3}}{4}}$ | R1 (AO2.1) | Rigorous argument including reason why that root corresponds to $\cos\left(\frac{11\pi}{12}\right)$ |13
\begin{enumerate}[label=(\alph*)]
\item Two of the solutions to the equation $\cos 6 \theta = 0$ are $\theta = \frac { \pi } { 4 }$ and $\theta = \frac { 3 \pi } { 4 }$\\
Find the other solutions to the equation $\cos 6 \theta = 0$ for $0 \leq \theta \leq \pi$
13
\item Use de Moivre's theorem to show that
$$\cos 6 \theta = 32 \cos ^ { 6 } \theta - 48 \cos ^ { 4 } \theta + 18 \cos ^ { 2 } \theta - 1$$
13
\item Use the fact that $\theta = \frac { \pi } { 4 }$ and $\theta = \frac { 3 \pi } { 4 }$ are solutions to the equation $\cos 6 \theta = 0$ to find a factor of $32 \cos ^ { 6 } \theta - 48 \cos ^ { 4 } \theta + 18 \cos ^ { 2 } \theta - 1$ in the form ( $a \cos ^ { 2 } \theta + b$ ), where $a$ and $b$ are integers.\\[0pt]
[4 marks]
\item Hence show that
$$\cos \left( \frac { 11 \pi } { 12 } \right) = - \sqrt { \frac { 2 + \sqrt { 3 } } { 4 } }$$
\includegraphics[max width=\textwidth, alt={}, center]{13abb93f-2fef-465c-980c-3b412de06618-25_2492_1721_217_150}
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 2 2021 Q13 [16]}}