Question

已知三角形有三个内角A.B.C,满足A+C=2B.1/cosA+1/cosC=(2)/cosB,

Answer 1

B=60,A+C=120

1/cosA+1/cosC=-√2/cosB =-2√2

cosA+cosC=-2√2cosAcosC

（左边用和差化积,右边用积化和差）

2cos[(A+C)/2]cos[(A-C)/2]=-√2[cos(A+C)+cos(A-C)]

cos[(A-C)/2]=-√2[-1/2+cos(A-C)]

=-√2{2[cos(A-C)/2]^2-3/2}

令：cos[(A-C)/2]=t

则：t=-√2{2t^2-3/2}

解得：t=√2/2