Question

一道简单的三角比证明题求证：4sin[(b-c)/2]sin[(a-b)/2]sin[(c-a)/2]=sin(b-c)

Answer 1

令(b-c)/2=x (a-b)/2=y

则x+y=(a-c)/2 (c-a)/2=-x-y

4sin[(b-c)/2]sin[(a-b)/2]sin[(c-a)/2]=sin(b-c)+sin(a-b)+sin(c-a)

即4sinxsinysin(-x-y)=sin2x+sin2y+sin(-2x-2y)

4sinxsinysin(x+y)=sin(2x+2y)-sin2x-sin2y

sin(2x+2y)-sin2x-sin2y

=sin2xcos2y+cos2xsin2y-sin2x-sin2y

=sin2x(cos2y-1)+sin2y(cos2x-1)

=-2sin2x(siny)^2-2sin2y(sinx)^2

4sinxsinysin(x+y)

=4sinxsiny(sinxcosy+cosxsiny)

=4(sinx)^2sinycosy+4(siny)^2sinxcosx

=2(sinx)^2sin2y+2(siny)^2sin2x

sin(2x+2y)-sin2x-sin2y=-4sinxsinysin(x+y)