求证:cos(cosx)>sin(sinx),
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/26 08:24:52
x){{f%$+|g^qf$A|"}ِfeӞ4*4*£YFE7h*<\UQ
(<
хhtB mO?7E
'XJTQT$1uO笈Ovt=ٱق=O{aE RѢ |OALt>~qAb(TAj
求证:cos(cosx)>sin(sinx),
求证:cos(cosx)>sin(sinx),
求证:cos(cosx)>sin(sinx),
首先sin(x)+cos(x) = √2·sin(x+π/4) ≤ √2 < π/2, 故sin(x) < π/2-cos(x).
同理可得sin(x) < π/2+cos(x), 于是-π/2 < sin(x) < π/2-|cos(x)| ≤ π/2.
由sin(x)在[-π/2,π/2]上严格单调递增, 有sin(sin(x)) < sin(π/2-|cos(x)|) = cos(|cos(x)|) = cos(cos(x)).
求证:cos(cosx)>sin(sinx),
已知sinαcosβ=cos^βsinx/2cosx+sin^2αcosx/2sinx,求证:tgx=sinα/cosβ
已知sinαcosβ=(cos^βsinx)/2cosx+(sin^2αcosx)/2sinx,求证:tanx=sinα/cosβ
求证:Sin^2 x / (sinx-cosx) - (sin x + cos x)/(ta求证:Sin^2 x / (sinx-cosx) - (sin x + cos x)/(tan^2-1) = sin x + cos x
cos(x+y)sinx-sin(x+y)cosx
求证 1+sinα+cosα+2sinαcosα/求证 (1+sinα+cosα+2sinαcosα)/(1+sinα+cosα)=sinα+cosα
求证:(1+cosθ+cosθ/2) /(sinθ+sinθ/2)=sinθ/1-cosθ
求证:sinα+sinβ=2sin(α+β)/2 *cos(α-β)/2
求证:(1+sinθ-cosθ)/(1+sinθ+cosθ)=sinθ/(cosθ+1)
(1)已知:(4sinx-2cosx)/(5cosx+3sinx)=6/11 求sinx乘以cosx的值(2)证明(cosx/1+sinx) - (sinx/1+cosx)=2(cosx-sinx)/1+sinx+cosx(3)求证:sin²x乘以tanx + cos²x乘以cotx+2sinx乘以cosx=tanx+cotx要用 sin²X+cos²X=1
求证:sin^2/(sin-cos) - (sin+cos)/(tan^2 -1) =sin+cos
求导Sin (CosX)
求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2)]
求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2)]
求证 sinx(cos^2 2x-sin^2 2x) + 2cosx cos2x sin2x= sin 5x
证明:cos(cos(cos(cosx)))>sin(sin(sin(sinx)))好的100分
若sin(cosx)*cos(sinx)>0 求x范围
求证:cos平方x+cos平方(x+a)-2cosa*cosx*cos(x+a)=sin平方a