Question

简单的数学问题

已知三角形ABC的内角B=60°，且 1/cosA + 1/cosC=-2倍根号2，则cos[（A-C)/2]=?

Answer 1

原式化简得 cosA+cosC+2√2cosAcosC=0
2cos（A+C)/2cos(A-C)/2+√2（cos(A+C)+cos(A-C)）=0
A+C=120°
原式为
cos(A-C)/2-√2/2+2√2[cos(A-C)/2]^2-√2=0
4[cos(A-C)/2]^2+√2cos(A-C)/2-3=0
cos(A-C)/2= √2/2