8 By considering \(\sum _ { r = 1 } ^ { n } z ^ { 2 r - 1 }\), where \(z = \cos \theta + \mathrm { i } \sin \theta\), show that, if \(\sin \theta \neq 0\),
$$\sum _ { r = 1 } ^ { n } \sin ( 2 r - 1 ) \theta = \frac { \sin ^ { 2 } n \theta } { \sin \theta }$$
Deduce that
$$\sum _ { r = 1 } ^ { n } ( 2 r - 1 ) \cos \left[ \frac { ( 2 r - 1 ) \pi } { 2 n } \right] = - \operatorname { cosec } \left( \frac { \pi } { 2 n } \right) \cot \left( \frac { \pi } { 2 n } \right)$$
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Question 8:
Answer Marks
Guidance
Working Marks
Notes
\(\sum_{r=1}^{n} z^{2r-1} = \frac{z(1-[z^2]^n)}{1-z^2} = \frac{z - z^{2n+1}}{1-z^2}\) M1A1
\(= \frac{1-z^{2n}}{z^{-1}-z} = \frac{1-(\cos 2n\theta + i\sin 2n\theta)}{\cos\theta - i\sin\theta - (\cos\theta + i\sin\theta)} = \frac{1-\cos 2n\theta - i\sin 2n\theta}{-2i\sin\theta}\) M1A1, A1
Equating imaginary parts: \(\sum_{r=1}^{n}\sin(2r-1)\theta = \frac{2\sin^2 n\theta}{2\sin\theta} = \frac{\sin^2 n\theta}{\sin\theta}\) M1A1
(7) (AG)
Differentiating: \(\sum_{r=1}^{n}\sin(2r-1)\cos(2r-1)\theta = 2n\sin n\theta\cos n\theta\operatorname{cosec}\theta - \sin^2 n\theta\operatorname{cosec}\theta\cot\theta\) M1A1
Putting \(\theta = \frac{\pi}{2n}\): \(\sum_{r=1}^{n}(2r-1)\cos(2r-1)\left(\frac{\pi}{2n}\right) = n\sin\frac{\pi}{2}\cos\frac{\pi}{2}\operatorname{cosec}\left(\frac{\pi}{2n}\right) - \sin^2\frac{\pi}{2}\operatorname{cosec}\left(\frac{\pi}{2n}\right)\cot\left(\frac{\pi}{2n}\right)\) dM1
\(\Rightarrow \sum_{r=1}^{n}(2r-1)\cos\left[\frac{(2r-1)\pi}{2n}\right] = -\operatorname{cosec}\left(\frac{\pi}{2n}\right)\cot\left(\frac{\pi}{2n}\right)\) A1
(4) (AG)
Total: 11
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## Question 8:
| Working | Marks | Notes |
|---------|-------|-------|
| $\sum_{r=1}^{n} z^{2r-1} = \frac{z(1-[z^2]^n)}{1-z^2} = \frac{z - z^{2n+1}}{1-z^2}$ | M1A1 | |
| $= \frac{1-z^{2n}}{z^{-1}-z} = \frac{1-(\cos 2n\theta + i\sin 2n\theta)}{\cos\theta - i\sin\theta - (\cos\theta + i\sin\theta)} = \frac{1-\cos 2n\theta - i\sin 2n\theta}{-2i\sin\theta}$ | M1A1, A1 | |
| Equating imaginary parts: $\sum_{r=1}^{n}\sin(2r-1)\theta = \frac{2\sin^2 n\theta}{2\sin\theta} = \frac{\sin^2 n\theta}{\sin\theta}$ | M1A1 | (7) (AG) |
| Differentiating: $\sum_{r=1}^{n}\sin(2r-1)\cos(2r-1)\theta = 2n\sin n\theta\cos n\theta\operatorname{cosec}\theta - \sin^2 n\theta\operatorname{cosec}\theta\cot\theta$ | M1A1 | |
| Putting $\theta = \frac{\pi}{2n}$: | | |
| $\sum_{r=1}^{n}(2r-1)\cos(2r-1)\left(\frac{\pi}{2n}\right) = n\sin\frac{\pi}{2}\cos\frac{\pi}{2}\operatorname{cosec}\left(\frac{\pi}{2n}\right) - \sin^2\frac{\pi}{2}\operatorname{cosec}\left(\frac{\pi}{2n}\right)\cot\left(\frac{\pi}{2n}\right)$ | dM1 | |
| $\Rightarrow \sum_{r=1}^{n}(2r-1)\cos\left[\frac{(2r-1)\pi}{2n}\right] = -\operatorname{cosec}\left(\frac{\pi}{2n}\right)\cot\left(\frac{\pi}{2n}\right)$ | A1 | (4) (AG) |
| | **Total: 11** | |
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8 By considering $\sum _ { r = 1 } ^ { n } z ^ { 2 r - 1 }$, where $z = \cos \theta + \mathrm { i } \sin \theta$, show that, if $\sin \theta \neq 0$,
$$\sum _ { r = 1 } ^ { n } \sin ( 2 r - 1 ) \theta = \frac { \sin ^ { 2 } n \theta } { \sin \theta }$$
Deduce that
$$\sum _ { r = 1 } ^ { n } ( 2 r - 1 ) \cos \left[ \frac { ( 2 r - 1 ) \pi } { 2 n } \right] = - \operatorname { cosec } \left( \frac { \pi } { 2 n } \right) \cot \left( \frac { \pi } { 2 n } \right)$$
\hfill \mbox{\textit{CAIE FP1 2015 Q8 [11]}}