3. trignometrical ratios,functions and identities — ગણિત ધોરણ 11 સાયન્સ — Question

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