Complex Numbers And Quadratic Equations question 211
Question: The following in the form of $ A+iB $ $ {{(\cos 2\theta +i\sin 2\theta )}^{-5}} $ $ {{(\cos 3\theta -i\sin 3\theta )}^{6}} $ $ {{(\sin \theta -i\cos \theta )}^{3}} $ in the form of $ A+iB $ is [MNR 1991]
Options:
A) $ (\cos 25\theta +i\sin 25\theta ) $
B) $ i(\cos 25\theta +i\sin 25\theta ) $
C) $ i(\cos 25\theta -i\sin 25\theta ) $
D) $ (\cos 25\theta -i\sin 25\theta ) $
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Answer:
Correct Answer: D
Solution:
$ \sin \theta -i\cos \theta =-i^{2}\sin \theta -i\cos \theta =-i(\cos \theta +i\sin \theta ) $ Given expression is $ {{(-i)}^{3}}[\cos (-10\theta -18\theta +3\theta )+i\sin (-25\theta )] $ = $ i(\cos 25\theta -i\sin 25\theta ] $ .
BETA
