A) \[\sin \frac{1}{2}(n+1)\text{ }\theta \sin \frac{1}{2}n\text{ }\theta /\sin \frac{\theta }{2}\]
B) \[\cos \frac{1}{2}(n+1)\text{ }\theta \sin \frac{1}{2}n\theta /\sin \frac{\theta }{2}\]
C) \[\sin \frac{1}{2}(n+1)\theta \cos \frac{1}{2}n\theta /\sin \frac{\theta }{2}\]
D) \[\cos \frac{1}{2}(n+1)\theta \cos \frac{1}{2}n\theta /\sin \frac{\theta }{2}\]
Correct Answer: A
Solution :
\[S=\sin \theta +\sin 2\theta +\sin 3\theta +.....+\sin n\theta \] We know, \[\sin \theta +\sin (\theta +\beta )+\sin (\theta +2\beta )+.......n\,\text{term}\] = \[\frac{\sin \frac{n\beta }{2}}{\sin \frac{\beta }{2}}\sin \left[ \frac{\theta +\theta +(n-1)\beta }{2} \right]\] Put\[\beta =\theta \], then \[S=\frac{\sin \frac{n\theta }{2}.\sin \frac{\theta (n+1)}{2}}{\sin \frac{\theta }{2}}\].
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