Complex Numbers And Quadratic Equations question 296
Question: The real part of $ {{\sin }^{-1}}({e^{i\theta }}) $ is [RPET 1997]
Options:
A) $ {{\cos }^{-1}}(\sqrt{\sin \theta }) $
B) $ {{\sinh }^{-1}}(\sqrt{\sin \theta }) $
C) $ {{\sin }^{-1}}(\sqrt{\sin \theta }) $
D) $ {{\sin }^{-1}}(\sqrt{\cos \theta }) $
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Answer:
Correct Answer: A
Solution:
$ {{\sin }^{-1}}({e^{i\theta }})=x+iy $
Þ $ \sin (x+iy)={e^{i\theta }} $
Þ $ \sin x\cosh y+i\cos x\sinh y=\cos \theta +i\sin \theta $ On comparison, $ \cosh y=\frac{\cos \theta }{\sin x} $ , $ \sinh y=\frac{\sin \theta }{\cos x}\therefore {{\cosh }^{2}}y-{{\sinh }^{2}}y=1 $
Þ $ \frac{1-{{\sin }^{2}}\theta }{1-{{\cos }^{2}}x}-\frac{{{\sin }^{2}}\theta }{{{\cos }^{2}}x}=1 $
Þ $ {{\cos }^{2}}x-{{\sin }^{2}}\theta {{\cos }^{2}}x-{{\sin }^{2}}\theta +{{\cos }^{2}}x{{\sin }^{2}}\theta $ = $ {{\cos }^{2}}x-{{\cos }^{4}}x $
Þ $ {{\cos }^{4}}x={{\sin }^{2}}\theta \Rightarrow x={{\cos }^{-1}}\sqrt{\sin \theta } $ .
BETA
