已知三角形ABC中,A B C依次成等差数列.且1/cosA+1/cosC=—(√2/cosB).求cos(A-C)/2

A B C依次成等差数列
=>2B=A+C
=>3B=180°
=>B=60°
=>A+C=120°
1/cosA +1/cosC=-√2/cosB
=>(cosA+cosC)/cosAcosC=√2cos(A+C)带入A+C=120°
=>(cosC+cosA)/cosCcosA=-2√2
=>2cos[(A+C)/2][cos(A-C)/2]=-√2[cos(A+C)+cos(A-C)]带入A+C=120°
=>
cos[(A-C)/2]=-√2[cos(A+C)+cos(A-C)]
化简cos(A-C)=2(cos[(A-C)/2])^2-1带入上式
化简全式
=>2(cos[(A-C)/2])^2 +cos[(A-C)/2] -(3√2)/2=0
把此方程看作是关于cos[(A-C)/2]的一元二次方程,可得到两个根.
cos[(A-C)/2]=(-3√2)/4
cos[(A-C)/2]=√2/2
因为A.C是锐角,(A-C)/2也是锐角,所以cos[(A-C)/2]>0
所以舍去第一个根,
所以,cos[(A-C)/2]=√2/2