Proof that in a triangle ABC a3sin(B-C)+b3sin(C-A)+c3sin(A-B)=0

Answer 1

Solution is:

for first part of LHS:

a³sin(B-C)

=(2RsinA)³.sin(B−C)

=2R³⋅2sin²A⋅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 for 2nd part

ie.

b³sin(C-A) = 2R³(cos2A−cos2C−cos2A⋅cos2B+cos2B⋅cos2C)

and 3rd part is

c³sin(A-B) = 2R³(cos2B−cos2A−cos2B⋅cos2C+cos2A⋅cos2C)

Now putting these value in the equation and adding and subtracting we get

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

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

=> 0 proved