Question

已知三角形ABC中，A.B.C的对应边分别为a.b.c，且bcosC＝（2a-c）cosB.若y＝cos^2A+cos^2C.求y的取值范围。

希望只有方法！不需要详细过程！

Answer 1

设a/sinA=b/sinB=c/sinC=k>0,则a=ksinA,b=ksinB,c=ksinC,代入bcosC＝(2a-c)cosB，除去k，
得sinBcosC＝(2sinA-sinC)cosB,
sinBcosC+cosBsinC=2sinAcosB,
sin(B+C)=2sinAcosB,
sin(π-A)=2sinAcosB,
sinA=2sinAcosB,(sinA不等于0）
cosB=1/2>0,
0<B<π/2,
B=π/3,
y＝cos^2A+cos^2C=(1+cos2A)/2+(1+cos2C)/2=1+(cos2A+cos2C)/2
=1+{cos[(A+C)+(A-C)]+cos[(A+C)-(A-C)]}/2
=1+cos(A+C)cos(A-C)
=1+cos(π-B)cos(A+C-2C)
=1-cosBcos(π-B-2C)
=1-[cos(π-π/3-2C)]/2
=1-[cos(2π/3-2C)]/2
因为B=π/3，A+C=2π/3,
0<C<2π/3,
-4π/3<-2C<0,-2π/3<2π/3-2C<2π/3,
-1/2=-cosπ/3=cos2π/3<cos(2π/3-2C)<=cos0=1
-1/4<[cos(2π/3-2C)]/2<=1/2
-1/2=<-[cos(2π/3-2C)]/2<1/4
1/2=<1-[cos(2π/3-2C)]/2<5/4,
即y的取值范围为[1/2,5/4)

Answer 2

角化边或边化角，这题很简单的