Complex Numbers And Quadratic Equations question 222
Question: We express $ \frac{{{(\cos 2\theta -i\sin 2\theta )}^{4}}{{(\cos 4\theta +i\sin 4\theta )}^{-5}}}{{{(\cos 3\theta +i\sin 3\theta )}^{-2}}{{(\cos 3\theta -i\sin 3\theta )}^{-9}}} $ in the form of $ x+iy $ , we get [Karnataka CET 2001]
Options:
A) $ \cos 49\theta -i\sin 49\theta $
B) $ \cos 23\theta -i\sin 23\theta $
C) $ \cos 49\theta +i\sin 49\theta $
D) $ \cos 21\theta +i\sin 21\theta $
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Answer:
Correct Answer: A
Solution:
$ \frac{{{(\cos 2\theta -i\sin 2\theta )}^{4}}{{(\cos 4\theta +i\sin 4\theta )}^{-5}}}{{{(\cos 3\theta +i\sin 3\theta )}^{-2}}{{(\cos 3\theta -i\sin 3\theta )}^{-9}}} $ $ =\frac{{{[{{(\cos \theta +i\sin \theta )}^{-2}}]}^{4}}{{[{{(\cos \theta +i\sin \theta )}^{4}}]}^{-5}}}{{{[{{(\cos \theta +i\sin \theta )}^{3}}]}^{-2}}{{[{{(\cos \theta +i\sin \theta )}^{-3}}]}^{-9}}} $ $ =\frac{{{(\cos \theta +i\sin \theta )}^{-8}}{{(\cos \theta +i\sin \theta )}^{-20}}}{{{(\cos \theta +i\sin \theta )}^{-6}}{{(\cos \theta +i\sin \theta )}^{27}}} $ $ ={{(\cos \theta +i\sin \theta )}^{-8-20+6-27}}={{(\cos \theta +i\sin \theta )}^{-49}} $ $ =\cos 49\theta -i\sin 49\theta . $
BETA
