设三角形ABC的内角A,B,C的对边分别为a,b,c已知b2+c2=a2+根号3bc求2sinBcosC-sin(B-C)的值

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2024-12-27 18:33:49
推荐回答(2个)
回答1:

由已知,余弦定理有:cosA=(b^2+c^2-a^2)/2bc=根号3bc/2bc=根号3/2
A=30度,
2sinBcosC-sin(B-C)
=2sinBcosC-sinBcosC+cosBsinC
=sinBcosC+cosBsinC
=sin(B+C)
=sin(180-A)
=sinA=1/2

回答2:

1/2