In DeltaAB prove that, a^3sin(B - C) + b^3sin(C - A) + c^3sin(A - B) = 0 ??

1 Answer
Jan 21, 2018

See below.

Explanation:

The first part of LHS=a^3sin(B-C)

=(2RsinA)^3sin(B-C)

=2R^3*2sin^2A*2sinA*sin(B-C)

=2R^3*(1-cos2A)*[cos(A-B+C)-cos(A+B-C)]

=2R^3*(1-cos2A)*[cos(pi-2B)-cos(pi-2C)]

=2R^3*(1-cos2A)*(cos2C-cos2B)

=2R^3*(cos2C-cos2B-cos2A*cos2C+cos2A*cos2B)

Similarly, the second part=2R^3(cos2A-cos2C-cos2A*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^3sin(C-A)+c^3sin(A-B)=0