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高中三角函数题求解!!!
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解:(1)∵∠DAB=∠DCB=π/2
则∠DAB+∠DCB=π
∴平面四边形ABCD四点共圆,且BD为圆的直径
在BD上取中点O,即点O为圆的圆心,连结OA,OC
∵∠ABC=2π/3
∴∠ADC=π/3
∴∠AOC=2π/3
又OA=OC
∴∠OAC=∠OCA=π/6
∴OA=(AC/2)/cos∠OAC
=(3√3/2)÷cos(π/6)
=3
∴BD=2OA=6
(2)设ΔADC面积为S,∠BDA=α (0<α<π/3)
则S=(1/2)DA*DCsin∠ADC
=(1/2)(BDcosα)*[BDcos(∠ADC-α)]sin∠ADC
=(1/2)×6cosα×6cos(π/3-α)×sin(π/3)
=9√3×cosα×cos(π/3-α)
=(9√3)×(1/2){cos[α+(π/3-α)]+cos[α-(π/3-α)]}
=(9√3/2)×[cos(π/3)+cos(2α-π/3)]
=(9√3/2)×[1/2+cos(2α-π/3)]
当(2α-π/3)=0即α=π/6时,cos(2α-π/3)有最大值1,此时S也取得最大值
∴Smax=(9√3/2)×(1/2+1)=27√3/4
则∠DAB+∠DCB=π
∴平面四边形ABCD四点共圆,且BD为圆的直径
在BD上取中点O,即点O为圆的圆心,连结OA,OC
∵∠ABC=2π/3
∴∠ADC=π/3
∴∠AOC=2π/3
又OA=OC
∴∠OAC=∠OCA=π/6
∴OA=(AC/2)/cos∠OAC
=(3√3/2)÷cos(π/6)
=3
∴BD=2OA=6
(2)设ΔADC面积为S,∠BDA=α (0<α<π/3)
则S=(1/2)DA*DCsin∠ADC
=(1/2)(BDcosα)*[BDcos(∠ADC-α)]sin∠ADC
=(1/2)×6cosα×6cos(π/3-α)×sin(π/3)
=9√3×cosα×cos(π/3-α)
=(9√3)×(1/2){cos[α+(π/3-α)]+cos[α-(π/3-α)]}
=(9√3/2)×[cos(π/3)+cos(2α-π/3)]
=(9√3/2)×[1/2+cos(2α-π/3)]
当(2α-π/3)=0即α=π/6时,cos(2α-π/3)有最大值1,此时S也取得最大值
∴Smax=(9√3/2)×(1/2+1)=27√3/4
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