锐角三角函数问题在锐角三角形AB
多了一条分数线,改正了!
问题 在锐角三角形ABC中,求证
1/[sin(2B)*sin(2C)]+1/[sin(2C)*sin(2A)]+1/[sin(2A)*sin(2B)]≥1/(sinA)^2+1/(sinB)^2+1/(sinC)^2
证明 ∵A,B,C为锐角,∴sin(2A)>0,sin(2B)>0,sin(2C)>0。
而 sin(2B)*sin(2C)=4*sinB*cosB*sinC*cosC
=4*(sinB*cosC)*(sinC*cosB)≤[sinB*cosC+sinC*cosB]^2
=[sin(B+C)]^2=(sinA)^2,
所以有
1/[sin(2B...全部
多了一条分数线,改正了!
问题 在锐角三角形ABC中,求证
1/[sin(2B)*sin(2C)]+1/[sin(2C)*sin(2A)]+1/[sin(2A)*sin(2B)]≥1/(sinA)^2+1/(sinB)^2+1/(sinC)^2
证明 ∵A,B,C为锐角,∴sin(2A)>0,sin(2B)>0,sin(2C)>0。
而 sin(2B)*sin(2C)=4*sinB*cosB*sinC*cosC
=4*(sinB*cosC)*(sinC*cosB)≤[sinB*cosC+sinC*cosB]^2
=[sin(B+C)]^2=(sinA)^2,
所以有
1/[sin(2B)*sin(2C)]≥1/(sinA)^2 (1)
同理可得:
1/[sin(2C)*sin(2A)]≥1/(sinB)^2 (2)
1/[sin(2A)*sin(2B)]≥1/(sinC)^2 (3)
(1)+(2)+(3)即得所证不等式。
根据已知代数不等式:
(x+y+z)^2≥3*(yz+zx+xy),及 3*(x^2+y^2+z^2)≥(x+y+z)^2
可得:
[1/sin(2A)+1/sin(2B)+1/sin(2C)]^2≥
3{1/[sin(2B)sin(2C)]+1/[sin(2C)sin(2A)]+1/[sin(2A)sin(2B)}≥
3*[1/(sinA)^2+1/(sinB)^2+1/(sinC)^2]≥
[1/sinA+1/sinB+1/sinC]^2,
故得:
1/sin(2A)+1/sin(2B)+1/sin(2C)≥1/sinA+1/sinB+1/sinC
。
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