化简:cos(A-π/2)·cot(-A-3π/2)·色彩(-A+5π/2)·tan(5π/2+A),
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/11/10 01:52:53
化简:cos(A-π/2)·cot(-A-3π/2)·色彩(-A+5π/2)·tan(5π/2+A),
那个“色彩”是sec吗?就当是吧……
因为:cot=1/tan sec=1/cos
所以:上式可写为:cos(A-π/2)·sec(-A+5π/2)·cot(-A-3π/2)·tan(5π/2+A)
=cos(A-π/2)·sec(-A+π/2) ·tan(π/2+A)·cot(-A-3π/2)
=cos(A-π/2)·sec(A-π/2) ·tan(π/2+A) ·cot(-A+π/2)
=1 ·tan(π/2+A) ·cot(-A+π/2)
=-tan(π/2+A) ·cot(A+π/2)
=-1
再问: 为什么sec(-A+5π/2=sec(-A+π/2) ???
再答: sec(-A+5π/2)=sec(-A+π/2+2π)=sec(-A+π/2)
因为:cot=1/tan sec=1/cos
所以:上式可写为:cos(A-π/2)·sec(-A+5π/2)·cot(-A-3π/2)·tan(5π/2+A)
=cos(A-π/2)·sec(-A+π/2) ·tan(π/2+A)·cot(-A-3π/2)
=cos(A-π/2)·sec(A-π/2) ·tan(π/2+A) ·cot(-A+π/2)
=1 ·tan(π/2+A) ·cot(-A+π/2)
=-tan(π/2+A) ·cot(A+π/2)
=-1
再问: 为什么sec(-A+5π/2=sec(-A+π/2) ???
再答: sec(-A+5π/2)=sec(-A+π/2+2π)=sec(-A+π/2)
化简:cos(A-π/2)·cot(-A-3π/2)·色彩(-A+5π/2)·tan(5π/2+A),
化简sin(3π+a)tan(a-π)cot(π+a)/tan(2π-a)cos(π-a)
tan(-a+3/2π)/cot(-a-π)
化简:(sin^2(a+π)cos(π+a)cot(-a-2π))/tan(π+a)cos^3(π+a)
化简sin(2π-a)tan(π+a)cot(-a-π)\cos(π-a)tan(3π-a)
化简 (sin^2(a+π)cos(2π-a)cot(a-2π))/(tan(π-a)cos^3(-a-π))
化简 cot(a+4π)cos(a+π)[sin(a+3π)]^2/tan(π+a)[cos(-π-a)]^2
化简:sin(a-π)cot(a-2π)/cos(a-π)tan(a-2π)
cos(3π/2+a)=3/5,a∈(π/2,π),tan(π+B)=1/2,求cot(a-2B)的值
sin(a-5π)/tan(3π-a)*cot(π/2-a)/tan(a-3π/2)*cos(8π-a)/sin(-a-
sin^2(a+π) cos(π+a) cot(-a-2π) /tan(π+a) cos^3(-a-π)
求证:cos^2a/[cot(a/2)-tan (a/2)]=1/4sin2a