Trigonometric Identities Question 251
Question: If $ \sin \alpha =1/\sqrt{5} $ and $ \sin \beta =3/5 $ ,then $ \beta -\alpha $ lies in the interval
[Roorkee Qualifying 1998]
Options:
A) $ [0,,\pi /4] $
B) $ [\pi /2,,3\pi /4] $
C) $ [3\pi /4,,\pi ] $
D) $ [\pi ,,5\pi /4] $
Show Answer
Answer:
Correct Answer: C
Solution:
We have $ \sin \alpha =1/\sqrt{5}\Rightarrow \cos \alpha =2/\sqrt{5} $ and $ \sin \beta =3/5\Rightarrow \cos \beta =4/5 $ $ \sin (\beta -\alpha )=\sin \beta \cos \alpha -\sin \alpha \cos \beta $ $ =\frac{3}{5}.\frac{2}{\sqrt{5}}-\frac{1}{\sqrt{5}}.\frac{4}{5}=\frac{2}{5\sqrt{5}}=0.1789 $ Now $ \sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}=0.7071=\sin \frac{3\pi }{4} $ Since $ 0<0.1789<0.7071 $
$ \therefore $ $ \sin 0<\sin (\beta -\alpha )<\sin \frac{\pi }{4}\Rightarrow 0<(\beta -\alpha )<\frac{\pi }{4} $ Also, $ \sin \pi <\sin (\beta -\alpha )<\sin \frac{3\pi }{4} $
$ \therefore $ $ (\beta -\alpha )\in [0,,\pi /4] $ and $ [3\pi /4,,\pi ] $ .
BETA
