Question

How do you prove that a triangle is equilateral if $\sin A + \sin B + \sin C = \dfrac{{3\sqrt 3 }}{2}?$

Answer 1

Hint : As we can see that the above question is related to trigonometry as sine is a trigonometric ratio. We know that a triangle whose all three sides are equal is called an equilateral triangle. To prove any triangle equilateral we have to show that the value of each and every angle is ${60^ \circ }$ . We will use trigonometric identities to solve this question.

Complete step-by-step answer :
As per the question we have to prove $\sin A + \sin B + \sin C = \dfrac{{3\sqrt 3 }}{2}$ .
We know the trigonometric formula which states that $\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right) \times \cos \left( {\dfrac{{A - B}}{2}} \right)$ .
Also we can write $\sin C = \sin \\{ 180 - (A + B)\\}$ and also that $\sin (180 - \theta ) = \sin \theta$ ,
Therefore $\sin C = \sin (A + B)$ . We will equate all the formula i.e. $\sin C = \sin (A + B) = 2\sin \left( {\dfrac{{A + B}}{2}} \right) \times \cos \left( {\dfrac{{A + B}}{2}} \right)$ .
Therefore, $\sin A + \sin B + \sin C = 2\sin \left( {\dfrac{{A + B}}{2}} \right) \times \cos \left( {\dfrac{{A - B}}{2}} \right) + 2\sin \left( {\dfrac{{A + B}}{2}} \right) \times \cos \left( {\dfrac{{A + B}}{2}} \right)$
We can take the common factor out and solve it now: 2\sin \left( {\dfrac{{A + B}}{2}} \right)\left\\{ {\cos \left( {\dfrac{{A - B}}{2}} \right)\cos \left( {\dfrac{{A + B}}{2}} \right)} \right\\}
= 2\sin \left( {\dfrac{{A + B}}{2}} \right)\left\\{ {2\cos \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{B}{2}} \right)} \right\\}
We can write it as $4\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{\pi }{2} - \dfrac{{B + C}}{2}} \right)\cos \left( {\dfrac{\pi }{2} - \dfrac{{A + B}}{2}} \right)$ .
It gives us the value $4\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{B + C}}{2}} \right)\sin \left( {\dfrac{{C + A}}{2}} \right)$ , because we know that $\sin A = \cos \left( {\dfrac{\pi }{2} - A} \right)$ .
So if $A + B = B + C = A + C$ , then $A = B = C = \dfrac{{A + B + C}}{3}$ , it is only possible when $\sin A + sinB + \sin C$ is maximum.
So we have 4\sin {\left\\{ {\left( {\dfrac{{180}}{3}} \right)} \right\\}^3} = 4{(\sin 60)^3} . We know that the value of $\sin 60 = \dfrac{{\sqrt 3 }}{2}$ , hence the required value is $4 \times {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^3} = \dfrac{{3\sqrt 3 }}{2}$ .
Hence $\sin A + \sin B + \sin C \leqslant \dfrac{{3\sqrt 3 }}{2}$ and also that the triangle is equilateral since each angle comes out to be ${60^ \circ }$ .

Note : We should note that there are trigonometric identities that we have used in the solution above such as $\sin 2x = 2\sin x\cos x$ . This is called the double angle formula. There is another formula used in terms of cosine i.e. $\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A + B}}{2}} \right)$ . Before solving this type of question we should be fully aware of the trigonometric formulas.