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.