Complex Numbers And Quadratic Equations question 223
Question: $ {{(\sin \theta +i\cos \theta )}^{n}} $ is equal to [RPET 2001]
Options:
A) $ \cos n\theta +i\sin n\theta $
B) $ \sin n\theta +i\cos n\theta $
C) $ \cos n( \frac{\pi }{2}-\theta )+i\sin n( \frac{\pi }{2}-\theta ) $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ {{(\sin \theta +i\cos \theta )}^{n}} $ $ ={{[ \cos ( \frac{\pi }{2}-\theta )+i\sin ( \frac{\pi }{2}-\theta ) ]}^{n}} $ = $ \cos n( \frac{\pi }{2}-\theta )+i\sin n( \frac{\pi }{2}-\theta ) $ .
BETA
