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三角不等式问题
在锐角△ABC中,求 sinAsinB/sinC+sinBsinC/sinA+sinAsinC/sinB的最小值
人气:203 ℃ 时间:2020-06-18 19:15:45
解答
原式=sinAsinB/sinC+sinBsin(A+B)/sinA+sinAsin(A+B)/sinB=sinAsinB/sinC+sinBsinBsin(A+B)/sinAsinB+sinAsinAsin(A+B)/sinBsinA (通分)=sinAsinB/sinC+sin(A+B)(sin²B+sin²A)/sinAsinB=sinAsinB/sinC+s...=sinAsinB/sinC+sin(A+B)(sin²B+sin²A)/sinAsinB =sinAsinB/sinC+sin(A+B)/sinAsinB这一步难道(sin²B+sin²A)=1额,不是,我想错了
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