Question

在三角形ABC中，a,b,c分别为角A，B，C的对边，且acosC,bcosB,ccosA成等差数列，求2sin²A+cos（A-C)的

急

Answer 1

2bcosB=a cosC+c cosA
代入正弦定理。 得 2sinBcosB=sinAcosC+sinCcosA
即 sin2B=sin(A+C) 2B=A+C=120°

Answer 2

(2)A-C=A-(2π/3-A)=2A-2π/3
故2sin²A+cos(A-C)=2sin²A+cos(2A-2π/3)=1-cos2A+cos(2A-2π/3)
=1+2sin(2A-π/3)sin(2π/3)
=1+√3sin(2A-π/3)
又因为A∈(0,2π/3),2A-π/3∈(-π/3,π)
2 sin²A+cos(A-C)∈(-0.5,1+√3]