[กิตติ]

ข้อความเดิมเขียนโดยคุณ C.Nontaya

$sinA + cosA = \frac{\sqrt{2}}{3}$
$(sinA + cosA)^2 = sin^2A + cos^2A + 2sinAcosA = 1 + 2sinAcosA = \frac{2}{9}$
$2sinAcosA = -\frac{7}{9}$
$|sinA - cosA| = \sqrt{(sinA - cosA)^2} = \sqrt{sin^2A - 2sinAcosA + cos^2A} = \sqrt{1 - 2\left(\,-\frac{7}{9}\right)} = \frac{\sqrt{23}}{3}$

บรรทัดสุดท้ายน่าจะเป็น
$|sinA - cosA| = \sqrt{(sinA - cosA)^2} = \sqrt{sin^2A - 2sinAcosA + cos^2A} = \sqrt{1 - \left(\,-\frac{7}{9}\right)} = \frac{4}{3}$