Question

在三角形ABC中，三内角A,B,C所对的边分别为a,b,c,且（2b-c)cosA=acosC.

(1)求角A的大小，（2）求cos平方B+cos平方C的取值范围

Answer 1

(1)2bcosA=ccosA+acosC=b
所以cosA=1/2
A=π/3

(2)B+C=π-π/3=2π/3
所以0<B<2π/3
cos²B+cos²C
=cos²B+cos²(2π/3-B)
=cos²B+(-1/2cosB+√3/2sinB)²
=5/4cos²B+3/4sin²-√3/2sinBcosB
=3/4+1/2cos²B-√3/2sinBcosB
=1+1/4cos2B-√3/4sin2B
=1+1/2(1/2cos2B-√3/2sin2B)
=1+1/2cos(2B+π/3)
0<B<2π/3
π/3<2B+π/3<5π/3
-1≤cos(2B+π/3)<1/2
∴1/2≤cos²B+cos²C<5/4