**Problem: **

(i) If \(A + B +C = n \pi \), show that \(\sin 2A + \sin 2B + \sin 2C = (-1)^{n-1} 4 \sin A \sin B \sin C \)

(ii) Let triangles ABC and DEF be inscribed in the same circle. If the triangles are of equal perimeter, then prove that \(\sin A + \sin B + \sin C = \sin D + \sin E + \sin F \)

(iii) State and prove the converse of (ii) above

**Discussion:**

(i) We know transformation formula from trigonometry \(\sin x + \sin y = 2 \sin \dfrac{x+y}{2} \cos \dfrac {x-y}{2} \)

Hence \(\sin 2A + \sin 2B + \sin 2C = 2 \sin \dfrac{2A+2B}{2} \cos \dfrac {2A-2B}{2} + sin 2C = 2 \sin (A+B) \cos (A-B) + \sin 2C \)

Now we know that \(A + B = n \pi – C \Rightarrow \sin (A+B) = \sin (n \pi – C) = (-1)^{n-1} \sin C \)

So \(2\sin (A+B) \cos (A-B) + \sin 2C = 2(-1)^{n-1} \sin C \cos (A-B) + 2 \sin C \cos C = 2(-1)^{n-1} \sin C \cos (A-B) + 2 \sin C \cos (n\pi -(A+B)) \)

\(= 2(-1)^{n-1} \sin C \cos (A-B) + 2 (-1)^n \sin C \cos (A+B) \)

\(= 2(-1)^{n-1} \sin C (\cos (A-B) – \cos (A+B)) \)

\(= 4(-1)^{n-1} \sin C \sin A \sin B \)

(ii)

Since the two triangles are inscribed in the same circle, they must have the same circumradius. Let the common circumradius be R. If a, b, c, d, e, f be the sides opposite to the sides BC, CA, AB, EF, DF, DE respectively, then using the rule of sines we can say,

\(\dfrac{\sin A}{a} = \dfrac { \sin B }{ b} = \dfrac {\sin C }{c} = \dfrac {1} {2R} \) and

\(\dfrac{\sin D}{d} = \dfrac { \sin E }{ e} = \dfrac {\sin F }{f} = \dfrac {1} {2R} \)

Hence \(\sin A = \dfrac {a}{2R}, sin B = \dfrac{b}{2R}, \sin C = \dfrac {c}{2R} \Rightarrow \sin A + \sin B + \sin C = \dfrac {a+b+c}{2R} \)

Similarly \(\sin D + \sin E + \sin F = \dfrac {d + e + f}{2R} \)

As the perimeter of the triangle are equal, hence a+b+c = d+e+f. This implies \(\sin A + \sin B + \sin C = \sin D + \sin E + \sin F \)

(iii)

We apply the sine rule in reverse order to get the converse.

## Chatuspathi:

**What is this topic:**Property of triangles**What are some of the associated concept:**Rule of sines**Where can learn these topics:**Cheenta**Book Suggestions:**Trigonometry by S.L. Loney