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
$$\sum _ { n = 1 } ^ { N } ( 2 n - 1 ) \sin \left[ \frac { ( 2 n - 1 ) \pi } { N } \right] = - N \operatorname { cosec } \frac { \pi } { N }$$
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\(\sum_{n=1}^N z^{2n-1} = (z - z^{2N+1})(1 - z^2)\) (Sum of G.P.) B1
\(S_N = \Re[(z - z^{2N+1})(1 - z^2)]\) (Real Part) M1
\(S_N = \Re[(z - z^{2N+1})\overline{1-z^2}/(2-2\cos2\theta)]\) (AEF with a real denominator) M1
\(= ... = (\cos(2N - 1)\theta - \cos(2N + 1)\theta)/(2 - 2\cos2\theta)\) (AEF in \(\theta\) and \(N\) only.) M1A1
Multiplying out numerator \(= ... = \sin 2N\theta/2\sin\theta\) (AG) A1
\(- \sum_{n=1}^N(2n - 1)\sin(2n - 1)\theta = N\cos 2N\theta\csc\theta - (1/2)\sin 2N\theta\csc\cot\theta\) (AEF) M1A1(LHS) A1(RHS)
Put \(\theta = \pi/N\) to obtain required result (AG) A1
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$\sum_{n=1}^N z^{2n-1} = (z - z^{2N+1})(1 - z^2)$ (Sum of G.P.) | B1 |
$S_N = \Re[(z - z^{2N+1})(1 - z^2)]$ (Real Part) | M1 |
$S_N = \Re[(z - z^{2N+1})\overline{1-z^2}/(2-2\cos2\theta)]$ (AEF with a real denominator) | M1 |
$= ... = (\cos(2N - 1)\theta - \cos(2N + 1)\theta)/(2 - 2\cos2\theta)$ (AEF in $\theta$ and $N$ only.) | M1A1 |
| Multiplying out numerator |
$= ... = \sin 2N\theta/2\sin\theta$ (AG) | A1 |
$- \sum_{n=1}^N(2n - 1)\sin(2n - 1)\theta = N\cos 2N\theta\csc\theta - (1/2)\sin 2N\theta\csc\cot\theta$ (AEF) | M1A1(LHS) A1(RHS) |
Put $\theta = \pi/N$ to obtain required result (AG) | A1 |
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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
$$\sum _ { n = 1 } ^ { N } ( 2 n - 1 ) \sin \left[ \frac { ( 2 n - 1 ) \pi } { N } \right] = - N \operatorname { cosec } \frac { \pi } { N }$$
\hfill \mbox{\textit{CAIE FP1 2008 Q10 [10]}}