问题描述: 已知三角形ABC,三内角满足A+B=2C,1/COSA+1/COSC=负根号2处以COSB,求COS(A-C)/2 1个回答 分类:数学 2014-11-11 问题解答: 我来补答 因A+B+C=π,又A+C=2B 得B=π/3 1/cosA+1/cosC=-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 展开全文阅读