Question

在△ABC中,证 (b^2-c^2)/a^2*sin2A+(c^2-a^2)/b^2*sin2B+(a^2-b^2)/c

Answer 1

这样由正弦定理有：

(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.