一道高一数学题,要过程。已知兀/4<贝塔<X<3兀/4,Cos(X-贝塔)=12/13...

发布网友 发布时间：2024-11-28 18:35

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热心网友 时间：2分钟前

由题设可知，
0<x-beta < 兀/2, sin(x-beta) = 5/13
兀<x+beta <3兀/2 cos(x+beta) = -4/5

sin2x=sin((x-beta)+(x+beta))
sin2beta = sin((x+beta)-(x-beta))
cos2x = cos((x-beta)+(x+beta))
上述式子用和角公式展开 代入即可求解。

热心网友 时间：9分钟前

兀/4<a<X<3兀/4,Cos(X-a)=12/13，Sin(x+a)=-3/5
0<X-a<兀/2,兀/2<x+a<3兀/2
sin(x-a)=5/13,cos(x+a)=-4/5
sin2x=sin[(x-a)+(x+a)]
=sin(x-a)cos(x+a)+cos(x-a)sin(x+a)
=(5/13)(-4/5)+(12/13)(-3/5)
=-56/65
sin2a=sin[(x+a)-(x-a)]
=sin(x+a)cos(x-a)-cos(x+a)sin(x-a)
=(-3/5)(12/13)-(-4/5)(5/13)
=-16/65
cos2x=cos[(x-a)+(x+a)]
=cos(x-a)cos(x+a)-sin(x-a)sin(x+a)
=(12/13)(-4/5)-(5/13)(-3/5)
=-33/65
tan2x=sin2x/cos2x=56/33

热心网友 时间：8分钟前

因为：π/4<x<3π/4 π/4<β<3π/4， -π/4>-β>-3π/4 x>β
所以：0<x-β<π/2 所以 sin(x-β)>0 sin(x-β)=√[1-(cos(x-β))^2]=√(1-144/169)=5/13
π/2 <x+β<π cos(x+β ) =√[1-(sin(x+β))^2]=4/5
sin(x-β)=5/13,cos(x-β)=12/13,sin(x+β)=-3/5,cos(x+β)=4/5
sin2x=sin[(x+β)+(x-β)]=sin(x+β)cos(x-β)+cos(x+β)sin(x-β)=-16/65
sin2β=sin[(x+β)-(x-β)]=sin(x+β)cos(x-β)-cos(x+β)sin(x-β)=-56/65
cos2x=cos[(x+β)+(x-β)]=cos(x+β)cos(x-β)-sin(x+β)sin(x-β)=63/65
tan2x=sin2x/cos2x=-16/63

热心网友 时间：9分钟前

∵π/4<β<x<3π/4
∴0<x-β<π/2,π/2<x+y<3π/2
∵cos(x-β)=12/13,sin(x+β)=-3/5
∴sin(x-β)=5/13,cos(x+β)=-4/5

∴sin2x=sin(x-β+x+β)=sin(x-β)cos(x+β)+cos(x-β)sin(x+β)=-56/65
cos2x=-33/65;
sin2β=sin(x+β-(x-β））=sin(x+β)cos(x-β)-cos(x+β)sin(x-β)=-16/65;
tan2x=sin2x/cos2x=56/33