JEE Advance - Mathematics (2009 - Paper 1 Offline - No. 1)
Let $$z = \,\cos \,\theta \, + i\,\sin \,\theta $$ . Then the value of $$\sum\limits_{m = 1}^{15} {{\mathop{\rm Im}\nolimits} } ({z^{2m - 1}})\,at\,\theta \, = {2^ \circ }$$ is
$${1 \over {\sin \,{2^ \circ }}}$$
$${1 \over {3\sin \,{2^ \circ }}}$$
$${1 \over {2\sin \,{2^ \circ }}}$$
$${1 \over {4\sin \,{2^ \circ }}}$$
Explanation
We have
$$X = \sin \theta + \sin 3\theta \, + \,...\, + \,\sin 29\theta $$
$$2(\sin \theta )X = 1 - \cos 2\theta + \cos 2\theta - \cos 4\theta \, + \,...\, + \,\cos 28\theta - \cos 30\theta $$
$$X = {{1 - \cos 30\theta } \over {2\sin \theta }} = {1 \over {4\sin 2^\circ }}$$
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