原題:
"已知三角形ABC,三內(nèi)角滿足A+B=2C,1/COSA+1/COSC=-根號2/COSB,求COS（A-C)/2"
因A+B+C=π,
又A+C=2B
得B=π/3
1/cosA+1/cosC=-√2cosB
=-√2/2
(cosA+cosC)=-2√2cosAcosC
2cos(A-C)/2cos(A+C)/2=-√2[cos(A+C)+cos(A-C)]
cos(A-C)/2=-√2[-1/2+cos(A-C)]
cos(A-C)/2=-√2[-1/2+2cos²(A-C)/2-1]
4cos²(A-C)/2+√2cos(A-C)/2-3=0
(|A-C|/2