sina+sinb=[（根3）/3](cosb-cosa).0<a<pai,0<b<pai. a-b等于？

sina+sinb=2sin(a/2+b/2)cos(a/2-b/2)
cosb-cosa=-2sin(b/2+a/2)sin(b/2-a/2)
代入,得 cos(a/2-b/2)=[(根)3/3]sin(a/2-b/2)
所以 tan(a/2-b/2)=(根)3
所以 (a - b)/2 = π/3 + kπ 即 a - b = 2π/3 + 2kπ (k属于整数)
又 0 < a < π, 0 < b < π， 所以 -π< a-b <π
得 a - b = 2π/3