FP2 June 2005 Q10
10.
(a) Given that \(z = \mathrm{e}^{\mathrm{i}\theta}\), show that \[z^n - \frac{1}{z^n} = 2\mathrm{i}\sin n\theta,\] where \(n\) is a positive integer. (2)
(b) Show that \[\sin^5\theta = \frac{1}{16}(\sin 5\theta - 5\sin 3\theta + 10\sin\theta).\] (5)
(c) Hence solve, in the interval \(0 \leqslant \theta \lt 2\pi\), \[\sin 5\theta - 5\sin 3\theta + 6\sin\theta = 0.\] (5)
| Scheme | Marks |
|---|---|
| \(z^n = \mathrm{e}^{\mathrm{i}n\theta} = (\cos n\theta + \mathrm{i}\sin n\theta),\ z^{-n} = \mathrm{e}^{-\mathrm{i}n\theta} = (\cos n\theta - \mathrm{i}\sin n\theta)\) | M1 |
| Completion (needs to be convincing) \(z^n - \dfrac{1}{z^n} = 2\mathrm{i}\sin n\theta\ (*)\) AG | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\left(z - \dfrac{1}{z}\right)^5 = z^5 - 5z^3 + 10z - \dfrac{10}{z} + \dfrac{5}{z^3} - \dfrac{1}{z^5}\) | M1A1 |
| \(= \left(z^5 - \dfrac{1}{z^5}\right) - 5\left(z^3 - \dfrac{1}{z^3}\right) + 10\left(z - \dfrac{1}{z}\right)\) | |
| \((2\mathrm{i}\sin\theta)^5 = 32\mathrm{i}\sin^5\theta = 2\mathrm{i}\sin 5\theta - 10\mathrm{i}\sin 3\theta + 20\mathrm{i}\sin\theta\) | M1A1 |
| \(\Rightarrow \sin^5\theta = \dfrac{1}{16}(\sin 5\theta - 5\sin 3\theta + 10\sin\theta)\ (*)\) AG | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| Finding \(\sin^5\theta = \tfrac{1}{4}\sin\theta\) | M1 |
| \(\theta = 0, \pi\) (both) | B1 |
| \((\sin^4\theta = \tfrac{1}{4}) \Rightarrow \sin\theta = \pm\dfrac{1}{\sqrt{2}}\) | M1 |
| \(\theta = \dfrac{\pi}{4}, \dfrac{3\pi}{4};\ \dfrac{5\pi}{4}, \dfrac{7\pi}{4}\) | A1;A1 |
| (5) | |
| (12 marks) |