三角形ABC中,向量m=(sinB+sinc,0),向量 n=(0,sinA)且(m+n)(m-n)=sinBsinC(1)求A; (2)求sinB+sinC的取值范围;
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三角形ABC中,向量m=(sinB+sinc,0),向量 n=(0,sinA)且(m+n)(m-n)=sinBsinC(1)求A; (2)求sinB+sinC的取值范围;
三角形ABC中,向量m=(sinB+sinc,0),向量 n=(0,sinA)且(m+n)(m-n)=sinBsinC
(1)求A; (2)求sinB+sinC的取值范围;
三角形ABC中,向量m=(sinB+sinc,0),向量 n=(0,sinA)且(m+n)(m-n)=sinBsinC(1)求A; (2)求sinB+sinC的取值范围;
(1)∵(m+n)(m-n)=sinBsinC
∴ m²-n²=sinBsinC
即 (sinB+sinC)²-sin²A=sinBsinC
∴ sin²B+sin²C-sin²A=-sinBsinC
由 正弦定理 a/sinA=b/sinB=c/sinC 和
余弦定理 cosA=(b²+c²-a²)/2bc 得
cosA=(sin²B+sin²C-sin²A)/2sinBsinC
∴ cosA=-sinBsinC/2sinBsinC=-1/2
又 角A为△ABC的内角
∴ A=2π/3
(2)由(1),可知
B+C=π/3
则 B=π/3-C
∴ sinB+sinC=sin(π/3-C)+sinC
=sinπ/3cosC-cosπ/3sinC+sinC
=sinπ/3cosC+cosπ/3sinC
=sin(C+π/3)
又 C∈(0,π/3)
∴ (C+π/3)∈(π/3,2π/3)
∴ sin(C+π/3)∈(√3/2,1]
∴ sinB+sinC∈(√3/2,1]
因此 sinB+sinC的取值范围为(√3/2,1]
1)由题意,m+n=(sinB+sinC,sinA),m-n=(sinB+sinC,-sinA),由(m+n)(m-n)=sinBsinC可知,(sinB+sinC)^2-(^2sinA)=sinBsinC
三角形中,A+B+C=π,且A、B、C都是大于0,小于π的角。sinA=sin(π-A)=sin(B+C)
(sinB+sinC)^2-(^2sinA)=(sinB)^2+2s...
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1)由题意,m+n=(sinB+sinC,sinA),m-n=(sinB+sinC,-sinA),由(m+n)(m-n)=sinBsinC可知,(sinB+sinC)^2-(^2sinA)=sinBsinC
三角形中,A+B+C=π,且A、B、C都是大于0,小于π的角。sinA=sin(π-A)=sin(B+C)
(sinB+sinC)^2-(^2sinA)=(sinB)^2+2sinBsinC+(sinC)^2-(sinBcosC+cosBsinC)^2=(sinB)^2+(sinC)^2+2sinBsinC-((sinBcosC)^2+2sinBcosBsinCcosC+(cosBsinC)^2)=(sinB)^2(1-(cosC)^2)+(sinC)^2(1-(cosB)^2)+2sinBsinC-2sinBsinCcosBcosC=2(sinB)^2(sinC)^2+2sinBsinC-2sinBsinCcosBcosC=sinBsinC
即2sinBsinC+2-2cosBcosC=1,即sinBsinC-cosBcosC=-.5,cos(B+C)=.5,即B+C=π/3,A=π-B-C
=2π/3(由ABC是三角形三角)
2)由1)sinB+sinC=sinB+sin(π/3-B)=sinB+sin(π/3)cosB-cos(π/3)sinB=.5sinB+sin(π/3)cosB=cos(π/3)sinB+sin(π/3)cosB=sin(B+π/3)
又B+C=π/3,0
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