FP2 June 2007 Q11
11.
(a) Given that \(z = \cos\theta + \mathrm{i}\sin\theta\), use de Moivre’s theorem to show that \[z^n + \frac{1}{z^n} = 2\cos n\theta.\] (2)
(b) Express \(32\cos^6\theta\) in the form \(p\cos 6\theta + q\cos 4\theta + r\cos 2\theta + s\), where \(p\), \(q\), \(r\) and \(s\) are integers. (5)
(c) Hence find the exact value of \[\int_0^{\frac{\pi}{3}} \cos^6\theta\,\mathrm{d}\theta.\] (4)
| Scheme | Marks |
|---|---|
| \(z^n = (\cos\theta + \mathrm{i}\sin\theta)^n = \cos n\theta + \mathrm{i}\sin n\theta\) \(z^{-n} = (\cos\theta + \mathrm{i}\sin\theta)^{-n} = \cos(-n\theta) + \mathrm{i}\sin(-n\theta) = \cos n\theta - \mathrm{i}\sin n\theta\) both | M1 |
| Adding \(z^n + \dfrac{1}{z^n} = 2\cos n\theta\ *\) cso | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\left(z + \dfrac{1}{z}\right)^6 = z^6 + 6z^4 + 15z^2 + 20 + 15z^{-2} + 6z^{-4} + z^{-6}\) | M1 |
| \(= z^6 + z^{-6} + 6(z^4 + z^{-4}) + 15(z^2 + z^{-2}) + 20\) | M1 |
| \(64\cos^6\theta = 2\cos 6\theta + 12\cos 4\theta + 30\cos 2\theta + 20\) | M1 |
| \(32\cos^6\theta = \cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10\) \((p = 1, q = 6, r = 15, s = 10)\) A1 any two correct | A1, A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \cos^6\theta\,\mathrm{d}\theta = \left(\frac{1}{32}\right)\int (\cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10)\,\mathrm{d}\theta\) \(= \left(\dfrac{1}{32}\right)\left[\dfrac{\sin 6\theta}{6} + \dfrac{6\sin 4\theta}{4} + \dfrac{15\sin 2\theta}{2} + 10\theta\right]\) | M1A1ft |
| \([\ldots\ldots]_0^{\frac{\pi}{3}} = \dfrac{1}{32}\left[-\dfrac{3}{2} \times \dfrac{\sqrt{3}}{2} + \dfrac{15}{2} \times \dfrac{\sqrt{3}}{2} + \dfrac{10\pi}{3}\right] = \dfrac{5\pi}{48} + \dfrac{3\sqrt{3}}{32}\) or exact equivalent | M1A1 |
| (4) | |
| (11 marks) |