Question

求数学大神给算下

设△ABC的三个内角为A,B,C,且tanA,tanB,tanC,2tanB成等差数列,则cos（B-A）=

Answer 1

∵A+B+C=π
∴tanB=-tan(A+C)=-(tanA+tanC)/(1-tanAtanC)
∵tanA,tanB,tanC,2tanB成等差数列
∴tanB=(tanA+tanC)/2
-(tanA+tanC)/(1-tanAtanC)=(tanA+tanC)/2
1-tanAtanC=-2
tanAtanC=3
2tanC=tanB+2tanB=3tanB
tanA=3/tanC=3/(3tanB/2)=2/tanB
tanAtanB=2 ①
(2tanB-tanB)=2(tanB-tanA)
2tanA=tanB ②
将②代入①：2tan^2A=2
tanA=1 A=π/4 tanB=2
tan(B-A)=1/(1+2)=1/3
sec^2(B-A)=1+1/9=10/9
cos^2(B-A)=9/10
cos(B-A)=3√10/10

Answer 2

答案是1

。