[กิตติ]

$x=cos A,y=sin A$
$x+y=\frac{\sqrt{2} }{3} $
$\left|\,y-x\right|=? $

$x^2+y^2=1$
$(x+y)^2=1+2xy=\frac{2}{9} $
$2xy=\frac{7}{9} $
$คิดเลขผิด$
$2xy=-\frac{7}{9} $

$(y-x)^2=1-2xy=1-\frac{7}{9} =\frac{2}{9} $
$\left|\,y-x\right| =\sqrt{(y-x)^2}=\sqrt{\frac{2}{9} } =\frac{\sqrt{2} }{3} $
$(y-x)^2=1-2xy=1+\frac{7}{9} =\frac{16}{9} $
$\left|\,y-x\right| =\sqrt{(y-x)^2}=\sqrt{\frac{16}{9} } =\frac{4}{3} $