Trigonometric Identities Question 268
Question: If $ \sin \alpha =\frac{-3}{5}, $ where $ \pi <\alpha <\frac{3\pi }{2}, $ then $ \cos \frac{1}{2}\alpha = $
[MP PET 1998]
Options:
A) $ \frac{-1}{\sqrt{10}} $
B) $ \frac{1}{\sqrt{10}} $
C) $ \frac{3}{\sqrt{10}} $
D) $ \frac{-3}{\sqrt{10}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \cos (\alpha /2)=-\sqrt{\frac{1+\cos \alpha }{2}} $ $ \cos \alpha =-\sqrt{1-{{\sin }^{2}}\alpha } $ [ $ \because \alpha $ lies in IIIrd Quadrant] $ =-\sqrt{1-\frac{9}{25}}=-\frac{4}{5} $
$ \therefore ,\cos (\alpha /2)=-\sqrt{\frac{1-\frac{4}{5}}{2}}=-\frac{1}{\sqrt{10}} $ .
BETA
