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°#.
