Question

求解一道数学题目 拜托大家了 谢谢。尽可能的多用几种方法，要过程。

三角形ABC，边a,b,c成等比数列，则cos(A-C)+cos2B+cosB=???

Answer 1

因为三边a、b、c成等比数列
所以b^2=ac即sin^2B=sinA*sinC
cos(A-C)-cos(A+C)=cos(A-C)-cos(180-B)=cos(A-C)+cos(B)

Cos(A-C)+Cos2B+CosB
=cosA*cosC+sinA*sinC+1-2(sinB)^2-cosA*cosC+sinA*sinC
=2sinA*sinC+1-2sinA*sinC
=1

Answer 2

因为三边a、b、c成等比数列
所以b^2=ac即sin^2B=sinA*sinC
cos(A-C)-cos(A+C)=cos(A-C)-cos(180-B)=cos(A-C)+cos(B)

Cos(A-C)+Cos2B+CosB
=cosA*cosC+sinA*sinC+1-2(sinB)^2-cosA*cosC+sinA*sinC
=2sinA*sinC+1-2sinA*sinC
=1
这样就可以了