Complex Numbers And Quadratic Equations question 217
Question: $ {{( \frac{\cos \theta +i\sin \theta }{\sin \theta +i\cos \theta } )}^{4}} $ equals [RPET 1996]
Options:
A) $ \sin 8\theta -i\cos 8\theta $
B) $ \cos 8\theta -i\sin 8\theta $
C) $ \sin 8\theta +i\cos 8\theta $
D) $ \cos 8\theta +i\sin 8\theta $
Show Answer
Answer:
Correct Answer: D
Solution:
$ {{( \frac{\cos \theta +i\sin \theta }{\sin \theta +i\cos \theta } )}^{4}}=\frac{{{(\cos \theta +i\sin \theta )}^{4}}}{i^{4}{{(\cos \theta -i\sin \theta )}^{4}}} $ $ =\frac{\cos 4\theta +i\sin 4\theta }{\cos 4\theta -i\sin 4\theta } $ $ =\frac{(\cos 4\theta +i\sin 4\theta )(\cos 4\theta +i\sin 4\theta )}{(\cos 4\theta -i\sin 4\theta )(\cos 4\theta +i\sin 4\theta )} $ $ \frac{{{(\cos 4\theta +i\sin 4\theta )}^{2}}}{{{\cos }^{2}}4\theta +{{\sin }^{2}}4\theta }=\cos 8\theta +i\sin 8\theta $
BETA
