a+b+c=2π 证明sina+sinb+sinc=4sina/2sinb/2sinc/2

sina+sinb+sinc
=sin[(a+b)/2+(a-b)/2]+sin[(a+b)/2-(a-b)/2]+sin[2π-(a+b)]
=2sin[(a+b)/2]*cos[(a-b)/2]-sin(a+b)
=2sin[(a+b)/2]*cos[(a-b)/2]-2sin[(a+b)/2]*cos[(a+b)/2]
=2sin[(a+b)/2]*{cos[(a-b)/2]-cos[(a+b)/2]}
=2sin[(2π-c)/2]*{[cos(a/2)cos(b/2)+sin(a/2)sin(b/2)]-[cos(a/2)cos(b/2)-sin(a/2)sin(b/2)]}
=2sin(c/2)*2sin(a/2)sin(b/2)
=4sin(a/2)sin(b/2)sin(c/2)

sinA+sinB =sin[(A+B)/2+(A-B)/2]+sin[(A+B)/2-(A-B)/2] 所以左边=2sin(90-C/2)cos(A-B)/2-2sinC/2cosC/2 =2cosC/2cos(A-B

(a+b)/2= π -c/2
sina+sinb+sinc
=2sin(a+b)cos(a-b)+2sin(c/2)cos(c/2)
=2sin[(a+b)/2]cos[(a-b)/2]-2sin[(a+b)/2]cos[(a+b)/2]
=2sin[(a+b)/2] [cos(a-b)/2-cos[(a+b)/2]
=2sinc/2 *[cos(a-b)/2-cos[(a+b)/2]
=4sina/2sinb/2sinc/2
得证.