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

Question

4758 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-21T07:02:37

See below.

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#