Complex Numbers And Quadratic Equations question 697
Question: What is the argument of $ (1-\sin \theta )+i\cos \theta $ ?
Options:
A) $ \frac{\pi }{2}-\frac{\theta }{2} $
B) $ \frac{\pi }{2}+\frac{\theta }{2} $
C) $ \frac{\pi }{4}-\frac{\theta }{2} $
D) $ \frac{\pi }{4}+\frac{\theta }{2} $
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Answer:
Correct Answer: D
Solution:
Given complex number is $ ( 1- sin\theta ) +i cos\theta = a + ib $ $ Argument = tan \theta = \frac{b}{a} $
$ \Rightarrow \tan \theta =\frac{\cos \theta }{1-\sin \theta } $ $ =\frac{{{\cos }^{2}}\frac{\theta }{2}-{{\sin }^{2}}\frac{\theta }{2}}{{{\sin }^{2}}\frac{\theta }{2}+{{\cos }^{2}}\frac{\theta }{2}-2\sin \frac{\theta }{2}\cos \frac{\theta }{2}} $ $ =\frac{( \cos \frac{\theta }{2}-\sin \frac{\theta }{2} )( \cos \frac{\theta }{2}+\sin \frac{\theta }{2} )}{{{( \sin \frac{\theta }{2}-\cos \frac{\theta }{2} )}^{2}}} $ $ =\frac{\cos \frac{\theta }{2}+\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}-\sin \frac{\theta }{2}} $ $ =\frac{1+\tan \frac{\theta }{2}}{1-\tan \frac{\theta }{2}}=\frac{\tan \frac{\pi }{4}+\tan \frac{\theta }{2}}{1-\tan \frac{\pi }{4}\tan \frac{\theta }{2}} $ $ \tan \theta =\tan ( \frac{\pi }{4}+\frac{\theta }{2} ) $ Hence, $ argument=\frac{\pi }{4}+\frac{\theta }{2} $
BETA
