Trigonometric Identities Question 313
Question: For $ A=133{}^\circ ,\ 2\cos \frac{A}{2} $ is equal to
[DCE 2001]
Options:
A) $ -\sqrt{1+\sin A}-\sqrt{1-\sin A} $
B) $ -\sqrt{1+\sin A}+\sqrt{1-\sin A} $
C) $ \sqrt{1+\sin A}-\sqrt{1-\sin A} $
D) $ \sqrt{1+\sin A}+\sqrt{1-\sin A} $
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Answer:
Correct Answer: C
Solution:
For $ A=133^{o},\frac{A}{2}={{66.5}^{o}} $
Þ $ \sin \frac{A}{2}>\cos \frac{A}{2}>0 $ Hence $ \sqrt{1+\sin A}=\sin \frac{A}{2}+\cos \frac{A}{2} $ ?..(i) and $ \sqrt{1-\sin A}=\sin \frac{A}{2}-\cos \frac{A}{2} $ ?..(ii) Subtract (ii) from (i), $ 2\cos \frac{A}{2}=\sqrt{1+\sin A}-\sqrt{1-\sin A} $ .
BETA
