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Let $$Z=\cos\theta+i\ \sin\theta$$. Then the value of $$\sum_{m=1}^{15}Im(z^{2m-1})$$ at $$\theta=2^{0}$$ is
$$\frac{1}{\sin2^{0}}$$
$$\frac{1}{3\sin2^{0}}$$
$$\frac{1}{2\sin2^{0}}$$
$$\frac{1}{4\sin2^{0}}$$
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