作业帮关于数学题（三角函数）.求救.回答多少算多少.
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/10 17:56:55
关于数学题（三角函数）.求救.回答多少算多少.
cos(x+y)=-1,tanx=2,求tany..
第2题.
tanx=4根号3,cos(x+y)=-11/14,xy为锐角.求cosy
第3题.
x+y=135度.则（1-tanx)(1-tany)..
看上去这些题目不难.可惜我是个地地道道的文科声音.
cos(x+y)=-1,tanx=2,求tany..
第2题.
tanx=4根号3,cos(x+y)=-11/14,xy为锐角.求cosy
第3题.
x+y=135度.则（1-tanx)(1-tany)..
看上去这些题目不难.可惜我是个地地道道的文科声音.
1)cos(x+y)=-1,说明x+y=180度,
sin(x+y)=√[1-cos^2(x+y)]=0
若,X在第一象限时,
令,sinx=2/r,cosx=1/r,则
r=√5,
sinx=2/√5,cosx=1/√5.
siny=sin(x+y-x)=sin(x+y)*cosx-cos(x+y)*sinx
=0-(-1)*2/√5
=2√5/5.
cosy=cos(x+y-x)=cos(x+y)cosx+sin(x+y)sinx
=(-1)*1/√5+0
=-√5/5.
tany=siny/cosy=-2.
2)tanx=4√3,xy为锐角,
sinx=4√3/r,cosx=1/r,
r=7.
sinx=4√3/7,cosx=1/7.
cos(x+y)=-11/14,
sin(x+y)=√[1-(-11/14)^2]=5√3/14.
cosy=cos(x+y-x)=cos(x+y)cosx+sin(x+y)sinx
=1/2.
3)x+y=135度,
因为:tan(x+y)=tan135=-1=(tanx+tany)/(1-tanxtany)
所以,
-(tanx+tany)=1-tanxtany.
（1-tanx)(1-tany)=1-(tanx+tany)+tanxtany
=1+(1-tanxtany)+tanxtany
=2.
sin(x+y)=√[1-cos^2(x+y)]=0
若,X在第一象限时,
令,sinx=2/r,cosx=1/r,则
r=√5,
sinx=2/√5,cosx=1/√5.
siny=sin(x+y-x)=sin(x+y)*cosx-cos(x+y)*sinx
=0-(-1)*2/√5
=2√5/5.
cosy=cos(x+y-x)=cos(x+y)cosx+sin(x+y)sinx
=(-1)*1/√5+0
=-√5/5.
tany=siny/cosy=-2.
2)tanx=4√3,xy为锐角,
sinx=4√3/r,cosx=1/r,
r=7.
sinx=4√3/7,cosx=1/7.
cos(x+y)=-11/14,
sin(x+y)=√[1-(-11/14)^2]=5√3/14.
cosy=cos(x+y-x)=cos(x+y)cosx+sin(x+y)sinx
=1/2.
3)x+y=135度,
因为:tan(x+y)=tan135=-1=(tanx+tany)/(1-tanxtany)
所以,
-(tanx+tany)=1-tanxtany.
（1-tanx)(1-tany)=1-(tanx+tany)+tanxtany
=1+(1-tanxtany)+tanxtany
=2.