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

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(D) Given $z = \cos 	heta + i \sin 	heta = e^{i 	heta}$.Using De Moivre's theorem,$z^{2m-1} = \cos((2m-1)	heta) + i \sin((2m-1)	heta)$.Therefore,

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

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(D) Given $z = \cos 	heta + i \sin 	heta = e^{i 	heta}$.Using De Moivre's theorem,$z^{2m-1} = \cos((2m-1)	heta) + i \sin((2m-1)	heta)$.Therefore,$	ext{Im}(z^{2m-1}) = \sin((2m-1)	heta)$.We need to calculate $S = \sum_{m=1}^{15} \sin((2m-1)	heta) = \sin 	heta + \sin 3	heta + \sin 5	heta + \dots + \sin 29	heta$.This is a sum of sines in arithmetic progression with first term $a = 	heta$,common difference $d = 2	heta$,and number of terms $n = 15$.The formula for the sum is $S = \frac{\sin(n d / 2)}{\sin(d / 2)} \sin(a + (n-1)d / 2)$.Substituting the values: $S = \frac{\sin(15 \cdot 2	heta / 2)}{\sin(2	heta / 2)} \sin(	heta + (15-1)2	heta / 2) = \frac{\sin(15	heta)}{\sin 	heta} \sin(	heta + 14	heta) = \frac{\sin^2(15	heta)}{\sin 	heta}$.At $	heta = 2^{\circ}$,$15	heta = 30^{\circ}$.$S = \frac{\sin^2(30^{\circ})}{\sin 2^{\circ}} = \frac{(1/2)^2}{\sin 2^{\circ}} = \frac{1/4}{\sin 2^{\circ}} = \frac{1}{4 \sin 2^{\circ}}$.

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

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