Show that (a^2sin(B-C))/(sinB+sinC)+(b^2sin(C-A))/(sinC+sinA)+(c^2sin(A-B))/(sinA+sinB)=0?

1 Answer
Jan 16, 2018

1st part

(a^2sin(B-C))/(sinB+sinC)

=(4R^2sinAsin(B-C))/(sinB+sinC)

=(4R^2sin(pi-(B+C))sin(B-C))/(sinB+sinC)

=(4R^2sin(B+C)sin(B-C))/(sinB+sinC)

=(4R^2(sin^2B-sin^2C))/(sinB+sinC)

=4R^2(sinB-sinC)

Similarly

2nd part

=(b^2sin(C-A))/(sinC+sinA)

=4R^2(sinC-sinA)

3rd part

=(c^2sin(A-B))/(sinA+sinB)

=4R^2(sinA-sinB)

Adding three parts we have

The given expression =0