在△ABC中,已知sin^2(A/2)+sin^2(B/2)+sin^2(C/2)=cos^2(B/2).求证cot(A/2).cot(B/2).cot(C/2)成等差数列
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![在△ABC中,已知sin^2(A/2)+sin^2(B/2)+sin^2(C/2)=cos^2(B/2).求证cot(A/2).cot(B/2).cot(C/2)成等差数列](/uploads/image/z/6879997-37-7.jpg?t=%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E5%B7%B2%E7%9F%A5sin%5E2%28A%2F2%29%2Bsin%5E2%28B%2F2%29%2Bsin%5E2%28C%2F2%29%3Dcos%5E2%28B%2F2%29.%E6%B1%82%E8%AF%81cot%28A%2F2%29.cot%EF%BC%88B%2F2%29.cot%EF%BC%88C%2F2%29%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97)
在△ABC中,已知sin^2(A/2)+sin^2(B/2)+sin^2(C/2)=cos^2(B/2).求证cot(A/2).cot(B/2).cot(C/2)成等差数列
在△ABC中,已知sin^2(A/2)+sin^2(B/2)+sin^2(C/2)=cos^2(B/2).求证cot(A/2).cot(B/2).cot(C/2)
成等差数列
在△ABC中,已知sin^2(A/2)+sin^2(B/2)+sin^2(C/2)=cos^2(B/2).求证cot(A/2).cot(B/2).cot(C/2)成等差数列
sin^2(A/2)+sin^2(B/2)+sin^2(C/2)=cos^2(B/2)
<=> ½* (cosA+cosC) - 2 sin^2(B/2)=0
<=> cos(A+C)/2 * cos(A-C)/2 - sin^2(B/2)= sin^2(B/2)
<=> sin(B/2)*[ cos(A-C)/2 - cos(A+C)/2] = sin^2(B/2)
<=> sin(B/2)*sin(A/2)*sin(C/2) = ½* sin^2(B/2)
<=> sin(B/2) = 2* sin(A/2)*sin(C/2)
cot(A/2) + cot(C/2)
=(cosA+1)/sinA + (cosC+1)/sinC
=[sin(A+C)+sinA+sinC]/(sinA*sinC)
=[sinA+sin B+sinC]/(sinA*sinC)
=4cos(A/2)*cos(B/2)*cos(C/2) /[4sin(A/2)*cos(A/2)*sin(C/2)*cos(C/2)]
=cos(B/2)/[sin(A/2)*sin(c/2)]
=2cos(B/2)/sin(B/2)
=2cot(B/2)
cot(A/2).cot(B/2).cot(C/2)成等差数列