[cot(θ+4π)·cos(θ+π)·sin^2(θ+3π)]/[tan(π+θ)·cos^3(-π-θ)]如题,望尽快予以解答
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/28 03:13:28
[cot(θ+4π)·cos(θ+π)·sin^2(θ+3π)]/[tan(π+θ)·cos^3(-π-θ)]如题,望尽快予以解答
[cot(θ+4π)·cos(θ+π)·sin^2(θ+3π)]/[tan(π+θ)·cos^3(-π-θ)]
如题,望尽快予以解答
[cot(θ+4π)·cos(θ+π)·sin^2(θ+3π)]/[tan(π+θ)·cos^3(-π-θ)]如题,望尽快予以解答
[cot(θ+4π)·cos(θ+π)·sin^2(θ+3π)]/[tan(π+θ)·cos^3(-π-θ)]
=[cotθ(-cosθsin²θ]/(tanθcos³θ)
=-[(cosθ/sinθ)sin²θ]/[(sinθ/cosθ)cos²θ]
=-sinθcosθ/sinθcosθ
=-1
化简:{cot(θ+4π)·cos(θ+π)·sin2次方(θ+3π)}/{tan(π+θ)·cos3次方(-π-化简:{cot(θ+4π)·cos(θ+π)·sin2次方(θ+3π)}/{tan(π+θ)·cos3次方(-π-θ)}
[cot(θ+4π)·cos(θ+π)·sin^2(θ+3π)]/[tan(π+θ)·cos^3(-π-θ)]如题,望尽快予以解答
已知cot(π/4+θ)=3则(2sinθ-cosθ)/(cosθ+2sinθ)=?
已知cot(π/4+θ)=3.则(2sinθ-cosθ)/(cosθ+2sinθ)=_
求证1-cosθ/sinθ=cotθ(1-cosθ)/sinθ=cot(θ/2)
(1-cos²a)·cot²a化简
化简[cot(θ+4π)cos(θ+π)sin²(θ+3π)]/[tan(π+θ)cos²(-π-θ)]
若COSα=4/5,α∈(0 π)则COtα的值
cot(π+θ)=2,则(3sinθ-cosθ)/(sinθ+cosθ)=?
cot(-π-θ)等于什么?等于-cot(π+θ)吗
化简【1+sinθ-cosθ/1+sinθ+cosθ】+cot(θ/2)
化简tan(π-x)×cot(x-π)+cos²y-2cos²x
若f(θ)=sinθ/(|sinθ|)+cos(θ)/|cosθ|+tanθ/|tanθ|+cotθ/|cotθ|,求y=f(θ)的值域
求计算题:cot(Θ+4π)?cos(Θ+π)?sin2(Θ+3π)/tan(π+Θ)?cos3(-π-Θ)
cot(-31/4 π)三角函数值
化简:cos(A-π/2)·cot(-A-3π/2)·色彩(-A+5π/2)·tan(5π/2+A),
已知sinθ+cosθ=1/5,θ∈(0,π),求cotθ?
tan(cosθ)·cot(sinθ)>0,试指出θ所在的象限