设A1A2A3.A7是圆内接正七边形,求证:1/(A1A2)等于1/(A1A3)+1/(A1A4)
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设A1A2A3.A7是圆内接正七边形,求证:1/(A1A2)等于1/(A1A3)+1/(A1A4)
设A1A2A3.A7是圆内接正七边形,求证:1/(A1A2)等于1/(A1A3)+1/(A1A4)
设A1A2A3.A7是圆内接正七边形,求证:1/(A1A2)等于1/(A1A3)+1/(A1A4)
等同于证AC*AD=AB*(AC+AD)
设圆的半径为1
所以
AB=2sin(π/7)
AC=2sin(2π/7)
AD=2sin(3π/7)
AC*AD
=4sin(2π/7)sin(3π/7)
=(-2)(cos(5π/7)-cos(π/7))
AB*(AC+AD)
=4sin(π/7)*(sin(2π/7)+sin(3π/7))
=(-2)(cos(3π/7)-cos(π/7)+cos(4π/7)-cos(2π/7))
=(-2)(cos(π-(4π/7))-cos(π/7)+cos(4π/7)-cos(π-(5π/7)))
=(-2)(cos(5π/7)-cos(π/7))
=AC*AD
故命题得证
A1A2=A3A4,转化到A1A3A4三角形里,通过计算三角之比为1:2:4,
则:A=π/7,B=2π/7,C=4π/7.
由正弦定理,a=2RsinA,b=2RsinB,C=2RsinC,
所以,1/b+1/c=1/2R(1/sinB+1/sinC).
而,1/sinB+1/sinC=(sinB+sinC)/sinBsinC
=2sin(3π/7)cos...
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A1A2=A3A4,转化到A1A3A4三角形里,通过计算三角之比为1:2:4,
则:A=π/7,B=2π/7,C=4π/7.
由正弦定理,a=2RsinA,b=2RsinB,C=2RsinC,
所以,1/b+1/c=1/2R(1/sinB+1/sinC).
而,1/sinB+1/sinC=(sinB+sinC)/sinBsinC
=2sin(3π/7)cos(π/7)/sin(2π/7)sin(4π/7)
=2sin(4π/7)cos(π/7)/sin(2π/7)sin(4π/7)
=2cos(π/7)/sin(2π/7)
=2cos(π/7)/2sin(π/7) cos(π/7)
=1/sin(π/7)
所以,1/b+1/c=1/2R*(1/sinπ/7).=1/a.
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