Trigonometric Identities Question 39
Question: The sum $ S=\sin \theta +\sin 2,\theta +….+\sin ,n\theta , $ equals
[AMU 2002]
Options:
A) $ \sin \frac{1}{2}(n+1)\theta \sin \frac{1}{2}n\theta /\sin \frac{\theta }{2} $
B) $ \cos \frac{1}{2}(n+1)\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} $
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Answer:
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,term $ = $ \frac{\sin \frac{n\beta }{2}}{\sin \frac{\beta }{2}}\sin [ \frac{\theta +\theta +(n-1)\beta }{2} ] $ Put $ \beta =\theta $ , then $ S=\frac{\sin \frac{n\theta }{2}.\sin \frac{\theta (n+1)}{2}}{\sin \frac{\theta }{2}} $ .
BETA
