cos(B+C)=cosBcosC-sinBsinC; sin(B-C)=sinBcosC-cosBsinC (trig identities).
So cos(B+C).sin(B-C)=sinBcosBcos^2C-cos^2BsinCcosC-sin^2BsinCcosC+sinBcosBsin^2C=
sinBcosB(cos^2C+sin^2C)-sinCcosC(cos^2B+sin^2B)=sinBcosB-sinCcosC=
(1/2)sin2B-(1/2)sin2C or -(1/2)(sin2C-sin2B). Therefore -2cos(B+C).sin(B-C)=sin2C-sin2B.