| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2014 |
| Session | June |
| Marks | 13 |
| Topic | Complex numbers 2 |
| Type | Express roots in trigonometric form |
| Difficulty | Challenging +1.8 This is a sophisticated multi-part question requiring de Moivre's theorem, algebraic manipulation to connect the identity to the equation, finding roots in trigonometric form, and using Vieta's formulas. While the techniques are standard for Further Maths, the question requires sustained reasoning across three connected parts with non-trivial algebraic insight to link cos 6θ to the given equation and extract the product of specific roots. |
| Spec | 1.05l Double angle formulae: and compound angle formulae4.02q De Moivre's theorem: multiple angle formulae |
(i) $\cos 6\theta = \mathrm{Re}(c + \mathrm{i}s)^6$ **M1**
Use of Binomial expansion for $(c + \mathrm{i}s)^6$, Re terms only required **M1**
$= c^6 - 15c^4s^2 + 15c^2s^4 - s^6$ **A1**
Use of $s^2 = 1 - c^2$ (and powers) **M1**
$\Rightarrow \cos 6\theta = c^6 - 15c^4(1-c^2) + 15c^2(1-2c^2+c^4) - (1-3c^2+3c^4-c^6)$
$\Rightarrow 2\cos 6\theta = 64c^6 - 96c^4 + 36c^2 - 2$ **Answer Given** **A1** [5]
(ii) Setting $x = 2\cos\theta$ in given equation **M1**
$\Rightarrow (2c)^6 - 6(2c)^4 + 9(2c)^2 - 3 = 2\cos 6\theta - 1 = 0$ **A1**
$\cos 6\theta = \frac{1}{2} \Rightarrow 6\theta = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}, \frac{13\pi}{3}, \frac{17\pi}{3}$ **M1**
$\Rightarrow \theta = \frac{\pi}{18}, \frac{5\pi}{18}, \frac{7\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}$ **A1**
$x = 2\cos\theta$ for each $\theta$, Must be 6 distinct $\theta$'s **B1** [5]
OR $x = \pm 2\cos\theta$ for $\theta = \frac{\pi}{18}, \frac{5\pi}{18}, \frac{7\pi}{18}$
(iii) Product of SIX roots $= -3 = \left(-4\cos^2\frac{\pi}{18}\right)\left(-4\cos^2\frac{5\pi}{18}\right)\left(-4\cos^2\frac{7\pi}{18}\right)$ **M1**
Signs justified; e.g. via $\cos(\pi - \theta) = -\cos\theta$ **M1**
$\Rightarrow \cos\frac{\pi}{18}\cos\frac{5\pi}{18}\cos\frac{7\pi}{18} = \frac{\sqrt{3}}{8}$ [+ve $\sqrt{\ }$ taken as read, since all angles acute] **A1** [3]10 (i) Use de Moivre's theorem to show that $2 \cos 6 \theta \equiv 64 \cos ^ { 6 } \theta - 96 \cos ^ { 4 } \theta + 36 \cos ^ { 2 } \theta - 2$.\\
(ii) Hence find, in exact trigonometric form, the six roots of the equation
$$x ^ { 6 } - 6 x ^ { 4 } + 9 x ^ { 2 } - 3 = 0$$
(iii) By considering the product of these six roots, determine the exact value of
$$\cos \left( \frac { 1 } { 18 } \pi \right) \cos \left( \frac { 5 } { 18 } \pi \right) \cos \left( \frac { 7 } { 18 } \pi \right) .$$
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2014 Q10 [13]}}