Principle Of Mathematical Induction Question 12
Question: If $ n\in N $ ,then $ \sin \theta +\sin 2\theta +\sin 3\theta +….+\sin n\theta = $
Options:
A) $ \sin [ \frac{n(n+1)\theta }{2} ] $
B) $ \frac{\sin [ \frac{(n+1)}{2}\theta ]\sin \theta }{\sin ( \frac{\theta }{2} )} $
C) $ \frac{\sin [ \frac{(n+1)}{2}\theta ]\sin ( \frac{n\theta }{2} )}{\sin ( \frac{\theta }{2} )} $
D) $ \frac{\sin [ \frac{(n+1)}{2}\theta ]{{\sin }^{n}}( \frac{\theta }{2} )}{\sin ( \frac{\theta }{2} )} $
Correct Answer: C
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Answer:
Solution:
BETA
