已知△ABC的三个锐角A,B,C满足A+C=2B,1/cosA +1/cosC=-√2/cosB,求cos(A/2-C/2)的值.

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已知△ABC的三个锐角A,B,C满足A+C=2B,1/cosA +1/cosC=-√2/cosB,求cos(A/2-C/2)的值.
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已知△ABC的三个锐角A,B,C满足A+C=2B,1/cosA +1/cosC=-√2/cosB,求cos(A/2-C/2)的值.
已知△ABC的三个锐角A,B,C满足A+C=2B,1/cosA +1/cosC=-√2/cosB,求cos(A/2-C/2)的值.

已知△ABC的三个锐角A,B,C满足A+C=2B,1/cosA +1/cosC=-√2/cosB,求cos(A/2-C/2)的值.
√2/2
A+C=2[180-(A+C)]
=>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)=2cos^2[(A-C)/2]-1带入上式
化简全式
=>2cos^2[(A-C)/2]+cos[(A-C)/2]-(3√2)/2=0
把cos[(A-C)/2]看为一项
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
所以取cos[(A-C)/2]=√2/2