Question

In ∆ ABC, prove that a^3 sin (B - C) + b^3 sin (C - A) + c^3 sin (A - B) = 0.

If you know then only answer plz..​

Answer 1

AⲛSⲱⲉꞅ⤵️

↬LHS=a³ sin(B−C)

↬(2RsinA)³ sin(B−C)

↬2R³⋅2sin² A⋅2sinA⋅sin(B−C)

2R³(1−cos2A)⋅[cos(A−B+C)−cos(A+B−C)]

=2R³ ⋅(1−cos2A)⋅[cos(π−2B)−cos(π−2C)]

=2R³⋅(1−cos2A)⋅(cos2C−cos2B)

=2R³⋅(cos2C−cos2B−cos2A⋅cos2C+cos2A⋅cos2B)

↬Similarly, the second part

=2R³(cos2A−cos2C−cos2⋅cos2B+cos2B⋅cos2C)

↬And the third part

=2R ³(cos2B−cos2A−cos2B⋅cos2C+cos2A⋅cos2C

↬Adding up all these three parts, we get,

↬a ³sin(B−C)+b ³ sin(C−A)+

↬c ³ sin(A−B)=0

Answer 2

Explanation:

In ∆ ABC, prove that a^3 sin (B - C) + b^3 sin (C - A) + c^3 sin (A - B) = 0.

ok