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1),
2cos(π/6-B)cos(π/6+B)
=2(1/2cosB+√3/2sinB)(1/2cosB-√3/2sinB)
=2(1/4cos²B-3/4sin²B)
=2[1/4▪(1+cos2B)/2-3/4▪(1-co2B)/2]
=cos2B-1/2
∵cos2A-cos2B+2cos(π/6-B)cos(π/6+B)=0
∴co2A=1/2
∵0<A<π,∴0<2A<2π
∴2A=π/3或2A=5π/3
∴A=π/6或A=5π/6
2),
∵b≤a
∴0<B≤π/6,或0<B≤5π/6
∵a/sinA=b/sinB
∴a=√3/2sinB
∵0<B≤π/6,或0<B≤5π/6
当0<B≤π/6时,√3/2sinB≥√3,∴a≥√3
当0<B≤5π/6时,√3/2sinB≥√3/2,∴a≥√3/2
2cos(π/6-B)cos(π/6+B)
=2(1/2cosB+√3/2sinB)(1/2cosB-√3/2sinB)
=2(1/4cos²B-3/4sin²B)
=2[1/4▪(1+cos2B)/2-3/4▪(1-co2B)/2]
=cos2B-1/2
∵cos2A-cos2B+2cos(π/6-B)cos(π/6+B)=0
∴co2A=1/2
∵0<A<π,∴0<2A<2π
∴2A=π/3或2A=5π/3
∴A=π/6或A=5π/6
2),
∵b≤a
∴0<B≤π/6,或0<B≤5π/6
∵a/sinA=b/sinB
∴a=√3/2sinB
∵0<B≤π/6,或0<B≤5π/6
当0<B≤π/6时,√3/2sinB≥√3,∴a≥√3
当0<B≤5π/6时,√3/2sinB≥√3/2,∴a≥√3/2
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