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
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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](/uploads/image/z/349158-30-8.jpg?t=%E5%B7%B2%E7%9F%A5%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E4%B8%89%E4%B8%AA%E5%86%85%E8%A7%92%E6%BB%A1%E8%B6%B3%EF%BC%9AA%2BC%3D2B%2C%281%2FcosA%29%2B%EF%BC%881%2FcosC%29%3D-%28%E6%A0%B9%E5%8F%B72%2FcosB%29+%E6%B1%82cos%28A-C%29%2F2%E7%9A%84%E5%80%BC)
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
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
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
cos (A-C)/2=√2/2
有题设知A+B+C=3B=π故B=π/3=60'因此cosB =1/2题设条件化为1/cosA +1/cosC=-2√2又由和差化积积化和差倍角公式1/cosA +1/cosC=(cosA+ cosC)/cosAcosC=4cos(A +C/2)cos(A-C/2)/(cos(A +C) cos(A-C))=2cos(A-C/2)/(-1/2) +cos(...
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cos (A-C)/2=√2/2
有题设知A+B+C=3B=π故B=π/3=60'因此cosB =1/2题设条件化为1/cosA +1/cosC=-2√2又由和差化积积化和差倍角公式1/cosA +1/cosC=(cosA+ cosC)/cosAcosC=4cos(A +C/2)cos(A-C/2)/(cos(A +C) cos(A-C))=2cos(A-C/2)/(-1/2) +cos(A-C)=-2√2
设(A-C)/2=x 那么上式化为cosx /(-1/2)+ cos2x=-√2 推出4cosx ^2 +√2cosx-3=0 解得cosx=√2/2 (另外一个值-3√2/4舍去,因为不可能是负值)即cos (A-C)/2=√2/2
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