11. (a) Given that \(z = \cos \theta + \mathrm { i } \sin \theta\), use de Moivre's theorem to show that
$$z ^ { n } + \frac { 1 } { z ^ { n } } = 2 \cos n \theta$$
(b) Express \(32 \cos ^ { 6 } \theta\) in the form \(p \cos 6 \theta + q \cos 4 \theta + r \cos 2 \theta + \mathrm { s }\), where \(p , q , r\) and \(s\) are integers.
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(a) \(z^n = (\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta\)M1
\(z^{-n} = (\cos\theta + i\sin\theta)^{-n} = \cos(-n\theta) + i\sin(-n\theta) = \cos n\theta - i\sin n\theta\) both
M1
Adding: \(z^n + \frac{1}{z^n} = 2\cos n\theta\) * cso
A12
(b) \(\left(z + \frac{1}{z}\right)^6 = z^6 + 6z^4 + 15z^2 + 20 + 15z^{-2} + 6z^{-4} + z^{-6}\)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
\(32\cos^6\theta = \cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10\) A1, A1
(p = 1, q = 6, r = 15, s = 10) Al any two correct
(c) \(\int\cos^6\theta d\theta = \left(\frac{1}{32}\right)\int(\cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10)d\theta\)M1A1ft
\(= \left(\frac{1}{32}\right)\left[\frac{\sin 6\theta}{6} + \frac{6\sin 4\theta}{4} + \frac{15\sin 2\theta}{2} + 10\theta\right]\) \(\left[....\right]_0^{\pi} = \frac{1}{32}\left[-\frac{3}{2} \times \frac{\sqrt{3}}{2} + \frac{2}{2} \times \frac{\sqrt{3}}{2} + \frac{10\pi}{3}\right] = \frac{5\pi}{48} - \frac{3\sqrt{3}}{32}\) M1A1ft
M1A14
or exact equivalent
Total: [11]
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**(a)** $z^n = (\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ | M1 |
$z^{-n} = (\cos\theta + i\sin\theta)^{-n} = \cos(-n\theta) + i\sin(-n\theta) = \cos n\theta - i\sin n\theta$ | both | M1 |
Adding: $z^n + \frac{1}{z^n} = 2\cos n\theta$ * | cso | A12 |
**(b)** $\left(z + \frac{1}{z}\right)^6 = z^6 + 6z^4 + 15z^2 + 20 + 15z^{-2} + 6z^{-4} + z^{-6}$ | 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 |
$32\cos^6\theta = \cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10$ | A1, A1 | (p = 1, q = 6, r = 15, s = 10) Al any two correct | 5 |
**(c)** $\int\cos^6\theta d\theta = \left(\frac{1}{32}\right)\int(\cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10)d\theta$ | M1A1ft |
$= \left(\frac{1}{32}\right)\left[\frac{\sin 6\theta}{6} + \frac{6\sin 4\theta}{4} + \frac{15\sin 2\theta}{2} + 10\theta\right]$ | |
$\left[....\right]_0^{\pi} = \frac{1}{32}\left[-\frac{3}{2} \times \frac{\sqrt{3}}{2} + \frac{2}{2} \times \frac{\sqrt{3}}{2} + \frac{10\pi}{3}\right] = \frac{5\pi}{48} - \frac{3\sqrt{3}}{32}$ | M1A1ft | M1A14 |
or exact equivalent | |
**Total: [11]**
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11. (a) Given that $z = \cos \theta + \mathrm { i } \sin \theta$, use de Moivre's theorem to show that
$$z ^ { n } + \frac { 1 } { z ^ { n } } = 2 \cos n \theta$$
(b) Express $32 \cos ^ { 6 } \theta$ in the form $p \cos 6 \theta + q \cos 4 \theta + r \cos 2 \theta + \mathrm { s }$, where $p , q , r$ and $s$ are integers.\\
(c) Hence find the exact value of
$$\int _ { 0 } ^ { \frac { \pi } { 3 } } \cos ^ { 6 } \theta \mathrm {~d} \theta$$
\hfill \mbox{\textit{Edexcel FP2 2007 Q11 [11]}}