Binomial Theorem And Its Simple Applications Question 145
Question: The product of $ \cos \theta \cos 2\theta \cos 3\theta …\cos ({2^{n-1}}\theta ) $ is
Options:
A) $ \frac{\sin (2^{n}\theta )}{2^{n}\sin \theta } $
B) $ \frac{cos(2^{n}\theta )}{2^{n}\cos \theta } $
C) $ \frac{cosn\theta }{2^{n}\cos \theta } $
D) $ \frac{sin(2^{n}\theta )}{\sin \theta } $
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Answer:
Correct Answer: A
Solution:
- [a] By setting n=1, given expression becomes $ \cos \theta $ .
By setting n=1, option (1) becomes $ \frac{\sin 2\theta }{2\sin \theta }=\frac{2\sin \theta \cos \theta }{2\sin \theta }=\cos \theta $ .
BETA
