已知△ABC的三个内角A、B、C满足A+C=2B,且1/cosA+1/cosC=-根号2/cosB,求cos[(A-c)

A+C=2B
3B=A+C+B=π
∴ B=π/3
1/cosA+1/cosC=-√2/(1/2)=-2√2
cosA+cosC=-2√2cosAcosC
2cos[(A+C)/2]cos[(A-C)/2]=-√2[cos(A+C)+cos(A-C)]
2cos60°cos[(A-C)/2]=-√2[cos120°+cos(A-C)]
cos[(A-C)/2]=-√2{-1/2+2cos²[(A-C)/2]-1}
cos[(A-C)/2]=3√2/2-2√2cos²[(A-C)/2]
2√2cos²[(A-C)/2]+cos[(A-C)/2]-3√2/2=0
4cos²[(A-C)/2]+√2cos[(A-C)/2]-3=0
解方程,cos[(A-C)/2]=√2/2或cos[(A-C)/2]=-3√2/4（舍）
所以 cos[(A-C)/2]=√2/2