abc属于(0,π/2)
sin(b-a)
=sinbcosa-sinacosb
=sinb(cosb+cosc)-(sinb-sinc)cosb
=sinbcosc+sinccosb
=sin(b+c)
b-a=b+c,-a=c舍去
或
b-a+b+c=π/2,即2b-a+c=π/2 ①
sin(b-a)
=sinbcosa-sinacosb
=(sina+sinc)cosa-sina(cosa-cosc)
=sinccosa+sinacosc
=sin(a+c)
同理,2a-b+c=π/2 ②
①-②,3(b-a)=0
所以 b-a=0