Question
Let $$z = \cos \theta + i\sin \theta .$$ Then the value of $$\sum\limits_{m = 1}^{15} {{\text{Im}}\left( {{z^{2m - 1}}} \right)} \,\,{\text{at }}\theta = {{\text{2}}^ \circ }$$ is
A.
$$\frac{1}{{\sin {2^ \circ }}}$$
B.
$$\frac{1}{{3\sin {2^ \circ }}}$$
C.
$$\frac{1}{{2\sin {2^ \circ }}}$$
D.
$$\frac{1}{{4\sin {2^ \circ }}}$$
Answer :
$$\frac{1}{{4\sin {2^ \circ }}}$$
Solution :
$$\eqalign{
& z = \cos \theta + i\sin \theta \cr
& \Rightarrow \,\,{z^{2m - 1}} = {\left( {\cos \theta + i\sin \theta } \right)^{2m - 1}} \cr} $$
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \cos \left( {2m - 1} \right)\theta + i\sin \left( {2m - 1} \right)\theta \,\,\,\,\,\,\,\,\left[ \begin{array}{l}
{\rm{using\,\, De\,\, Moiver's\,\, theorem}}\\
{\left( {\cos \theta + i\sin \theta } \right)^n} = \cos n\theta + i\sin n\theta
\end{array} \right]\]
$$\eqalign{
& \therefore \,\,{\text{Im}}\left( {{z^{2m - 1}}} \right) = \sin \left( {2m - 1} \right)\theta \cr
& \therefore \,\,\sum\limits_{m = 1}^{15} {{\text{Im}}\left( {{z^{2m - 1}}} \right)} = \sum\limits_{m = 1}^{15} {\sin \left( {2m - 1} \right)\theta } \cr
& = \sin \theta + \sin 3\theta + \sin 5\theta + ..... + {\text{upto 15 terms}} \cr
& {\text{ = }}\frac{{\sin \left[ {15\left( {\frac{{2\theta }}{2}} \right)} \right].\sin \left[ {\theta + 14 \times \theta } \right]}}{{\sin \theta }} \cr} $$
\[\left[ \begin{array}{l}
{\rm{Using\,\, sin}}\alpha + \sin \left( {\alpha + \beta } \right) + \sin \left( {\alpha + 2\beta } \right) + ..... + n\,{\rm{terms}}\\
{\rm{ = }}\frac{{\sin \left( {\frac{{n\beta }}{2}} \right).\sin \left[ {\alpha + \frac{{\left( {n - 1} \right)\beta }}{2}} \right]}}{{\sin \left( {\frac{\beta }{2}} \right)}}
\end{array} \right]\]
$$\eqalign{
& = \frac{{\sin 15\theta .\sin 15\theta }}{{\sin \theta }} \cr
& = \frac{{\sin {{30}^ \circ }.\sin {{30}^ \circ }}}{{\sin {2^ \circ }}} \cr
& = \frac{1}{{4\sin {2^ \circ }}} \cr} $$