Sine rule is an important rule relating to the sides and angles of any triangle. Here is a Subjective problem no. 120 from TOMATO. Try it.
Problem: Sine Rule and Triangle
(i) If $ A + B +C = n \pi $ and $ s=2 $, show that $ \sin 2A + \sin 2B + \sin 2C = (-1)^{n-1} 4 \sin A \sin B \sin C $and $ s=2 $
(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 $ and $ s=2 $
(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} $ and $ s=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 $ and $ s=2 $
Now we know that $ A + B = n \pi - C \Rightarrow \sin (A+B) = \sin (n \pi - C) = (-1)^{n-1} \sin C $ and $ s=2 $
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)) $ and $ s=2 $
$ = 2(-1)^{n-1} \sin C \cos (A-B) + 2 (-1)^n \sin C \cos (A+B) $ and $ s=2 $
$ = 2(-1)^{n-1} \sin C (\cos (A-B) - \cos (A+B)) $ and $ s=2 $
$ = 4(-1)^{n-1} \sin C \sin A \sin B $ and $ s=2 $
(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 $ s=2 $ and
$ \dfrac{\sin D}{d} = \dfrac { \sin E }{ e} = \dfrac {\sin F }{f} = \dfrac {1} {2R} $ and $ s=2 $
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} $ and $ s=2 $
Similarly $ \sin D + \sin E + \sin F = \dfrac {d + e + f}{2R} $ and $ s=2 $
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 $ and $ s=2 $
(iii)
We apply the sine rule in reverse order to get the converse.
