[tongkub]

$sinA + sinB = 2sin(\dfrac{A+B}{2})cos(\dfrac{A-B}{2}) = \dfrac{\sqrt{2}}{2}$ -- *

$cosA + cosB = 2cos(\dfrac{A+B}{2})cos(\dfrac{A-B}{2}) = \dfrac{\sqrt{2}}{2}$ -- **

$ *\div ** = tan(\dfrac{A+B}{2}) = 1$

จาก $cos(A+B) = \dfrac{1 - tan^2(\dfrac{A+B}{2})}{1 + tan^2(\dfrac{A+B}{2})}$

$cos(A+B) = 0 \therefore sin(A+B) = 1$