If \[\sin \alpha =1/\sqrt{5}\]and \[\sin \beta =3/5\],then \[\beta -\alpha \]lies in the interval [Roorkee Qualifying 1998]

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