问题: 谢谢帮助,HELP
已知△ABC的三边a,b,c成等比数列,
求证:cos(A-C)+cosB+cos2B=1.
解答:
b^2=ac--->(2RsinB)^2=2RsinA*sinC--->(sinB)^2=sinAsinC.
cos(A-C)+cosB+cos2B
=cos(A-C)-cos(A+C)+cos2B
=(cosAcosC+sinAsinC)-(cosAcosC-sinAsinC)+[1-2(sinB)^2]
=2sinAsinC-2(sinB)^2+1
=2sinAsinC-2sinAsinC+1
=1.
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