已知三角形ABC的三个内角A,B,C满足A+C=2B,1/cosA+1/cosC=负的根号2/cosB,求cos(A-C

cos （A-C）/2=√2/2有题设知A＋B＋C=3B=π故B=π/3=60'因此cosB =1/2题设条件化为1/cosA ＋1/cosC=-2√2又由和差化积积化和差倍角公式1/cosA ＋1/cosC＝（cosA＋ cosC）／cosAcosC＝4cos(A ＋C/2)cos(A-C/2)/(cos(A ＋C) cos(A-C))=2cos(A-C/2)/(-1/2) ＋cos(A-C)=-2√2设（A-C）/2＝x 那么上式化为cosx /(-1/2)＋ cos2x=-√2 推出4cosx ^2 ＋√2cosx-3=0 解得cosx=√2/2 (另外一个值－3√2/4舍去,因为不可能是负值)即cos （A-C）/2=√2/2