[Hutchjang]

ข้อความเดิมเขียนโดยคุณ กิตติ

$\left(\dfrac{\cos A-\cos B}{a-b}\right)c+\cos C=\dfrac{(\cos A-\cos B)\sin C}{\sin A-\sin B}+\cos C$
$\left(\dfrac{\cos A-\cos B}{a-b}\right)c=\left(\dfrac{\cos A-\cos B}{\frac{a}{c} -\frac{b}{c} }\right)$
$\frac{\sin A}{a}=\frac{\sin B}{b} =\frac{\sin C}{c} $
$\frac{a}{c}=\frac{\sin A}{\sin B} ,\frac{b}{c}=\frac{\sin B}{\sin B} $

$\dfrac{(\cos A-\cos B)\sin C}{\sin A-\sin B}+\cos C=\dfrac{\left(\,\cos A \sin C+\sin A \cos C\right) -\left(\,\cos B\sin C+\sin B\cos C\right) }{\sin A-\sin B} $

$=\dfrac{\sin(A+C)-\sin(B+C)}{\sin A-\sin B}$

$A+B+C=180^o$

$\sin(A+C)-\sin(B+C) =\sin (180^o-B)-\sin (180^o-A)=\sin B-\sin A$

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