已知向量m=(sinB,-cosB),且与向量n=(2,0)所成角为π/3,其中A,B,C是三角形ABC的内角.⑴求角B的大小
2个回答
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(1) 因为 cosπ/3=m*n/(|m|*|n|),
且|m|=√[(sinB)^2+(1-cosB)^2]=√[2(1-cosB)]=2sin(B/2),
|n|=2, m*n=2sinB,
所以 2sinB/[4sin(B/2)]=1/2,
因此 2*2sin(B/2)cos(B/2)/[4sin(B/2)]=1/2,
化简得 cos(B/2)=1/2,
B/2=π/3,
B=2π/3.
(2) sinA+sinC=sin(B+C)+sinC
=sinB×cosC+cosB×sinC+sinC
=√3/2cosC-1/2sinC+sinC
=√3/2cosC+1/2sinC
=sin(C+π/3)
∵C∈(0,π/3)
∴C+π/3∈(π/3,2π/3)
∴sin(C+π/3)∈(√3/2,1].
即sinA+sinC∈(√3/2,1].
且|m|=√[(sinB)^2+(1-cosB)^2]=√[2(1-cosB)]=2sin(B/2),
|n|=2, m*n=2sinB,
所以 2sinB/[4sin(B/2)]=1/2,
因此 2*2sin(B/2)cos(B/2)/[4sin(B/2)]=1/2,
化简得 cos(B/2)=1/2,
B/2=π/3,
B=2π/3.
(2) sinA+sinC=sin(B+C)+sinC
=sinB×cosC+cosB×sinC+sinC
=√3/2cosC-1/2sinC+sinC
=√3/2cosC+1/2sinC
=sin(C+π/3)
∵C∈(0,π/3)
∴C+π/3∈(π/3,2π/3)
∴sin(C+π/3)∈(√3/2,1].
即sinA+sinC∈(√3/2,1].
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