已知三角形ABC中,角A,B,C成等差数列,求证1/a+b+1/b+c=3/a+b+c
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已知三角形ABC中,角A,B,C成等差数列,求证1/a+b+1/b+c=3/a+b+c
已知三角形ABC中,角A,B,C成等差数列,求证1/a+b+1/b+c=3/a+b+c
已知三角形ABC中,角A,B,C成等差数列,求证1/a+b+1/b+c=3/a+b+c
因为(a+b+c)*(1/(a+b) + 1/(b+c))
= (a+b+c)/(a+b) + (a+b+c)/(b+c)
= 1 + c/(a+b) + 1 + a/(b+c)
= 2 + c/(a+b) + a/(b+c)
= 2 + sinC/(sinA+sinB) + sinA/(sinB+sinC)
因为A,B,C成等差数列,所以A-B = B-C,即A+C=2B,易得B=60度.
所以
(a+b+c)*(1/(a+b) + 1/(b+c))
= 2 + sinC/(sinA+sinB) + sinA/(sinB+sinC)
= 2 + sinC/(2sin((A+B)/2)cos((A-B)/2)) + sinA/(2sin((B+C)/2)cos((B-C)/2))
= 2 + sinC/(2*cos(C/2)*cos((A-B)/2)) + sinA/(2*cos(A/2)*cos((B-C)/2))
= 2 + sin(C/2) / cos((A-B)/2) + sin(A/2) / cos((B-C)/2)
= 2 + (sin(C/2) + sin(A/2)) / cos((A-B)/2)
= 2 + 2sin(B/2)cos((A-B)/2) / cos((A-B)/2)
= 2 + 2sin(B/2)
= 2 + 2sin30
= 3
故1/(a+b) + 1/(b+c) = 3/(a+b+c)