Let  $Z=\mathrm{c}is\theta$. Then the value of $\sum _{m=1}^{15}\mathrm{Im}\left(Z^{2m-1}\right)$ at $\theta =2°$ is 

$Z=\cos \theta +i\sin \theta$

$\Rightarrow Z^{2m-1}=\cos \left(2m-1\right)\theta +i\sin \left(2m-1\right)\theta$

$\therefore I_m\left(Z^{2m-1}\right)=\sin \left(2m-1\right)\theta$

$\therefore \sum _{m=1}^{15}{I}_m\left(Z^{2m-1}\right)=\sum _{m=1}^{15}\sin \left(2m-1\right)\theta$

$=\sin \theta +\sin 3\theta +\sin 5\theta +\mathrm{....}+$upto 15 terms

$=\frac{\sin \left[15\left(\frac{2\theta }{2}\right)\right].\sin \left(\theta +14\times \theta \right)}{\sin \theta }$

$=\frac{\sin 15\theta .\sin 15\theta }{\sin \theta }=\frac{\sin 30°.\sin 30°}{\sin 2°}$

$\left\{\because \sin \alpha +\sin \left(\alpha +\beta \right)+\sin \left(\alpha +2\beta \right)+\mathrm{.....}+nterms=\frac{\sin \left(\frac{n\beta }{2}\right).\sin \left[\alpha +\left(n-1\right)\frac{\beta }{2}\right]}{\left(\sin \frac{\beta }{2}\right)}\right\}$

$=\frac{1}{2}.\frac{1}{2}.\frac{1}{\sin 2°}=\frac{1}{4\sin 2°}$