โจทย์ตรีโกณ

[TaPaew]

ถ้า $\sin A+\cos A=\frac{\sqrt{2}}{3}$
จงหาค่าของ $\left|\,\sin A-\cos A\right| $

[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} $

[C.Nontaya]

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

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