| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive trigonometric identities |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring de Moivre's theorem and algebraic manipulation, which places it above average difficulty. However, part (i) is a standard textbook application, part (ii) is straightforward algebra, and parts (iii)-(iv) involve routine polynomial analysis. The multi-step nature and FP3 content elevate it, but the techniques are well-practiced for Further Maths students with no novel insights required. |
| Spec | 1.05l Double angle formulae: and compound angle formulae4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Re\((c+is)^4 = \cos4\theta = c^4-6c^2s^2+s^4\) | M1* A1 | For expanding \((c+is)^4\): at least 2 terms and 1 binomial coefficient needed; For 3 correct terms |
| \(\cos 4\theta = c^4-6c^2(1-c^2)+(1-c^2)^2\) | M1 | For using \(s^2=1-c^2\) |
| \(\Rightarrow \cos 4\theta = 8\cos^4\theta-8\cos^2\theta+1\) | A1 4 | For correct expression for cos4θ CAO |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos 4\theta\cos 2\theta = \left(8c^4-8c^2+1\right)\left(2c^2-1\right)\) | For multiplying by \((2c^2-1)\) | |
| \(= 16\cos^6\theta-24\cos^4\theta+10\cos^2\theta-1\) | B1 1 | to obtain AG WWW |
| Answer | Marks | Guidance |
|---|---|---|
| \(16c^6-24c^4+10c^2-2 = 0\) | M1 | For factorising sextic |
| \(\Rightarrow (c^2-1)(8c^4-4c^2+1) = 0\) | with \((c-1), (c+1)\) or \((c^2-1)\) | |
| For quartic, \(b^2-4ac = 16-32 < 0\) | A1 | For justifying no other roots CWO |
| \(\Rightarrow c = \pm 1\) only \(\Rightarrow \theta = n\pi\) | A1 3 | For obtaining \(\theta = n\pi\) AG; Note that M1 A0 A1 is possible |
| Answer | Marks | Guidance |
|---|---|---|
| \(16c^6-24c^4+10c^2 = 0\) | M1 | For factorising sextic with \(c^2\) |
| \(\Rightarrow c^2(8c^4-12c^2+5) = 0\) | ||
| For quartic, \(b^2-4ac = 144-160 < 0\) | A1 | For justifying no other roots CWO |
| \(\Rightarrow \cos\theta = 0\) only | A1 3 | For correct condition obtained AG; Note that M1 A0 A1 is possible |
## (i)
Re$(c+is)^4 = \cos4\theta = c^4-6c^2s^2+s^4$ | M1* A1 | For expanding $(c+is)^4$: at least 2 terms and 1 binomial coefficient needed; For 3 correct terms
$\cos 4\theta = c^4-6c^2(1-c^2)+(1-c^2)^2$ | M1 | For using $s^2=1-c^2$
$\Rightarrow \cos 4\theta = 8\cos^4\theta-8\cos^2\theta+1$ | A1 4 | For correct expression for cos4θ CAO
## (ii)
$\cos 4\theta\cos 2\theta = \left(8c^4-8c^2+1\right)\left(2c^2-1\right)$ | | For multiplying by $(2c^2-1)$
$= 16\cos^6\theta-24\cos^4\theta+10\cos^2\theta-1$ | B1 1 | to obtain AG WWW
## (iii)
$16c^6-24c^4+10c^2-2 = 0$ | M1 | For factorising sextic
$\Rightarrow (c^2-1)(8c^4-4c^2+1) = 0$ | | with $(c-1), (c+1)$ or $(c^2-1)$
For quartic, $b^2-4ac = 16-32 < 0$ | A1 | For justifying no other roots CWO
$\Rightarrow c = \pm 1$ only $\Rightarrow \theta = n\pi$ | A1 3 | For obtaining $\theta = n\pi$ AG; Note that M1 A0 A1 is possible
SR For verifying $\theta = n\pi$ by substituting $c=\pm 1$ into $16c^6-24c^4+10c^2-2=0$ B1
## (iv)
$16c^6-24c^4+10c^2 = 0$ | M1 | For factorising sextic with $c^2$
$\Rightarrow c^2(8c^4-12c^2+5) = 0$ | |
For quartic, $b^2-4ac = 144-160 < 0$ | A1 | For justifying no other roots CWO
$\Rightarrow \cos\theta = 0$ only | A1 3 | For correct condition obtained AG; Note that M1 A0 A1 is possible
SR For verifying $\cos\theta = 0$ by substituting $c=0$ into $16c^6-24c^4+10c^2=0$ B1
SR For verifying $\theta = \frac{1}{2}\pi$ and $\theta = -\frac{1}{2}\pi$ satisfy $\cos 4\theta\cos 2\theta = -1$ B1\begin{enumerate}[label=(\roman*)]
\item Use de Moivre's theorem to express $\cos 4\theta$ as a polynomial in $\cos \theta$. [4]
\item Hence prove that $\cos 4\theta \cos 2\theta \equiv 16 \cos^6 \theta - 24 \cos^4 \theta + 10 \cos^2 \theta - 1$. [1]
\item Use part (ii) to show that the only roots of the equation $\cos 4\theta \cos 2\theta = 1$ are $\theta = n\pi$, where $n$ is an integer. [3]
\item Show that $\cos 4\theta \cos 2\theta = -1$ only when $\cos \theta = 0$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 2011 Q8 [11]}}