Question

sina+sinb=[（根3）/3](cosb-cosa).0

Answer 1

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-