Tomando $S = \displaystyle{\sum^{15}_{k=1}~\pu{cis}^{2k-1} \left(\dfrac{\pi}{36}\right)}$, temos$$S~=~\pu{cis} \left(\dfrac{\pi}{36}\right)+\pu{cis}^3 \left(\dfrac{\pi}{36}\right)+\pu{cis}^5 \left(\dfrac{\pi}{36}\right)+ \cdots + \pu{cis}^{29} \left(\dfrac{\pi}{36}\right)$$$$S~=~\dfrac{\pu{cis}\left(\dfrac{\pi}{36}\right) \cdot \left(\pu{cis}^{30}\left(\dfrac{\pi}{36}\right)-1\right)}{\pu{cis}^2\left(\dfrac{\pi}{36}\right)-1}$$Da identidade $\pu{cis}^n (\theta) - 1 = 2i\cdot \sin (\frac{n\theta}{2}) \cdot \pu{cis} (\frac{n\theta}{2})$, e pelas Leis de Moivre, determina-se$$S=\dfrac{\cancel{\pu{cis}\left(\dfrac{\pi}{36}\right)} \cdot \cancel{2i} \cdot \sin \left(\dfrac{5\pi}{12}\right) \cdot \pu{cis} \left(\dfrac{5\pi}{12}\right)}{\cancel{2i} \cdot \sin \left(\dfrac{\pi}{36}\right) \cdot \cancel{\pu{cis} \left(\dfrac{\pi}{36}\right)}}=\dfrac{\sin \left(\dfrac{5\pi}{12}\right)}{\sin \left(\dfrac{\pi}{36}\right)} \cdot \left[\cos \left(\dfrac{5\pi}{12}\right) + i \sin \left(\dfrac{5\pi}{12}\right)\right]$$Portanto$$\pu{Img} (S)~=~\dfrac{\sin^2 \left(\dfrac{5\pi}{12}\right)}{\sin \left(\dfrac{\pi}{36}\right)}~=~\dfrac{1-\cos \left(\frac{5\pi}{6}\right)}{2\sin \left(\frac{\pi}{36}\right)}~=~\dfrac{1-\left(-\frac{\sqrt{3}}{2}\right)}{2\sin \left(\frac{\pi}{36}\right)}$$$$\boxed{\pu{Img} (S)~=~\dfrac{2+\sqrt{3}}{4\sin \left(\frac{\pi}{36}\right)}}$$$$\bf{Alternativa~(A)}$$