sin（x+y)=1\2,sin(x—y）=1\3，求[tan（x+y）-tanx-tany]\[tany的平方tan（x+y)]

显示全部楼层 · 2009-5-24 22:51:47

sin(x+y)=sinxcosy+cosxsiny=1/2; sin(x-y)=sinxcosy-cosxsiny=1/3; 所以sinxcosy=5/12,cosxsiny=1/12,两式相比得tanx/tany=5,即tanx=5tany 注意到tan(x+y)=(tanx+tany)/(1-tanxtany) 所以tan(x+y)-tanx-tany=(tanx+tany)(tanxtany)/(1-tanxtany) (tany)^2tan(x+y)= (tany)^2(tanx+tany)/(1-tanxtany)从而[tan(x+y)-tanx-tany]/[(tany)^2tan(x+y)]= (tanx+t...

千问 · 2009-5-24 22:51:47

sin(x+y)=sinxcosy+cosxsiny=1/2sin(x-y)=sinxconsy-cosxsiny=1/3sinxcosy=5/12, cosxsiny=1/12tanx/tany=sinxcosy/cosxsiny=5.[tan（x+y）-tanx-tany]/[tany的平方tan（x+y)]=[(tanx+tan...