作业帮已知三角形ABC的三个内角A,B,C满足:A+C=2B,1/cosA+1/cosC=-√2/cosB,求cos(A-C)
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已知三角形ABC的三个内角A,B,C满足:A+C=2B,1/cosA+1/cosC=-√2/cosB,求cos(A-C)/2的值.
A+C=2B,
B=60,A+C=120
1/cosA+1/cosC=-√2/cosB =-2√2
cosA+cosC=-2√2cosAcosC
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}
cos[(A-C)/2]=t
t=-√2{2t^2-3/2}
t=√2/2
B=60,A+C=120
1/cosA+1/cosC=-√2/cosB =-2√2
cosA+cosC=-2√2cosAcosC
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}
cos[(A-C)/2]=t
t=-√2{2t^2-3/2}
t=√2/2
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