设△ABC的内角A、B、C所对的边长分别为a、b、c,且acosB-bcosA=(3/5)c.(1)试求
设△ABC的内角A、B、C所对的边长分别为a、b、c,且acosB-bcosA=(3/5)c.
(1)试求tanA与tanB的关系
(2)求tan(A-B)的最大值
人气:451 ℃ 时间:2020-03-22 19:19:14
解答
(1)∵acosB-bcosA=(3/5)c∴sinAcosB-sinBcosA=(3/5)sinC=(3/5)sin(A+B)=(3/5)(sinAcosB+cosAsinB)∴(2/5)sinAcosB=(8/5)cosAsinB∴(sinAcosB)/(cosAsinB)=(8/5)/(2/5)=4∴tanAcotB=(sinA/cosA)/(cosB/sinB)=(sinAcos...
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