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Let $z=\cos \theta+i \sin \theta$. Then, the value of $\sum_{m=1}^{15} \operatorname{Im}\left(z^{2 m-1}\right)$ at $\theta=2^{\circ}$ is
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Verified Answer
The correct answer is:
$\frac{1}{4 \sin 2^{\circ}}$
$\frac{1}{4 \sin 2^{\circ}}$
Given that $z=\cos \theta+i \sin \theta=e^{i \theta}$
$$
\begin{aligned}
\therefore \sum_{m=1}^{15} \operatorname{Im}\left(z^{2 m-1}\right) & =\sum_{m=1}^{15} \operatorname{Im}\left(e^{i \theta}\right)^{2 m-1}=\sum_{m=1}^{15} \operatorname{Im} e^{i(2 m-1) \theta} \\
& =\sin \theta+\sin 3 \theta+\sin 5 \theta+\ldots+\sin 29 \theta \\
& =\frac{\sin \left(\frac{\theta+29 \theta}{2}\right) \sin \left(\frac{15 \times 2 \theta}{2}\right)}{\sin \left(\frac{2 \theta}{2}\right)}=\frac{\sin (15 \theta) \sin (15 \theta)}{\sin \theta}=\frac{1}{4 \sin 2^{\circ}}
\end{aligned}
$$
$$
\begin{aligned}
\therefore \sum_{m=1}^{15} \operatorname{Im}\left(z^{2 m-1}\right) & =\sum_{m=1}^{15} \operatorname{Im}\left(e^{i \theta}\right)^{2 m-1}=\sum_{m=1}^{15} \operatorname{Im} e^{i(2 m-1) \theta} \\
& =\sin \theta+\sin 3 \theta+\sin 5 \theta+\ldots+\sin 29 \theta \\
& =\frac{\sin \left(\frac{\theta+29 \theta}{2}\right) \sin \left(\frac{15 \times 2 \theta}{2}\right)}{\sin \left(\frac{2 \theta}{2}\right)}=\frac{\sin (15 \theta) \sin (15 \theta)}{\sin \theta}=\frac{1}{4 \sin 2^{\circ}}
\end{aligned}
$$
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