已知cos(α+π⼀4)=-4⼀5,0<α<π⼀2，则cosα=

已知cos(α+π/4)=-4/5,0＜α＜π/2
所以π/4＜α+π/4＜3π/4
所以sin(α+π/4)=√[1-(-4/5)²]=3/5

所以cosα=cos[(α+π/4)-π/4]=cos(α+π/4)cos(π/4)+sin(α+π/4)sin(π/4)
=(-4/5)*(√2/2)+(3/5)*(√2/2)
=-√2/10

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0<α<π/2

则：π/4<α+π/4<3π/4
又cos(α+π/4)=-4/5
所以，sin²(α+π/4)=1-cos²(α+π/4)=9/25
sin(α+π/4)>0
所以，sin(α+π/4)=3/5
cosα=cos[(α+π/4)-π/4]
=cos(α+π/4)cos(π/4)+sin(α+π/4)sin(π/4)
=(-4/5)(√2/2)+(3/5)(√2/2)
=-√2/10

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楼主的问题本身有问题！

解：
因为：0＜α＜π/2
所以：α是第一象限角，sinα＞0、cosα＞0
(sinα)^2+(cosα)^2=1
sinα=√[1-(cosα)^2]
cos(α+π/4)=-4/5
cosαcos(π/4)-sinαsin(π/4)=-4/5
cosα(√2)/2-sinα(√2)/2=-4/5
cosα-sinα=-(4/5)√2
cosα-√[1-(cosα)^2]=-(4/5)√2
cosα+(4/5)√2=√[1-(cosα)^2]
(cosα)^2+[(8√2)/5]cosα+32/25=1-(cosα)^2
2(cosα)^2+[(8√2)/5]cosα+7/25=0
cosα={-[(8√2)/5±√{[(8√2)/5]^2-56/25}}/4
={-[(8√2)/5]±6(√2)/5}/4
=-(√2)/10
cosα1=-(√2)/10
cosα2=-(7√2)/10
可见：cosα＜0，这与已知0＜α＜π/2相矛盾
因此：楼主的问题本身有问题！