[JEE Advanced 2009] let z cos theta i sin theta then sum m 1 15 im z
Let $z=\cos\theta+i\sin\theta$. Then $\sum_{m=1}^{15}\operatorname{Im}(z^{2m-1})$ at $\theta=2^\circ$ is
(a) $\frac1{\sin2^\circ}$
(b) $\frac1{3\sin2^\circ}$
(c) $\frac1{2\sin2^\circ}$
(d) $\frac1{4\sin2^\circ}$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 2d ago
▲ 0
Correct answer: (d) $\frac1{4\sin2^\circ}$
$\sum_{m=1}^{15}\sin((2m-1)\theta)=\frac{\sin^2(15\theta)}{\sin\theta}=\frac{\sin^230^\circ}{\sin2^\circ}=\frac{1/4}{\sin2^\circ}=\frac1{4\sin2^\circ}.$
JEE Advanced 2009 · Complex Numbers — verified solution by the Vidaara Team.
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