To prove sinA+sinB+sinC = (3sqrt3)/2.
We know that color(red)(sinA+sinB) = 2 sin((A+B)/2) * cos((A-B)/2)
sinC= sin (180-(A+B)) =sin(A+B) [because "sin(180-x)=sinx)]
color(blue)(sinC)= sin(A+B) = 2sin((A+B)/2) cos((A+B)/2) [because "(sin2x = 2sinxcosx)]
Therefore,
color(red)(sinA+sinB) +color(blue)(sinC)= color(red)(2sin((A+B)/2)× cos((A-B)/2))+ color(blue)(2sin((A+B)/2) ×cos ((A+B)/2)
color(white)(ddddd
= 2 sin ((A+B)/2) * {cos((A-B)/2) +cos ((A+B)/2)}
color(white)(ddddd
= 2 sin((A+B)/2) {2Cos (A/2 )* cos(B/2)}
[because " cosA + cosB= 2cos((A+B)/2)cos((A+B)/2)]
color(white)(ddddd
= 4sin((A+B)/2)×cos(π/2-(B+C)/2) cos(π/2-(A+B)/2)
color(white)(ddddd
= 4 sin ((A+B)/2) sin ((B+C)/2) sin((C+A)/2).
[because " sinA = cos(pi/2 - A)]
color(white)(ddddd
So if x+y+z =1 and xyz is maximum when x=y=z =1/3
So A+B = B+C = A+C.
=>A =B= C = (A+B+C) /3, when sinA+sinB+ sinC is maximum = 4 sin {(180/3)}^3 = 4(sin60)^3 = (sqrt3/2)^3 = 4*3sqrt3/2^3 = (3sqrt3)/2.
Therefore sinA+sinB+sinC ≤(3sqrt3)/2 and also the triangle is equilateral, since each angle comes out to be 60°.