• # question_answer If $\sin \alpha =1/\sqrt{5}$and $\sin \beta =3/5$,then $\beta -\alpha$lies in the interval [Roorkee Qualifying 1998] A) $[0,\,\pi /4]$ B) $[\pi /2,\,3\pi /4]$ C) $[3\pi /4,\,\pi ]$ D) $[\pi ,\,5\pi /4]$

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$[3\pi /4,\,\pi ]$.