在三角形ABC中,cosB是sinA,sinC的等比中项,sinB是cosA,cosC的等差中项,则角B=?

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在三角形ABC中,cosB是sinA,sinC的等比中项,sinB是cosA,cosC的等差中项,则角B=?
在三角形ABC中,cosB是sinA,sinC的等比中项,sinB是cosA,cosC的等差中项,则角B=?

在三角形ABC中,cosB是sinA,sinC的等比中项,sinB是cosA,cosC的等差中项,则角B=?
由sinB为cosA,cosC的等差中项得
2sinB=cosA+cosC
→4·sinB/2·cosB/2=2·cos(A+C)/2·cos(A-C)/2
→4·sinB/2·cosB/2=2·cos(π-B)/2·cos(A-C)/2
→4·sinB/2·cosB/2=2·sinB/2·cos(A-C)/2
→2cosB/2=cos(A-C)/2
→8(cosB/2)^2-1=2[cos(A-C)/2]^2-1
→cos(A-C)=4cosB+3…………………①
又由cosB为sinA,sinC的等比中项得
(cosB)^2=sinA·sinB
→(cosB)^2=-1/2×[cos(A+C)-cos(A-C)]
→(cosB)^2=-1/2×[cos(π-B)-cos(A-C)]
→(cosB)^2=1/2×[cosB+cos(A-C)]……………②
把①式代入②式得
(cosB)^2=1/2×[cosB+4cosB+3]
→2×(cosB)^2-5cosB-3=0
→cosB=-1/2 (舍弃cosB=3)
所以,B=120° 很高兴为你解答,