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

Mar 28, 2018

To prove $\sin A + \sin B + \sin C = \frac{3 \sqrt{3}}{2.}$

We know that $\textcolor{red}{\sin A + \sin B} = 2 \sin \left(\frac{A + B}{2}\right) \cdot \cos \left(\frac{A - B}{2}\right)$

$\sin C = \sin \left(180 - \left(A + B\right)\right) = \sin \left(A + B\right)$ [because "sin(180-x)=sinx)]

$\textcolor{b l u e}{\sin C} = \sin \left(A + B\right) = 2 \sin \left(\frac{A + B}{2}\right) \cos \left(\frac{A + B}{2}\right)$ [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

$= 2 \sin \left(\frac{A + B}{2}\right) \cdot \left\{\cos \left(\frac{A - B}{2}\right) + \cos \left(\frac{A + B}{2}\right)\right\}$

color(white)(ddddd

$= 2 \sin \left(\frac{A + B}{2}\right) \left\{2 C o s \left(\frac{A}{2}\right) \cdot \cos \left(\frac{B}{2}\right)\right\}$
[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

$= 4 \sin \left(\frac{A + B}{2}\right) \sin \left(\frac{B + C}{2}\right) \sin \left(\frac{C + A}{2}\right) .$
[because " sinA = cos(pi/2 - A)]

color(white)(ddddd

So if $x + y + z = 1$ and $x y z$ is maximum when $x = y = z = \frac{1}{3}$

So $A + B = B + C = A + C$.

$\implies A = B = C = \frac{A + B + C}{3}$, when $\sin A + \sin B + \sin C$ is maximum $= 4 \sin {\left\{\left(\frac{180}{3}\right)\right\}}^{3} = 4 {\left(\sin 60\right)}^{3} = {\left(\frac{\sqrt{3}}{2}\right)}^{3} = 4 \cdot 3 \frac{\sqrt{3}}{2} ^ 3 = \frac{3 \sqrt{3}}{2.}$

Therefore sinA+sinB+sinC ≤(3sqrt3)/2 and also the triangle is equilateral, since each angle comes out to be 60°.