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共回答了22个问题采纳率:77.3% 举报
(Ⅰ)由a、b、c成等比数列,得b2=ac,sin2B=sinAsinC----(2分)
又cosB=[1/3],得sinB=
2
2
3(0<B<π)-----------------------------(3分)
[1/tanA]+[1/tanC]=[cosA/sinA]+[cosC/sinC]---------------------------------------(4分)
=
sin(A+C)
sinAsinC-------------------------------------(5分)
=[1/sinB]=
3
2
4------------------------------------------(6分)
(Ⅱ)cosB=
a2+c2−b2
2ac≥[ac/2ac]=[1/2],-------------------------(7分)
∴B∈(0,[π/3]],∴sinB≤
3
2--------------------------------(8分)
又6=a+b+c=
ac+a+c≥3
1年前
1