Question

If $sin\alpha =\frac{1}{\sqrt{5}}andsin\beta =\frac{3}{5},then\beta -\alpha$ lies in the interval

• A. $[0,\frac{\pi }{4}]$
  
• B. $[\frac{\pi }{2},\frac{3\pi }{4}]$
• C. $[\frac{3\pi }{4},\pi ]$
  
• D. $[\pi ,\frac{5\pi }{4}]$

Answer 1

Answer: (A), (C)

$[\frac{3\pi }{4},\pi ]$


We have $sin\alpha =\frac{1}{\sqrt{5}}\Rightarrow cos\alpha =\frac{2}{\sqrt{5}}$
and $sin\beta =\frac{3}{5}\Rightarrow cos\beta =\frac{4}{5}$
$sin(\beta -\alpha )=sin\beta cos\alpha -sin\alpha cos\beta$
$\frac{3}{5}.\frac{2}{\sqrt{5}}-\frac{1}{\sqrt{5}}.\frac{4}{5}=\frac{2}{5\sqrt{5}}=0.1789$
Now $sin\frac{\pi }{4}=\frac{1}{\sqrt{2}}=0.7071=sin\frac{3\pi }{4}$
Since 0 < 0.1789 < 0.7071
$\therefore sin0$ < $sin(\beta -\alpha )$ < $sin\frac{\pi }{4}\Rightarrow 0$ < $(\beta -\alpha )$ < $\frac{\pi }{4}$
Also, $sin\pi$ < $sin(\beta -\alpha )$ < $sin\frac{3\pi }{4}$
$\therefore (\beta -\alpha )ϵ[0,\frac{\pi }{4}]and[\frac{3\pi }{4},\pi ].$