| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Derive triple angle formula only |
| Difficulty | Challenging +1.2 This is a structured Further Maths question with clear signposting through multiple parts. While it involves De Moivre's theorem and requires several algebraic manipulations, each step is heavily guided. Part (a) is standard derivation of triple angle formulae, part (b) applies these in a straightforward way, and part (c) uses Vieta's formulas. The question requires competent algebra and knowledge of FP2 content, but the path is clearly laid out with no novel insights needed, making it moderately above average difficulty. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) \(\cos 3\theta + i\sin 3\theta = (\cos \theta + i\sin \theta)^3\) | M1 | |
| \(= \cos^3 \theta + 3i\cos^2 \theta \sin \theta + 3i^2 \cos \theta \sin^2 \theta + i^3 \sin^3 \theta\) | A1 | |
| Real parts: \(\cos 3\theta = \cos^3 \theta - 3\cos \theta \sin^2 \theta\) | A1 | 3 marks; AG |
| (ii) Imaginary parts: \(\sin 3\theta = 3\cos^2 \theta \sin \theta - \sin^3 \theta\) | A1F | 1 mark |
| (iii) \(\tan 3\theta = \frac{\sin 3\theta}{\cos 3\theta}\) | M1 | Used |
| \(= \frac{3\cos^2 \theta \sin \theta - \sin^3 \theta}{\cos^3 \theta - 3\sin^2 \theta \cos \theta}\) | A1F | Error in \(\sin 3\theta\) |
| \(= \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta}\) | A1 | 3 marks; AG |
| (b)(i) \(\tan \frac{3\pi}{12} = 1\) | B1 | Used (possibly implied) |
| \(\tan \frac{\pi}{12}\) is a root of \(1 = \frac{x^3 - 3x}{3x^2 - 1}\) | M1 | Must be hence |
| \(x^3 - 3x^2 - 3x + 1 = 0\) | A1 | 3 marks |
| (ii) Other roots are \(\tan \frac{5\pi}{12}\), \(\tan \frac{9\pi}{12}\) | B1B1 | 2 marks |
| (c) \(\tan \frac{\pi}{12} + \tan \frac{5\pi}{12} + \tan \frac{9\pi}{12} = 3\) | M1 | Must be hence |
| \(\tan \frac{\pi}{12} + \tan \frac{5\pi}{12} = 4\) | A1 | 2 marks |
**(a)(i)** $\cos 3\theta + i\sin 3\theta = (\cos \theta + i\sin \theta)^3$ | M1 |
$= \cos^3 \theta + 3i\cos^2 \theta \sin \theta + 3i^2 \cos \theta \sin^2 \theta + i^3 \sin^3 \theta$ | A1 |
Real parts: $\cos 3\theta = \cos^3 \theta - 3\cos \theta \sin^2 \theta$ | A1 | 3 marks; AG
**(ii)** Imaginary parts: $\sin 3\theta = 3\cos^2 \theta \sin \theta - \sin^3 \theta$ | A1F | 1 mark
**(iii)** $\tan 3\theta = \frac{\sin 3\theta}{\cos 3\theta}$ | M1 | Used
$= \frac{3\cos^2 \theta \sin \theta - \sin^3 \theta}{\cos^3 \theta - 3\sin^2 \theta \cos \theta}$ | A1F | Error in $\sin 3\theta$
$= \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta}$ | A1 | 3 marks; AG
**(b)(i)** $\tan \frac{3\pi}{12} = 1$ | B1 | Used (possibly implied)
$\tan \frac{\pi}{12}$ is a root of $1 = \frac{x^3 - 3x}{3x^2 - 1}$ | M1 | Must be hence
$x^3 - 3x^2 - 3x + 1 = 0$ | A1 | 3 marks
**(ii)** Other roots are $\tan \frac{5\pi}{12}$, $\tan \frac{9\pi}{12}$ | B1B1 | 2 marks
**(c)** $\tan \frac{\pi}{12} + \tan \frac{5\pi}{12} + \tan \frac{9\pi}{12} = 3$ | M1 | Must be hence
$\tan \frac{\pi}{12} + \tan \frac{5\pi}{12} = 4$ | A1 | 2 marks
**Question 6 Total: 14 marks**
---6
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item By applying De Moivre's theorem to $( \cos \theta + \mathrm { i } \sin \theta ) ^ { 3 }$, show that
$$\cos 3 \theta = \cos ^ { 3 } \theta - 3 \cos \theta \sin ^ { 2 } \theta$$
\item Find a similar expression for $\sin 3 \theta$.
\item Deduce that
$$\tan 3 \theta = \frac { \tan ^ { 3 } \theta - 3 \tan \theta } { 3 \tan ^ { 2 } \theta - 1 }$$
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Hence show that $\tan \frac { \pi } { 12 }$ is a root of the cubic equation
$$x ^ { 3 } - 3 x ^ { 2 } - 3 x + 1 = 0$$
\item Find two other values of $\theta$, where $0 < \theta < \pi$, for which $\tan \theta$ is a root of this cubic equation.
\end{enumerate}\item Hence show that
$$\tan \frac { \pi } { 12 } + \tan \frac { 5 \pi } { 12 } = 4$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2008 Q6 [14]}}