同角三角基本关系式3

$\mathrm{已}\mathrm{知}\sin \alpha +\cos \alpha =\frac{1}{5},\alpha \in (0,\pi )，\mathrm{求}：$

$(1)\sin \alpha \cos \alpha$
$(2)\sin \alpha -\cos \alpha$

解析：
$(1)(\sin \alpha +\cos \alpha {)}^2=1+2\sin \alpha \cos \alpha =\frac{1}{25}$
$\sin \alpha \cos \alpha =-\frac{12}{25}$
$(2)(\sin \alpha -\cos \alpha {)}^2=1-2\sin \alpha \cos \alpha =\frac{49}{25}$
$\because \sin \alpha \cos \alpha <0,\alpha \in (0,\pi )$
$\therefore \sin \alpha >0,\cos \alpha <0$
$\sin \alpha -\cos \alpha =\frac{7}{5}$