((b*b-c*c)/a*a)sin(2A)+((c*c-a*a)/b*b)sin(2B)+((a*a-b*b)/c*c)sin(2C)=

Question

((b*b-c*c)/a*a)sin(2A)+((c*c-a*a)/b*b)sin(2B)+((a*a-b*b)/c*c)sin(2C)=

Answer 1

[(b^2-c^2)/a^2]sin2A+[c^2-a^2)/b^2]sin2B+[(a^2-b^2)/c^2]sin2C
由正弦定理可得：
(b^2-c^2)/a^2=(sin^2B-sin^2c)/sin^2A
=(sinB+sinC)(sinB-sinC)/sinA*sinA
=[4sin(B+C)/2*sin(B-C)/2*sin(B-c)/2*cos(B-C)/2]/sin^2A
=sin(B+C)*sin(B-C)/sin^2A
=sin(B-C)/sinA,
于是
[(b^2-c^2)/a^2]*sin2A
=2sinAcosA*sin(B-C)/sinA
=sin(B-C)cosA
=-sin(B-C)cos(B+C)
=sin2C-sin2B；
同理可得
[(c^2-a^2)/b^2]sin2B=sin2A-sin2C；
[(a^2-b^2)/c^2]sin2C=sin2B-sin2A.
于是
[(b^2-c^2)/a^2 ]*sin2A+[(c^2-a^2)/b^2]sin2B+[(a^2-b^2)/c^2]sin2C=0