sin A+sin B = 2 sin((A+B)/2)cos((A-B)/2)

sinC = sin(π-(A+B))= sin(A+B)= 2 sin((A+B)/2)cos((A+B)/2)

cos A+cos B = 2cos((A+B)/2)cos((A-B)/2)

Then 2 sin ((a+b)/2) cos ((a-b)/2) = 2 sin ((a+b)/2) cos ((a+b)/2) 2 cos.

Obviously, if sin ((A+B)/2)=0, cos ((A-B)/2) or (cos((A+B)/2))? = 1/2

If sin ((A+B)/2)=0, then sin((π-C)/2)=cos(C/2)=0. There is no value in the reasonable range.

If cos ((A-B)/2)=0, then (A-B)/2=π/2+nπ, that is, A-b = π+2nπ. No answer

If (cos((A+B)/2))？ = 1/2, that is, (sin(C/2))? = 1/2。 In a reasonable range, only C=π/2, which is a right triangle.

To sum up, a triangle is a right triangle.