Question

In any ∆ABC, Prove that - T-114

a^3sin(B-C)+b^3sin(C-A)+c^3sin

Answer 1

$\Large{\textbf{\underline{\underline{According\;to\;the\;Questions}}}}$

Correct Question

a³sin(B - C) + b³sin(C - A) + c³sin(A - B) = 0

Here,

Assume

$\rm \: \: \: \: \: \: \: \: \:=\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}=Z$

a = ZsinA

b = ZsinB

c = ZsinC

Therefore,

⇒ a³sin(B - C) = Z³sin³A × sin(B - C)

= Z³sin²A × sinA × sin(B - C)

= Z³sin²A[sin{π - (B + C)}sin(B - C)]

Therefore,

⇒ A + (B + C) = π

= Z³sin²A[sin(B + C)sin(B - C)]

= Z³sin²A × (sin²B - sin²C)

Similarly,

b³sin(C - A) = Z³sin²B(sin²C - sin²A)

And,

c³sin(A - B) = Z³sin²C(sin²A - sin²B)

Therefore,

a³sin(B - C) + b³sin(C - A) + c³sin(A - B)

= Z³sin²A(sin²B - sin²C) + Z³sin²B(sin²C - sin²A) + Z³sin²C(sin²A - sin²B)

= 0