6 Use de Moivre's theorem to express \(\cot 7 \theta\) in terms of \(\cot \theta\).
Use the equation \(\cot 7 \theta = 0\) to show that the roots of the equation
$$x ^ { 6 } - 21 x ^ { 4 } + 35 x ^ { 2 } - 7 = 0$$
are \(\cot \left( \frac { 1 } { 14 } k \pi \right)\) for \(k = 1,3,5,9,11,13\), and deduce that
$$\cot ^ { 2 } \left( \frac { 1 } { 14 } \pi \right) \cot ^ { 2 } \left( \frac { 3 } { 14 } \pi \right) \cot ^ { 2 } \left( \frac { 5 } { 14 } \pi \right) = 7$$
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Question 6:
Answer Marks
Guidance
Working/Answer Marks
Guidance
\((c+is)^7 = c^7 + 7c^6(is) + \ldots + (is)^7\) M1
\(\frac{\cos 7\theta}{\sin 7\theta} = \frac{c^7 - 21c^5s^2 + 35c^3s^4 - 7cs^6}{7c^6s - 35c^4s^3 + 21c^2s^5 - s^7}\) A1, M1
\(\cot 7\theta = \frac{\cot^7\theta - 21\cot^5\theta + 35\cot^3\theta - 7\cot\theta}{7\cot^6\theta - 35\cot^4\theta + 21\cot^2\theta - 1}\) A1 [4]
\(\cot 7\theta = 0\) and \(\cot\theta \neq 0 \Rightarrow x^6 - 21x^4 + 35x^2 - 7 = 0\), where \(x = \cot\theta\) and \(\theta = k\frac{\pi}{14}\) where \(k = 1, 3, 5, 9, 11, 13\) M1
Product of roots \(\Rightarrow \cot\frac{\pi}{14}\cot\frac{3\pi}{14}\cot\frac{5\pi}{14}\cot\frac{9\pi}{14}\cot\frac{11\pi}{14}\cot\frac{13\pi}{14} = -7\) M1 A1
But \(\cot\frac{\pi}{14} = -\cot\frac{13\pi}{14}\), \(\cot\frac{3\pi}{14} = -\cot\frac{11\pi}{14}\), \(\cot\frac{5\pi}{14} = -\cot\frac{9\pi}{14}\) M1
Hence \(\cot^2\frac{1}{14}\pi\cot^2\frac{3}{14}\pi\cot^2\frac{5}{14}\pi = 7\) (AG) (Penultimate line must be seen) A1 [5]
SC Award B1 if product of roots mentioned, without proper pairing seen
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## Question 6:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $(c+is)^7 = c^7 + 7c^6(is) + \ldots + (is)^7$ | M1 | |
| $\frac{\cos 7\theta}{\sin 7\theta} = \frac{c^7 - 21c^5s^2 + 35c^3s^4 - 7cs^6}{7c^6s - 35c^4s^3 + 21c^2s^5 - s^7}$ | A1, M1 | |
| $\cot 7\theta = \frac{\cot^7\theta - 21\cot^5\theta + 35\cot^3\theta - 7\cot\theta}{7\cot^6\theta - 35\cot^4\theta + 21\cot^2\theta - 1}$ | A1 [4] | |
| $\cot 7\theta = 0$ and $\cot\theta \neq 0 \Rightarrow x^6 - 21x^4 + 35x^2 - 7 = 0$, where $x = \cot\theta$ and $\theta = k\frac{\pi}{14}$ where $k = 1, 3, 5, 9, 11, 13$ | M1 | |
| Product of roots $\Rightarrow \cot\frac{\pi}{14}\cot\frac{3\pi}{14}\cot\frac{5\pi}{14}\cot\frac{9\pi}{14}\cot\frac{11\pi}{14}\cot\frac{13\pi}{14} = -7$ | M1 A1 | |
| But $\cot\frac{\pi}{14} = -\cot\frac{13\pi}{14}$, $\cot\frac{3\pi}{14} = -\cot\frac{11\pi}{14}$, $\cot\frac{5\pi}{14} = -\cot\frac{9\pi}{14}$ | M1 | |
| Hence $\cot^2\frac{1}{14}\pi\cot^2\frac{3}{14}\pi\cot^2\frac{5}{14}\pi = 7$ **(AG)** (Penultimate line must be seen) | A1 [5] | SC Award **B1** if product of roots mentioned, without proper pairing seen |
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6 Use de Moivre's theorem to express $\cot 7 \theta$ in terms of $\cot \theta$.
Use the equation $\cot 7 \theta = 0$ to show that the roots of the equation
$$x ^ { 6 } - 21 x ^ { 4 } + 35 x ^ { 2 } - 7 = 0$$
are $\cot \left( \frac { 1 } { 14 } k \pi \right)$ for $k = 1,3,5,9,11,13$, and deduce that
$$\cot ^ { 2 } \left( \frac { 1 } { 14 } \pi \right) \cot ^ { 2 } \left( \frac { 3 } { 14 } \pi \right) \cot ^ { 2 } \left( \frac { 5 } { 14 } \pi \right) = 7$$
\hfill \mbox{\textit{CAIE FP1 2016 Q6 [9]}}