已知三角形ABC的三个内角满足:A+C=2B,(1/cosA)+(1/cosC)=-(根号2/cosB) 求cos(A-
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已知三角形ABC的三个内角满足:A+C=2B,(1/cosA)+(1/cosC)=-(根号2/cosB) 求cos(A-C)/2的值
A+C=2B,B=60度
(1/cosA)+(1/cosC)=-(根号2/cosB) =-2√2
(cosA+cosC)/(cosAcosC)
2cos[(A+C)/2]*cos[(A-C)/2]=-√2[cos(A+C)+cos(A-C)]
令x=cos(A-C)/2
x=-√2{-1/2+2x^2-1]
4x^2+√2x-3=0
x=√2/2,另一负根舍去
cos(A-C)/2=√2/2
(1/cosA)+(1/cosC)=-(根号2/cosB) =-2√2
(cosA+cosC)/(cosAcosC)
2cos[(A+C)/2]*cos[(A-C)/2]=-√2[cos(A+C)+cos(A-C)]
令x=cos(A-C)/2
x=-√2{-1/2+2x^2-1]
4x^2+√2x-3=0
x=√2/2,另一负根舍去
cos(A-C)/2=√2/2
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