已知cos(x-y/2)=-1/9,sin(x/2-y)=2/3,0<x<pi(圆周率),0<y<pi/2.求 cos(x+y)的值
来源:百度知道 编辑:UC知道 时间:2024/05/18 04:02:56
y=cosx/(2cosx+1)
=(2cosx+1)y=cosx
=2ycosx+y=cosx
y=(1-2y)cosx
y=1/2时
0=1/2
y不等于1/2
0<x<π,0<y<π/2→0<x/2<π/2,0<y/2<π/4→-π/4<x-y/2<π,-π/2<x/2-y<π/2
又cos(x-y/2)=-1/9<0,sin(x/2-y)=2/3>0,所以π/2<x-y/2<π,0<x/2-y<π/2,sin(x-y/2)>0,cos(x/2-y)>0,
sin(x-y/2)=√(1-cos(x-y/2)^2)=4√5/9,cos(x/2-y)=√(1-sin(x/2-y)^2=√5/3.
cos(x+y)=2[cos(x/2+y/2)]^2-1=2{cos[x-y/2-(x/2-y)]}^2-1=2[cos(x-y/2)cos(x/2-y)+sin(x-y/2)sin(x/2-y)]^2-1=2(7√5/27)^2-1=490/729-1=-239/729
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已知cos(x-y/2)=-1/9,sin(x/2-y)=2/3,0<x<pi(圆周率),0<y<pi/2.求 cos(x+y)的值
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