Question

高三数学第16题

Answer 1

【答案】16

【解析】
2sinCcosA=sinA(sinC-2cosC)
∴2(sinCcosA+sinAcosC)=sinAsinC
∴2/tanA+2/tanC=1
【两边同时除以sinAsinC可得】

∴1/2=1/tanA+1/tanC
≥2·√(1/tanA·1/tanC)
=2/√(tanA·tanC)
∴√(tanA·tanC)≥4
∴tanA·tanC≥16
【tanA=tanC=4时取得】

Answer 2

如图

Answer 3

2c.cosA=a(sinC -2cosC)
cosA = [a/(2c)] (sinC -2cosC)
cosA = [sinA/(2.sinC)] (sinC -2cosC)
tanA = 2sinC/(sinC -2cosC)
= 2tanC/(tanC -2 )
S = tanA. tanC
=2(tanC)^2 /(tanC -2 )
= 2(tanC+2) +8/(tanC -2 )
dS/dtanC = 2 - 8/(tanC -2 )^2
dS/dtanC = 0
(tanC-2)^2 -4 =0
tanC .(tanC -4)=0
tanC =0(rej) or 4
dS/dtanC |tanC=4+ >0
dS/dtanC |tanC=4- <0
tanC =4 (min)
S = tanA. tanC
=2(tanC)^2 /(tanC -2 )
min S = S(tanC=4) =2(4)^2 /(4 -2 ) = 32/2 =16