已知阿尔法,贝塔属于(3π/4,π),sin(阿尔法+贝塔)=-3/5,sin(贝塔-π/4)=12/13,则sin(阿尔法+π/4)=
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已知阿尔法,贝塔属于(3π/4,π),sin(阿尔法+贝塔)=-3/5,sin(贝塔-π/4)=12/13,则sin(阿尔法+π/4)=
已知阿尔法,贝塔属于(3π/4,π),sin(阿尔法+贝塔)=-3/5,sin(贝塔-π/4)=12/13,则sin(阿尔法+π/4)=
已知阿尔法,贝塔属于(3π/4,π),sin(阿尔法+贝塔)=-3/5,sin(贝塔-π/4)=12/13,则sin(阿尔法+π/4)=
3π/4所以cos(a+b)>0
sin²(a+b)+cos²(a+b)=1
cos(a+b)=4/5
同理
cos(b-π/4)=-5/13
原式= cos[(a+b)-(b-π/4)]
=cos(a+b)cos(b-π/4)+sin(a+b)sin(b-π/4)
=-56/65
a,b∈(3π/4,π)
a+b∈(3π/2,2π)
sin(a+b)=-3/5
cos(a+b)>0
cos(a+b)=4/5
b∈(3π/4,π)
b-π/4∈(π/2,3π/4)
sin(b-π/4)=12/13
cos(b-π/4)<0
cos(b-π/4)=-5/13
sin(a+π/4)
=sin[...
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a,b∈(3π/4,π)
a+b∈(3π/2,2π)
sin(a+b)=-3/5
cos(a+b)>0
cos(a+b)=4/5
b∈(3π/4,π)
b-π/4∈(π/2,3π/4)
sin(b-π/4)=12/13
cos(b-π/4)<0
cos(b-π/4)=-5/13
sin(a+π/4)
=sin[(a+b)-(b-π/4)]
=sin(a+b)cos(b-π/4)-cos(a+b)sin(b-π/4)
=-3/5*(-5/13)-4/5*12/13
=15/65-48/65
=-33/65
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