7 Use de Moivre's theorem to show that
$$\tan 5 \theta = \frac { 5 t - 10 t ^ { 3 } + t ^ { 5 } } { 1 - 10 t ^ { 2 } + 5 t ^ { 4 } }$$
where \(t = \tan \theta\).
Deduce that the roots of the equation \(t ^ { 4 } - 10 t ^ { 2 } + 5 = 0\) are \(\pm \tan \frac { 1 } { 5 } \pi\) and \(\pm \tan \frac { 2 } { 5 } \pi\).
Hence show that \(\tan \frac { 1 } { 5 } \pi \tan \frac { 2 } { 5 } \pi = \sqrt { } 5\).
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Question 7:
Answer Marks
Guidance
Working/Answer Mark
Guidance
\((\cos\theta + \text{i}\sin\theta)^5 = \cos 5\theta + \text{i}\sin 5\theta\) M1
\(= c^5 + 5c^4(\text{i}s) + 10c^2(\text{i}s)^3 + 5c(\text{i}s)^4 + (\text{i}s)^5\) A1
\(\tan 5\theta = \frac{\sin 5\theta}{\cos 5\theta} = \frac{5c^4s - 10c^2s^3 + s^5}{c^5 - 10c^3s^2 + 5cs^4} = \frac{5t - 10t^3 + t^5}{1 - 10t^2 + 5t^4}\) M1A1
AG
\(\tan 5\theta = 0 \Rightarrow t(t^4 - 10t^2 + 5) = 0\) M1
Allow non-factorised form
Rejecting \(t = 0\) (implies \(\theta = k\pi\)); roots of \(t^4 - 10t^2 + 5 = 0\) are: A1
\(\pm\tan\frac{1}{5}\pi\) and \(\pm\tan\frac{2}{5}\pi\), since \(5\theta = \pm 2\pi\) or \(5\theta = \pm 4\pi\) A1
AG
Product of roots \(= \tan^2\frac{1}{5}\pi\tan^2\frac{2}{5}\pi = 5 \Rightarrow \tan\frac{1}{5}\pi\tan\frac{2}{5}\pi = \sqrt{5}\) M1A1
S.C. Award B1\(\checkmark\) if result deduced from information given in the question
Total: [4]+[3]+[2] = [9]
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## Question 7:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $(\cos\theta + \text{i}\sin\theta)^5 = \cos 5\theta + \text{i}\sin 5\theta$ | M1 | |
| $= c^5 + 5c^4(\text{i}s) + 10c^2(\text{i}s)^3 + 5c(\text{i}s)^4 + (\text{i}s)^5$ | A1 | |
| $\tan 5\theta = \frac{\sin 5\theta}{\cos 5\theta} = \frac{5c^4s - 10c^2s^3 + s^5}{c^5 - 10c^3s^2 + 5cs^4} = \frac{5t - 10t^3 + t^5}{1 - 10t^2 + 5t^4}$ | M1A1 | AG |
| $\tan 5\theta = 0 \Rightarrow t(t^4 - 10t^2 + 5) = 0$ | M1 | Allow non-factorised form |
| Rejecting $t = 0$ (implies $\theta = k\pi$); roots of $t^4 - 10t^2 + 5 = 0$ are: | A1 | |
| $\pm\tan\frac{1}{5}\pi$ and $\pm\tan\frac{2}{5}\pi$, since $5\theta = \pm 2\pi$ or $5\theta = \pm 4\pi$ | A1 | AG |
| Product of roots $= \tan^2\frac{1}{5}\pi\tan^2\frac{2}{5}\pi = 5 \Rightarrow \tan\frac{1}{5}\pi\tan\frac{2}{5}\pi = \sqrt{5}$ | M1A1 | S.C. Award B1$\checkmark$ if result deduced from information given in the question |
**Total: [4]+[3]+[2] = [9]**
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7 Use de Moivre's theorem to show that
$$\tan 5 \theta = \frac { 5 t - 10 t ^ { 3 } + t ^ { 5 } } { 1 - 10 t ^ { 2 } + 5 t ^ { 4 } }$$
where $t = \tan \theta$.
Deduce that the roots of the equation $t ^ { 4 } - 10 t ^ { 2 } + 5 = 0$ are $\pm \tan \frac { 1 } { 5 } \pi$ and $\pm \tan \frac { 2 } { 5 } \pi$.
Hence show that $\tan \frac { 1 } { 5 } \pi \tan \frac { 2 } { 5 } \pi = \sqrt { } 5$.
\hfill \mbox{\textit{CAIE FP1 2014 Q7 [9]}}