Question

已知三角形ABC中三个内角ABC满足A+C=2B,求cos²A+cos²C的取值范围

Answer 1

三角形内角和180°，即：A+B+C=180°
又因为：A+C=2B
所以解得：B=60°，A+C=120°，C=120°-A
cos2A=2cos²A-1，cos2C=2cos²C-1
所以：cos²A=（cos2A+1）/2
cos²C=（cos2C+1）/2
所以：cos²A+cos²C=（cos2A+1）/2+（cos2C+1）/2
=(cos2A+cos2C)/2+1
={2cos[(2A+2C)/2]cos[(2A-2C)/2]}/2+1
=cos(A+C)cos(A-C)+1
=1-[cos(A-C)]/2
=1-[cos(A-120°+A)]/2
=1-cos(2A-120°)/2
0<A<120 0<2A<240 -120<2A-120<120
-1/2<cos(2A-120)<=1 -1/2=< -[cos(A-120°+A)]/2<1/4
1/2=<1-cos(2A-120°)/2<5/4
所以：
1/2=<cos²A+cos²C<5/4