sin(x+y)=1\2,sin(x—y)=1\3,求[tan(x+y)-tanx-tany]\[tany的平方tan(x+y)]
人气:170 ℃ 时间:2019-10-17 02:14:37
解答
sin(x+y)=sinxcosy+cosxsiny=1/2 sin(x-y)=sinxconsy-cosxsiny=1/3 sinxcosy=5/12,cosxsiny=1/12 tanx/tany=sinxcosy/cosxsiny=5.[tan(x+y)-tanx-tany]/[tany的平方tan(x+y)] =[(tanx+tany)/(1-tanxtany)-(tanx+t...
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