If\[n\in N\],then \[\sin \theta +\sin 2\theta +\sin 3\theta +....+\sin n\theta =\]

\[\frac{\sin \left[ \frac{(n+1)}{2}\theta \right]{{\sin }^{n}}\left( \frac{\theta }{2} \right)}{\sin \left( \frac{\theta }{2} \right)}\]

\[\sin \left[ \frac{n(n+1)\theta }{2} \right]\]

\[\frac{\sin \left[ \frac{(n+1)}{2}\theta \right]\sin \theta }{\sin \left( \frac{\theta }{2} \right)}\]

\[\frac{\sin \left[ \frac{(n+1)}{2}\theta \right]\sin \left( \frac{n\theta }{2} \right)}{\sin \left( \frac{\theta }{2} \right)}\]

If\[n\in N\],then \[\sin \theta +\sin 2\theta +\sin 3\theta +....+\sin

1.	If\[n\in N\],then \[\sin \theta +\sin 2\theta +\sin 3\theta +....+\sin n\theta =\]
A.	\[\sin \left[ \frac{n(n+1)\theta }{2} \right]\]
B.	\[\frac{\sin \left[ \frac{(n+1)}{2}\theta \right]\sin \theta }{\sin \left( \frac{\theta }{2} \right)}\]
C.	\[\frac{\sin \left[ \frac{(n+1)}{2}\theta \right]\sin \left( \frac{n\theta }{2} \right)}{\sin \left( \frac{\theta }{2} \right)}\]
D.	\[\frac{\sin \left[ \frac{(n+1)}{2}\theta \right]{{\sin }^{n}}\left( \frac{\theta }{2} \right)}{\sin \left( \frac{\theta }{2} \right)}\]
Answer» D. \[\frac{\sin \left[ \frac{(n+1)}{2}\theta \right]{{\sin }^{n}}\left( \frac{\theta }{2} \right)}{\sin \left( \frac{\theta }{2} \right)}\]

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If\[n\in N\],then \[\sin \theta +\sin 2\theta +\sin 3\theta +....+\sin n\theta =\]

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