作业帮已知cos(x+y)=1/3,cos(x-y)+2/3,且0
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已知cos(x+y)=1/3,cos(x-y)+2/3,且0
已知cos(x+y)=1/3,cos(x-y)+2/3,且0
已知cos(x+y)=1/3,cos(x-y)+2/3,且0
cos(x+y)=1/3,cos(x-y)=2/3.
0＜x＜90º,60º＜y＜90º
∴可得:0＜x＜30＜60＜y＜90º
sin(x+y)=(2√2)/3.cos(x+y)=1/3.
sin(y-x)=(√5)/3,cos(y-x)=2/3.
[[[1]]]
cos2x=cos[(y+x)-(y-x)]
=cos(x+y)cos(y-x)+sin(x+y)sin(y-x)
=(2/9)+[(2√10)/9]
=2(1+√10)/9
[[[2]]]
1=cos(x+y)+cos(x-y)=2cosxcosy
1/3=cos(x-y)-cos(x+y)=-2sinxsin(-y)
∴sinxsiny=1/6
cosxcosy=1/2
∴两式相除,可得
tanxtany=1/3
cos(x+y)=1/3,所以x+y<π/2..
cos(x+y)=1/3,cos(x-y)+2/3 可知sin(x+y)=2√2/3,sin(x-y)=√5/3.
cos2x=cos[(x+y)-(x-y)]=cos(x+y)cos(x-y)+sin(x+y)sin(x-y)=1/3x2/3+2√2/3x√5/3=(2+2√10)/9
cos(x-y)/cos(x+y)=...
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cos(x+y)=1/3,所以x+y<π/2..
cos(x+y)=1/3,cos(x-y)+2/3 可知sin(x+y)=2√2/3,sin(x-y)=√5/3.
cos2x=cos[(x+y)-(x-y)]=cos(x+y)cos(x-y)+sin(x+y)sin(x-y)=1/3x2/3+2√2/3x√5/3=(2+2√10)/9
cos(x-y)/cos(x+y)=2，用和差化积公式展开分子分母，再同时除以cosx*cosy有：
(1+tanx*tany)/(1-tanx*tany)=2.
最后得tanx*tany=1/3
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