| Exam Board | CAIE |
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| Module | FP1 (Further Pure Mathematics 1) |
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| Year | 2008 |
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| Session | June |
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| Topic | Complex numbers 2 |
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10 By considering \(\sum _ { n = 1 } ^ { N } z ^ { 2 n - 1 }\), where \(z = \mathrm { e } ^ { \mathrm { i } \theta }\), show that
$$\sum _ { n = 1 } ^ { N } \cos ( 2 n - 1 ) \theta = \frac { \sin ( 2 N \theta ) } { 2 \sin \theta }$$
where \(\sin \theta \neq 0\).
Deduce that
$$\sum _ { n = 1 } ^ { N } ( 2 n - 1 ) \sin \left[ \frac { ( 2 n - 1 ) \pi } { N } \right] = - N \operatorname { cosec } \frac { \pi } { N }$$