三角形三边成等差数列且最大角与最
设最小角为A=α,最大角为C=90+α ,
另一角B=90-2α
sinB=sin(90-2α)=cos2α
sinC=sin(90+α)=cosα
a=2RsinA=2Rsinα
b=2RsinB=2Rcos2α
c=2RsinC=2Rcosα
三角形三边成等差数列:a+c=2b→
2RsinA+2RsinC=2*2RsinB→
2Rsinα+2Rcosα=2*2Rcos2α→
sinα+cosα=2cos2α→
sinα+cosα=2[(cosα)^2-(sinα)^2]→
sinα+cosα=2(cosα+sinα)(cosα-sinα)
最小角为a0
1=2(cosα-sinα)→...全部
设最小角为A=α,最大角为C=90+α ,
另一角B=90-2α
sinB=sin(90-2α)=cos2α
sinC=sin(90+α)=cosα
a=2RsinA=2Rsinα
b=2RsinB=2Rcos2α
c=2RsinC=2Rcosα
三角形三边成等差数列:a+c=2b→
2RsinA+2RsinC=2*2RsinB→
2Rsinα+2Rcosα=2*2Rcos2α→
sinα+cosα=2cos2α→
sinα+cosα=2[(cosα)^2-(sinα)^2]→
sinα+cosα=2(cosα+sinα)(cosα-sinα)
最小角为a0
1=2(cosα-sinα)→
cosα-sinα=1/2→
(cosα-sinα)^2=1/4→
1-2sinαcosα=1/4→
-2sinαcosα=-3/4→
2sinαcosα=3/4→
sin2α=3/4→
cos2α=±√[1-(3/4)^2]=±√7/4
又sinB=cos2α>0,
∴cos2α=√7/4
2(cosα)^2-1=√7/4
2(cosα)^2=(4+√7)/4
(cosα)^2=(4+√7)/8
(cosα)^2=(8+2√7)/16=[(√7+1)/4]^2
sinC=cosα>0
∴cosα=(√7+1)/4
sinα=√[1-(cosα)^2]=√[1-(4+√7)/8]
=√(4-√7)/8]=√(8-2√7)/16=[(√7-1)/4]^2
∴sinα=(√7-1)/4
∴三边比a:b:c=sinA:sinB:sinC=sinα:cos2α:cosα=
=(√7-1)/4:√7/4:=(√7+1)/4
=(√7-1):√7:=(√7+1)
。
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