Let $z=\cos heta+i \sin heta$. Then, the value of $\sum_{m=1}^{15} \operatorname{Im}\left(z^{2 m-1} ight)$ at $ heta=2^{\circ}$ is

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Let $z=\cos 	heta+i \sin 	heta$. Then, the value of $\sum_{m=1}^{15} \operatorname{Im}\left(z^{2 m-1}ight)$ at $	heta=2^{\circ}$ is

Aicharya · Accepted Answer

Given that $z=\cos 	heta+i \sin 	heta=e^{i 	heta}$ $ \begin{aligned} 	herefore \sum_{m=1}^{15} \operatorname{Im}\left(z^{2 m-1}ight) & =\sum_{m=1}^{15} \operatorname{Im}\left(e^{i 	heta}ight)^{2 m-1}=\sum_{m=1}^{15} \operatorname{Im} e^{i(2 m-1) 	heta} \ & =\sin 	heta+\sin 3 	heta+\sin 5 	heta+\ldots+\sin 29 	heta \ & =\frac{\sin \left(\frac{	heta+29 	heta}{2}ight) \sin \left(\frac{15 	imes 2 	heta}{2}ight)}{\sin \left(\frac{2 	heta}{2}ight)}=\frac{\sin (15 	heta) \sin (15 	heta)}{\sin 	heta}=\frac{1}{4 \sin 2^{\circ}} \end{aligned} $

Let $z=\cos \theta+i \sin \theta$. Then, the value of $\sum_{m=1}^{15} \operatorname{Im}\left(z^{2…

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