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△ABC中,角ABC对边为abc,已知sinB/2=sinA/2*sinC/2求tanA/2*tanC/2
还求证a+c=3b
人气:193 ℃ 时间:2019-08-20 23:11:26
解答
tanA/2 * tanC/2=(sinA/2*sinC/2)/(cosA/2*cosC/2)
=(sinB/2)/(cos((A+C)/2)+(sinA/2*sinC/2))
=(sinB/2)/(cos((A+C)/2)+(sinB/2))=(sinB/2)/(cos((π-B)/2)+(sinB/2))
=(sinB/2)/(sinB/2+sinB/2)=1/2
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