Question

Prove that LHS=RHS Hence prove it that

Answer 1

From sine rule and 3A trig identity a³ = 8R³sin³(A) = 2R³(3sin(A)−sin(3A))

So X = 2R³ ∑( 3sin(A)−sin(3A) )cos(B−C) = R³Y where ∑ is over cycles

2sin(A)cos(B−C) = sin(A+B−C) + sin(A−B+C) = sin(2C) + sin(2B)

2sin(3A)cos(B−C) = sin(3A+B−C) + sin(3A−B+C)

with similar expressions for B,C,A & C,A,B

∴ Y = 3∑sin(2C) + 3∑sin(2B) − ∑sin(3A+B−C) − ∑sin(3A−B+C)

∑sin(3A+B−C) + ∑sin(3A−B+C) = sin(3A+B−C) + sin(3B+C−A) + sin(3C+A−B) +
sin(3A−B+C) + sin(3B−C+A) + sin(3C−A+B) … (i)

But (3A+B−C) + (3C−A+B) = 2(A+B+C) = 2π so sin(3A+B−C) + sin(3C−A+B) = 0 with similar for other pairs of terms in (i). So (i) is zero.

∴ Y = 6∑sin(2A) = 24sin(A)sin(B)sin(C) (an easily proved identity for a triangle)

Y = 24abc/(8R³) = 3abc/R³ → X = 3abc