Let z = cos theta + i sin theta. Then the val | vedclass

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(D) Given $z = \cos 	heta + i \sin 	heta$,by De Moivre's Theorem,$z^n = \cos(n	heta) + i \sin(n	heta)$.Thus,$\operatorname{Im}(z^{2m-1}) = \sin((2

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$\frac{1}{4 \sin 2^{\circ}}$

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(D) Given $z = \cos 	heta + i \sin 	heta$,by De Moivre's Theorem,$z^n = \cos(n	heta) + i \sin(n	heta)$.Thus,$\operatorname{Im}(z^{2m-1}) = \sin((2m-1)	heta)$.The sum is $S = \sum_{m=1}^{15} \sin((2m-1)	heta) = \sin 	heta + \sin 3	heta + \sin 5	heta + \dots + \sin 29	heta$.This is an arithmetic progression of angles with $n = 15$ terms,first term $a = 	heta$,and common difference $d = 2	heta$.The sum of the series $\sum_{k=0}^{n-1} \sin(a + kd) = \frac{\sin(nd/2)}{\sin(d/2)} \sin(a + (n-1)d/2)$.Substituting $n=15, a=	heta, d=2	heta$:$S = \frac{\sin(15 	imes 2	heta / 2)}{\sin(2	heta / 2)} \sin(	heta + (14 	imes 2	heta / 2)) = \frac{\sin(15	heta)}{\sin 	heta} \sin(	heta + 14	heta) = \frac{\sin^2(15	heta)}{\sin 	heta}$.At $	heta = 2^{\circ}$,$15	heta = 30^{\circ}$.$S = \frac{\sin^2(30^{\circ})}{\sin 2^{\circ}} = \frac{(1/2)^2}{\sin 2^{\circ}} = \frac{1}{4 \sin 2^{\circ}}$.

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

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