How do you prove that a triangle is equilateral iff sinA+sinB+sinC=(3sqrt3)/2?

1 Answer
Mar 28, 2018

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°.