FP2 June 2008 Q11
11. De Moivre’s theorem states that \[(\cos\theta + \mathrm{i}\sin\theta)^n = \cos n\theta + \mathrm{i}\sin n\theta \quad \text{for } n \in \Re\]
(a) Use induction to prove de Moivre’s theorem for \(n \in \mathbb{Z}^{+}\). (5)
(b) Show that \(\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta\) (5)
(c) Hence show that \(2\cos\dfrac{\pi}{10}\) is a root of the equation \[x^4 - 5x^2 + 5 = 0\] (3)
| Scheme | Marks |
|---|---|
| \((\cos\theta + \mathrm{i}\sin\theta)^1 = \cos\theta + \mathrm{i}\sin\theta\ \therefore\) true for \(n = 1\) | B1 |
| Assume true for \(n = k\), \((\cos\theta + \mathrm{i}\sin\theta)^k = \cos k\theta + \mathrm{i}\sin k\theta\) | |
| \((\cos\theta + \mathrm{i}\sin\theta)^{k+1} = (\cos k\theta + \mathrm{i}\sin k\theta)(\cos\theta + \mathrm{i}\sin\theta)\) | M1 |
| \(= \cos k\theta\cos\theta - \sin k\theta\sin\theta + \mathrm{i}(\sin k\theta\cos\theta + \cos k\theta\sin\theta)\) (Can be achieved either from the line above or the line below) | M1 |
| \(= \cos(k + 1)\theta + \mathrm{i}\sin(k + 1)\theta\) | A1 |
| Requires full justification of \((\cos\theta + \mathrm{i}\sin\theta)^{k+1} = \cos(k + 1)\theta + \mathrm{i}\sin(k + 1)\theta\) (\(\therefore\) true for \(n = k + 1\) if true for \(n = k\)) \(\therefore\) true for \(n \in \mathbb{Z}^{+}\) by induction | A1cso |
| (5) |
Alternative
For the 2nd M mark: \((\mathrm{e}^{\mathrm{i}k\theta})(\mathrm{e}^{\mathrm{i}\theta}) = \mathrm{e}^{\mathrm{i}\theta(k+1)}\)
| Scheme | Marks |
|---|---|
| \(\cos 5\theta = \mathrm{Re}[(\cos\theta + \mathrm{i}\sin\theta)^5]\) | |
| \(= \cos^5\theta + 10\cos^3\theta\,\mathrm{i}^2\sin^2\theta + 5\cos\theta\,\mathrm{i}^4\sin^4\theta\) | M1A1 |
| \(= \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta\) | M1 |
| \(= \cos^5\theta - 10\cos^3\theta(1 - \cos^2\theta) + 5\cos\theta(1 - \cos^2\theta)^2\) | M1 |
| \(\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta\) (*) | A1cso |
| (5) |
Alternative
| Scheme | Marks |
|---|---|
| \(\left(z + \dfrac{1}{z}\right)^5 = z^5 + 5z^4\left(\dfrac{1}{z}\right) + 10z^3\left(\dfrac{1}{z}\right)^2 + 10z^2\left(\dfrac{1}{z}\right)^3 + 5z\left(\dfrac{1}{z}\right)^4 + \left(\dfrac{1}{z}\right)^5\) | M1 |
| \(= 2\cos 5\theta + 10\cos 3\theta + 20\cos\theta\) | A1 |
| \((2\cos\theta)^5 = \)……..and attempt to put \(\cos 3\theta\) in powers of \(\cos\theta\) | M1 |
| Correct method (or formula) for putting \(\cos 3\theta\) in powers of \(\cos\theta\) | M1 |
| \(\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta\) | A1cso |
| Scheme | Marks |
|---|---|
| \(\dfrac{\cos 5\theta}{\cos\theta} = 0 \Rightarrow \cos 5\theta = 0\) | M1 |
| \(5\theta = \dfrac{\pi}{2}\) \(\theta = \dfrac{\pi}{10}\) | A1 |
| \(x = 2\cos\theta,\ x = 2\cos\dfrac{\pi}{10}\) is a root (*) | A1 |
| (3) | |
| (13 marks) |
Alternatives
Alternative (i)
| Scheme | Marks |
|---|---|
| Substitute given root into \(x^4 - 5x^2 + 5\): \(\left(2\cos\dfrac{\pi}{10}\right)^4 - 5\left(2\cos\dfrac{\pi}{10}\right)^2 + 5 = 2^4\left(\cos\dfrac{\pi}{10}\right)^4 - 5 \times 2^2\left(\cos\dfrac{\pi}{10}\right)^2 + 5\) | M1 |
| ‘Multiply by \(\cos\theta\)’ and use result from part (b): … \(= \cos\dfrac{5\pi}{10}\) | A1 |
| \(= 0\) and conclusion | A1 |
Alternative (ii)
| Scheme | Marks |
|---|---|
| Use \(5\theta = \dfrac{\pi}{2}\) in result from part (b) | M1 |
| \(16\left(\cos\dfrac{\pi}{10}\right)^5 - 20\left(\cos\dfrac{\pi}{10}\right)^3 + 5\left(\cos\dfrac{\pi}{10}\right)\) and divide by \(\cos\theta\) | A1 |
| \(= 0\) and conclusion | A1 |