A) \[\frac{1}{\sin 2{}^\circ }\]
B) \[\frac{1}{2\sin 2{}^\circ }\]
C) \[\frac{1}{3\sin 2{}^\circ }\]
D) \[\frac{1}{4\sin 2{}^\circ }\]
Correct Answer: D
Solution :
| We have, \[z=\cos \theta +isin\theta \] |
| \[{{z}^{2m-1}}={{(\cos \theta +i\sin \theta )}^{2m-1}}\] |
| \[{{z}^{2m-1}}=\cos \,(2m-1)\theta +isin\,(2m-1)\theta \] |
| \[{{I}_{m}}({{z}^{2m-1}})=\sum\limits_{m=1}^{15}{\sin \,(2m-1)\theta }\] |
| \[\sum\limits_{m=1}^{15}{{{I}_{m}}({{z}^{2m-1}})=\sum\limits_{m=1}^{15}{\sin \,(2m-1)\,\theta }}\]\[=\frac{1}{2\sin \theta }\sum\limits_{m=1}^{15}{2\sin \theta \sin (2m-1)\,\theta }\] |
| \[=\frac{1-\cos 30{}^\circ \cdot \theta }{2\sin \theta }\] |
| \[=\frac{1-\cos 60{}^\circ }{2\sin 2{}^\circ }\] \[[\theta =2{}^\circ ]\] |
| \[=\frac{1}{4\sin 2{}^\circ }\] |
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