Question

In Δ ABC, prove that a³ sin(B – C) + b³sin(C – A) + c³sin(A – B) = 0

Answer 1

By the sine rule,
`a/"sinA" = b/"sinB" = c/"sinC"` = k
∴ a = k sinA, b = k sinB, c = k sinC
L.H.S. = a³sin(B − C) + b³sin(C − A) + c³sin(A − B)
= a³(sinB cosC − cosB sinC) + b³(sinC cosA − cosC sinA) + c³(sinA cosB − cosA sinB)

`= a^3(b/kcos"C" − c/k cos"B") + b^3(c/kcos"A" − a/kcos"C") + c^3(a/kcos"B" − b/kcos"A")`

`= (1)/k[a^3bcos"C" − a^3"c"cos"B" + b^3"c"cos"A" − b^3"a"cos"C" + c^3"a"cos"B" - c^3"b"cos"A"]`

`= (1)/k[a^3b((a^2 + b^2 - c^2)/(2ab)) - a^3"c"((c^2 + a^2 - b^2)/(2ca)) + b^3"c"((b^2 + c^2 - a^2)/(2bc)) - ab^3((a^2 + b^2 - c^2)/(2ab)) + ac^3((c^2 + a^2 - b^2)/(2ca)) - bc^3((b^2 + c^2 - a^2)/(2bc))]`  ...[By cosine rule]

`= (1)/(2k)[a^2(a^2 + b^2 - c^2) - a^2(a^2 + c^2 - b^2) + b^2(b^2 + c^2 - a^2) - b^2(a^2 + b^2 - c^2) + c^2(c^2 + a^2 - b^2) - c^2(b^2 + c^2 - a^2)]`

`= (1)/(2k)[a^4 + a^2b^2 - a^2c^2 - a^4 - a^2c^2 + a^2b^2 + b^4 + b^2c^2 - a^2b^2 - a^2b^2 - b^4 + b^2c^2 + c^4 + a^2c^2 - b^2c^2 - b^2c^2 - c^4 + a^2c^2]`

`= (1)/(2k)(0)`
= 0
= R.H.S.