Question

设三角形ABC的内角A,B,C的对边分别为a,b,c,已知b^2+c^2=a^2+√3bc,求2sinBcosC-sin

Answer 1

^2+c^2=a^2+√3bc

==>cosA=cos(b,c)=(b^2+c^2-a^2)/(2bc)=根号3/2

==>sinA=sin(b,c)=1/2

2sinBcosC-sin(B-C)=2sinBcosC-sinBcosC+cosBsinC

=sinBcosC+cosBsinC

=sin(B+C)

=sinA

=1/2