Question

一道三角函数的题，求详细的解题过程，谢谢

Answer 1

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°

追问

请问这个是怎么得到的？